The water analogy seems to imply that power = current. Why is this incorrect?












20














Many people think of the water analogy to try to explain how electromagnetic energy is delivered to a device in a circuit. Using that analogy, in a DC circuit, one could imagine the device is like a water wheel being pushed by the current.



In the case of an actual water wheel, the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time: power = current, but in electric circuits power = voltage x current.



Why is this?










share|cite|improve this question
























  • Can you make your post a little more exact? Are you saying Power = Current? Please define all terms. And make sure units match up.
    – ggcg
    Dec 16 at 19:02










  • Thanks. Is it more clear now?
    – lyndon
    Dec 16 at 19:08






  • 1




    en.wikipedia.org/wiki/Hydraulic_analogy#Equation_examples
    – BowlOfRed
    Dec 16 at 19:29






  • 8




    In the case of a water wheel, it's the amount of water flowing times how hard the water pushes the wheel.
    – immibis
    Dec 16 at 23:16






  • 1




    Pressure does matter to hydraulic power, that's why water is piped from mountain dams to valley power stations.
    – Aetol
    Dec 17 at 10:18
















20














Many people think of the water analogy to try to explain how electromagnetic energy is delivered to a device in a circuit. Using that analogy, in a DC circuit, one could imagine the device is like a water wheel being pushed by the current.



In the case of an actual water wheel, the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time: power = current, but in electric circuits power = voltage x current.



Why is this?










share|cite|improve this question
























  • Can you make your post a little more exact? Are you saying Power = Current? Please define all terms. And make sure units match up.
    – ggcg
    Dec 16 at 19:02










  • Thanks. Is it more clear now?
    – lyndon
    Dec 16 at 19:08






  • 1




    en.wikipedia.org/wiki/Hydraulic_analogy#Equation_examples
    – BowlOfRed
    Dec 16 at 19:29






  • 8




    In the case of a water wheel, it's the amount of water flowing times how hard the water pushes the wheel.
    – immibis
    Dec 16 at 23:16






  • 1




    Pressure does matter to hydraulic power, that's why water is piped from mountain dams to valley power stations.
    – Aetol
    Dec 17 at 10:18














20












20








20


1





Many people think of the water analogy to try to explain how electromagnetic energy is delivered to a device in a circuit. Using that analogy, in a DC circuit, one could imagine the device is like a water wheel being pushed by the current.



In the case of an actual water wheel, the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time: power = current, but in electric circuits power = voltage x current.



Why is this?










share|cite|improve this question















Many people think of the water analogy to try to explain how electromagnetic energy is delivered to a device in a circuit. Using that analogy, in a DC circuit, one could imagine the device is like a water wheel being pushed by the current.



In the case of an actual water wheel, the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time: power = current, but in electric circuits power = voltage x current.



Why is this?







fluid-dynamics electric-circuits electric-current power flow






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 at 20:16









Qmechanic

101k121821140




101k121821140










asked Dec 16 at 18:58









lyndon

12715




12715












  • Can you make your post a little more exact? Are you saying Power = Current? Please define all terms. And make sure units match up.
    – ggcg
    Dec 16 at 19:02










  • Thanks. Is it more clear now?
    – lyndon
    Dec 16 at 19:08






  • 1




    en.wikipedia.org/wiki/Hydraulic_analogy#Equation_examples
    – BowlOfRed
    Dec 16 at 19:29






  • 8




    In the case of a water wheel, it's the amount of water flowing times how hard the water pushes the wheel.
    – immibis
    Dec 16 at 23:16






  • 1




    Pressure does matter to hydraulic power, that's why water is piped from mountain dams to valley power stations.
    – Aetol
    Dec 17 at 10:18


















  • Can you make your post a little more exact? Are you saying Power = Current? Please define all terms. And make sure units match up.
    – ggcg
    Dec 16 at 19:02










  • Thanks. Is it more clear now?
    – lyndon
    Dec 16 at 19:08






  • 1




    en.wikipedia.org/wiki/Hydraulic_analogy#Equation_examples
    – BowlOfRed
    Dec 16 at 19:29






  • 8




    In the case of a water wheel, it's the amount of water flowing times how hard the water pushes the wheel.
    – immibis
    Dec 16 at 23:16






  • 1




    Pressure does matter to hydraulic power, that's why water is piped from mountain dams to valley power stations.
    – Aetol
    Dec 17 at 10:18
















Can you make your post a little more exact? Are you saying Power = Current? Please define all terms. And make sure units match up.
– ggcg
Dec 16 at 19:02




Can you make your post a little more exact? Are you saying Power = Current? Please define all terms. And make sure units match up.
– ggcg
Dec 16 at 19:02












Thanks. Is it more clear now?
– lyndon
Dec 16 at 19:08




Thanks. Is it more clear now?
– lyndon
Dec 16 at 19:08




1




1




en.wikipedia.org/wiki/Hydraulic_analogy#Equation_examples
– BowlOfRed
Dec 16 at 19:29




en.wikipedia.org/wiki/Hydraulic_analogy#Equation_examples
– BowlOfRed
Dec 16 at 19:29




8




8




In the case of a water wheel, it's the amount of water flowing times how hard the water pushes the wheel.
– immibis
Dec 16 at 23:16




In the case of a water wheel, it's the amount of water flowing times how hard the water pushes the wheel.
– immibis
Dec 16 at 23:16




1




1




Pressure does matter to hydraulic power, that's why water is piped from mountain dams to valley power stations.
– Aetol
Dec 17 at 10:18




Pressure does matter to hydraulic power, that's why water is piped from mountain dams to valley power stations.
– Aetol
Dec 17 at 10:18










9 Answers
9






active

oldest

votes


















37














Power to a water-wheel depends both on the current (amount of water delivered)
and the head (vertical drop of water as it turns the wheel). So, the
water analogy does have TWO variables that multiply together to make
power: current, measuring (for instance) the water flow at Niagara,
and vertical drop (like the height of Niagara Falls).



Current is NOT the same as power, in a river, because long stretches of
moving water in a channel don't dissipate energy as much as a waterfall does.
Siting a hydroelectric power plant at Niagara Falls makes sense.
In the analogy to electricity, a wire can deliver current at little voltage
drop (and has tiny power dissipation) but a resistor which has that same
current will be warmed (it has a substantial terminal-to-terminal voltage drop).






share|cite|improve this answer





























    22














    Here is a simple way to keep this stuff straight.



    Power is always the product of an effort variable and a flow variable. In hydraulic systems, the effort variable is pressure and the flow variable is the flow rate.



    For flow in open channels, the effort variable is typically very small (but not zero) and the flow variable is very large. BTW power exchange which occurs at low effort and large flow represents the low-impedance regime.






    share|cite|improve this answer





















    • Very small compared to... flow in pipes? In open channels the effort variable is predominately height rather than pressure, but in any flow it's always the combination (could be evaluated via bernoulli's equation)
      – Rick
      Dec 17 at 13:00










    • flow in pipes, or in hydraulic systems. the driving pressure in open channel flow is so small it is measured in inches of water column rather than pounds per square inch, but it is still pressure.
      – niels nielsen
      Dec 17 at 16:49










    • For a water wheel (as pictured in Cort's answer) the effort from pressure is rather irrelevant compared to the effort from the height difference. If this height effort is included in your driving pressure that wasn't clear to me.
      – Rick
      Dec 17 at 17:13






    • 2




      Ahh I think I see where we miscommunicated. I think you're saying that the effort variable is measured in units of pressure (which could be calculated as $P_0=rho,g,h+frac12rho,v^2+p$) regardless of which term is dominant in bernoulli's equation. I was just thrown off by your comment that open channels have low effort and thought you were neglecting to consider the height portion of the equation.
      – Rick
      Dec 17 at 18:23






    • 1




      And in this specific case, the "stuff" that flows is a charge, so the effort per unit stuff is a voltage, the stuff per unit time is a current but the dimension of (effort per unit stuff) × (stuff per unit time) is independent of what "stuff" is.
      – Alexis
      Dec 18 at 19:58



















    12














    In your water example power cannot be equal to current because they have different units (power is an energy per unit time, while current would be something like a number of particles passing through a surface per unit time).




    ...the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time




    What you have noticed here through your analogy is that power is proportional to current (as an example, the more force you apply to an object, the larger its acceleration, but this does not mean that force and acceleration are equal). In a circuit element this proportionality is the voltage, since it tells you how much energy is associated with a "unit of current". You would need a similar way to convert your water current to the power generated by that current (although this might be a simplistic model of how power is generated using a water wheel).



    You also have to keep in mind that it is an analogy, and all analogies have imperfections. With the water analogy, power is generated by water actually pushing on a wheel. In circuits, $P=IV$ is much more general and applies to any charges undergoing a potential difference.






    share|cite|improve this answer































      11














      One of the challenges with using analogies like the water analogy is making sure you use the correct objects to draw your analogy from. Many answers here argue that hydraulic power depends on volume and pressure. This is true if your water wheel looks something like this:



      Turbine



      This is a drawing of what a modern hydroelectric turbine looks like. They are designed to be able to efficiently draw power from large volumes of water with large pressure drops, such as the pressure drop from the bottom of a lake out to atmospheric pressure. In these cases, the water analogy works rather well, as expected.



      However, when I think of a "water wheel," I have a different image. I imagine something much older:



      Water wheel



      These operate differently, and lead to the conclusion you drew, which is that water wheels generate power from current alone. The reason you end up drawing this conclusion is because this sort of water wheel wastes any power from pressure or velocity. The only potential it is effective at generating power from is the gravitational energy of the water from a high altitude to a low altitude. If you were to spray high pressure water at one of these water wheels, most of the energy would be wasted as the water splashes off of the blades. Some energy would indeed be transferred to the wheel, but it would be tremendously wasteful.



      The water wheel is most efficient at handling cases where the vast majority of the energy of the water is stored up as gravitational potential energy -- energy from being up high. And it's best at converting water which is exactly at the height of the wheel. If you drop water down from high above the wheel it will turn, but most of the energy will be wasted in splashing and sloshing.



      Thus we would treat this water wheel as a "constant voltage" device, in our electrical analogy. The setup around the wheel has the effect of guaranteeing that most of the potential of the water is its fixed height as it enters the wheel. Any energy above this is wasted as heat. And, indeed, if you look at the math, when your voltage is constant, your power is indeed proportional to the current. It's the special case where this is true.



      We do indeed have devices that operate this way, but you have to enter the world of semiconductors to do so. Diodes are small semiconductor junctions used to make current flow only one way. Try to flow against it, and they're like a check valve that stops the water flow.



      Well, almost like a check valve. They operate like a check valve up to a point, called the "breakdown voltage." If you put a voltage higher than this across the diode in the wrong direction, it starts to let current through. It will dissipate any energy that comes from running current across this voltage drop as heat.



      So what the old fashioned water wheel is most like is a motor with a reverse-biased diode on it. Any potential of the water beyond the gravitational potential energy the wheel can handle is lost to splashing and sloshing. Any potential from higher voltages applied to the diode and motor circuit is lost to heat as the current flows through the diode. The remaining gravitational potential of the water multiplied by the volume of water sent through the wheel tells you how much mechanical power is generated by the water wheel. The voltage across that motor (after the diode limits it) multiplied by the current going through the motor is how much mechanical power the motor generates. The analogy holds, you just need a more complex circuit to model the 6000 year old device!



      Incidentally, we do actually design circuits like this. In modern circuitry, we often have "zener diodes" which have a carefully tuned breakdown voltage to be a "voltage reference," and we have voltage regulators which are designed to resist the flow of electricity just enough to ensure a specific voltage across the remaining circuit.






      share|cite|improve this answer























      • The water wheel actually does benefit from velocity. During the drop the water gets accelerated by gravity. This energy gets absorbed by the wheel's blades as the fluid de-accelerates.
        – Muhlemmer
        Dec 17 at 16:19










      • @Muhlemmer Actually, if you think about it, it doesn't really get accelerated. It does have the force of gravity pulling on it, but if it accelerated on the way down, the bottom of the wheel would have to be spinning faster than the top of the wheel. (And it does benefit a little, but its limited) There's also angular acceleration from bending around the wheel, but that's also pretty slight.
        – Cort Ammon
        Dec 17 at 16:26












      • +1 for emphasizing the point that while the amount of useful energy a water wheel harvests for a unit quantity of water may be roughly constant, the amount of energy removed per unit quantity is proportional to the pressure difference before and after, with increases in pressure differential leading to increases in waste.
        – supercat
        Dec 17 at 18:30



















      2














      Water and current flow as well as other “mechanical” analogies (pipe resistance vs. electrical resistance, voltage vs. pressure, etc.) are useful for introducing electrical circuit concepts at an elementary level. This is because mechanical concepts are easier to visualize whereas electrical concepts are more abstract. The analogies can only go so far without a deeper understanding.



      Current does not equal electrical power and neither does water flow equal mechanical power.



      Electrical current ($frac {Coul}{s}$) times (in phase) voltage ($frac {J}{Coul}$) equals power ($frac {J}{s}$ = watts).



      Current flow ($frac {m^3}{s}$) times pressure ($frac {N}{m^2}$) equals power ($frac {J}{s}$ = watts).
      The water pressure for the water wheel can come from water dropping from a height above the wheel (potential energy) or even from a hose directed horizontally, depending on the orientation of the water wheel.



      To complete the analogy between current and water flow with respect to electrical an mechanical power, you need in addition the analogy between voltage (electrical potential) and pressure (mechanical potential).



      Hope this helps.






      share|cite|improve this answer































        2














        The person who said that is wrong. Current doesn't push. Voltage or pressure pushes. Current is how much actually flows.



        Useful power is the force of the push (voltage, pressure) multiplied by the current (amps, CFM).



        Obviously with no push, you have no power. But with no flow, you also have no power.



        In water generation, pressure also equates to height, called "head".



        For instance, a PC power supply is a lot like run-of-river hydro -- not much pressure but lots of flow.



        A very tall, but low-flow dam like Oroville is a lot like an LED light on a 240V line - lots of pressure, not much flow. Not much flow needed.






        share|cite|improve this answer








        New contributor




        Harper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.


























          1














          The analogy with water actually holds really nicely if you consider a water wheel, or other hydroelectric system.



          But what you're missing is that the power produced does not only depend on the amount of water going past - it also depends on the speed at which it does so. (this makes sense for the hydro system because kinetic energy depends on velocity and mass)



          To make the analogy better, rather than thinking of the speed of the flow, think about how far it has fallen to aquire that speed. At this point you have a volume per second of water - the current - and you have a loss of height, which is literally a potential difference.






          share|cite|improve this answer





























            1














            I'd like to put what others said already into equations:



            A river has mass flow rate $dot m$ (kg/s) (the "current"). The water flows with velocity $v$ (m/s). Power is the kinetic energy that is carried per unit time:



            $$
            dot W = frac{dot m v^2}{2}
            $$

            and notice that this has proper units of watt. If we take "voltage" to be $v^2/2$, then we get
            $$
            text{(power)} = text{(current)}timestext{(voltage)}
            $$



            Notice that the mass flow rate is not enough to give you high power. When the river is wide the velocity is low. If you want the water wheel to run fast you should build it at a narrow passage.





            Variation



            The river flows with fixed flow rate $dot m$, the width of the river is constant throughout, but the river is made to go over a waterfall that drops by distance $Delta h$. The velocity at the top of the waterfall is $v_1$ and the bottom $v_2$. By energy balance,
            $$
            frac{v_2^2-v_1^2}{2}=g Delta h
            $$

            The power that is available to us is
            $$
            dot W = dot m frac{v_2^2-v_1^2}{2}
            $$

            We have to subtract $v_1^2$ because the river must continue to flow with its steady velocity after it hits the wheel, or else we will flood. Combining with the energy balance:
            $$
            dot W = dot m g Delta h
            $$

            Here now we have a one-to-one analogy with electric current: $dot m$ is the current and $gDelta h$ is the voltage. Increase the voltage and the same current will produce more power.






            share|cite|improve this answer































              0














              Power is defined as work done per unit time. So a mass of water moving from one potential to some other lower potetial can do work when it hits the wheel. How much work per unit time? It depends how much mass falls times height times gravitational constant g and all of that divided by time. Water current, on the other hand is just total volume or mass that flows per unit time. Sure, it is connected to the work done or power but is not the same thing. If you define gravitational potential difference as gH and water flow as dm/dt then to have power you have to have curent times this potential difference: gHdm/dt...
              In a conductor, work is done also and energy per unit charge or potential difference is given by U (voltage) and current by dQ/dt so it seems to me that everything is same...






              share|cite|improve this answer





















                Your Answer





                StackExchange.ifUsing("editor", function () {
                return StackExchange.using("mathjaxEditing", function () {
                StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
                StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
                });
                });
                }, "mathjax-editing");

                StackExchange.ready(function() {
                var channelOptions = {
                tags: "".split(" "),
                id: "151"
                };
                initTagRenderer("".split(" "), "".split(" "), channelOptions);

                StackExchange.using("externalEditor", function() {
                // Have to fire editor after snippets, if snippets enabled
                if (StackExchange.settings.snippets.snippetsEnabled) {
                StackExchange.using("snippets", function() {
                createEditor();
                });
                }
                else {
                createEditor();
                }
                });

                function createEditor() {
                StackExchange.prepareEditor({
                heartbeatType: 'answer',
                autoActivateHeartbeat: false,
                convertImagesToLinks: false,
                noModals: true,
                showLowRepImageUploadWarning: true,
                reputationToPostImages: null,
                bindNavPrevention: true,
                postfix: "",
                imageUploader: {
                brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                allowUrls: true
                },
                noCode: true, onDemand: true,
                discardSelector: ".discard-answer"
                ,immediatelyShowMarkdownHelp:true
                });


                }
                });














                draft saved

                draft discarded


















                StackExchange.ready(
                function () {
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f448724%2fthe-water-analogy-seems-to-imply-that-power-current-why-is-this-incorrect%23new-answer', 'question_page');
                }
                );

                Post as a guest















                Required, but never shown

























                9 Answers
                9






                active

                oldest

                votes








                9 Answers
                9






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                37














                Power to a water-wheel depends both on the current (amount of water delivered)
                and the head (vertical drop of water as it turns the wheel). So, the
                water analogy does have TWO variables that multiply together to make
                power: current, measuring (for instance) the water flow at Niagara,
                and vertical drop (like the height of Niagara Falls).



                Current is NOT the same as power, in a river, because long stretches of
                moving water in a channel don't dissipate energy as much as a waterfall does.
                Siting a hydroelectric power plant at Niagara Falls makes sense.
                In the analogy to electricity, a wire can deliver current at little voltage
                drop (and has tiny power dissipation) but a resistor which has that same
                current will be warmed (it has a substantial terminal-to-terminal voltage drop).






                share|cite|improve this answer


























                  37














                  Power to a water-wheel depends both on the current (amount of water delivered)
                  and the head (vertical drop of water as it turns the wheel). So, the
                  water analogy does have TWO variables that multiply together to make
                  power: current, measuring (for instance) the water flow at Niagara,
                  and vertical drop (like the height of Niagara Falls).



                  Current is NOT the same as power, in a river, because long stretches of
                  moving water in a channel don't dissipate energy as much as a waterfall does.
                  Siting a hydroelectric power plant at Niagara Falls makes sense.
                  In the analogy to electricity, a wire can deliver current at little voltage
                  drop (and has tiny power dissipation) but a resistor which has that same
                  current will be warmed (it has a substantial terminal-to-terminal voltage drop).






                  share|cite|improve this answer
























                    37












                    37








                    37






                    Power to a water-wheel depends both on the current (amount of water delivered)
                    and the head (vertical drop of water as it turns the wheel). So, the
                    water analogy does have TWO variables that multiply together to make
                    power: current, measuring (for instance) the water flow at Niagara,
                    and vertical drop (like the height of Niagara Falls).



                    Current is NOT the same as power, in a river, because long stretches of
                    moving water in a channel don't dissipate energy as much as a waterfall does.
                    Siting a hydroelectric power plant at Niagara Falls makes sense.
                    In the analogy to electricity, a wire can deliver current at little voltage
                    drop (and has tiny power dissipation) but a resistor which has that same
                    current will be warmed (it has a substantial terminal-to-terminal voltage drop).






                    share|cite|improve this answer












                    Power to a water-wheel depends both on the current (amount of water delivered)
                    and the head (vertical drop of water as it turns the wheel). So, the
                    water analogy does have TWO variables that multiply together to make
                    power: current, measuring (for instance) the water flow at Niagara,
                    and vertical drop (like the height of Niagara Falls).



                    Current is NOT the same as power, in a river, because long stretches of
                    moving water in a channel don't dissipate energy as much as a waterfall does.
                    Siting a hydroelectric power plant at Niagara Falls makes sense.
                    In the analogy to electricity, a wire can deliver current at little voltage
                    drop (and has tiny power dissipation) but a resistor which has that same
                    current will be warmed (it has a substantial terminal-to-terminal voltage drop).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 16 at 23:00









                    Whit3rd

                    6,67121326




                    6,67121326























                        22














                        Here is a simple way to keep this stuff straight.



                        Power is always the product of an effort variable and a flow variable. In hydraulic systems, the effort variable is pressure and the flow variable is the flow rate.



                        For flow in open channels, the effort variable is typically very small (but not zero) and the flow variable is very large. BTW power exchange which occurs at low effort and large flow represents the low-impedance regime.






                        share|cite|improve this answer





















                        • Very small compared to... flow in pipes? In open channels the effort variable is predominately height rather than pressure, but in any flow it's always the combination (could be evaluated via bernoulli's equation)
                          – Rick
                          Dec 17 at 13:00










                        • flow in pipes, or in hydraulic systems. the driving pressure in open channel flow is so small it is measured in inches of water column rather than pounds per square inch, but it is still pressure.
                          – niels nielsen
                          Dec 17 at 16:49










                        • For a water wheel (as pictured in Cort's answer) the effort from pressure is rather irrelevant compared to the effort from the height difference. If this height effort is included in your driving pressure that wasn't clear to me.
                          – Rick
                          Dec 17 at 17:13






                        • 2




                          Ahh I think I see where we miscommunicated. I think you're saying that the effort variable is measured in units of pressure (which could be calculated as $P_0=rho,g,h+frac12rho,v^2+p$) regardless of which term is dominant in bernoulli's equation. I was just thrown off by your comment that open channels have low effort and thought you were neglecting to consider the height portion of the equation.
                          – Rick
                          Dec 17 at 18:23






                        • 1




                          And in this specific case, the "stuff" that flows is a charge, so the effort per unit stuff is a voltage, the stuff per unit time is a current but the dimension of (effort per unit stuff) × (stuff per unit time) is independent of what "stuff" is.
                          – Alexis
                          Dec 18 at 19:58
















                        22














                        Here is a simple way to keep this stuff straight.



                        Power is always the product of an effort variable and a flow variable. In hydraulic systems, the effort variable is pressure and the flow variable is the flow rate.



                        For flow in open channels, the effort variable is typically very small (but not zero) and the flow variable is very large. BTW power exchange which occurs at low effort and large flow represents the low-impedance regime.






                        share|cite|improve this answer





















                        • Very small compared to... flow in pipes? In open channels the effort variable is predominately height rather than pressure, but in any flow it's always the combination (could be evaluated via bernoulli's equation)
                          – Rick
                          Dec 17 at 13:00










                        • flow in pipes, or in hydraulic systems. the driving pressure in open channel flow is so small it is measured in inches of water column rather than pounds per square inch, but it is still pressure.
                          – niels nielsen
                          Dec 17 at 16:49










                        • For a water wheel (as pictured in Cort's answer) the effort from pressure is rather irrelevant compared to the effort from the height difference. If this height effort is included in your driving pressure that wasn't clear to me.
                          – Rick
                          Dec 17 at 17:13






                        • 2




                          Ahh I think I see where we miscommunicated. I think you're saying that the effort variable is measured in units of pressure (which could be calculated as $P_0=rho,g,h+frac12rho,v^2+p$) regardless of which term is dominant in bernoulli's equation. I was just thrown off by your comment that open channels have low effort and thought you were neglecting to consider the height portion of the equation.
                          – Rick
                          Dec 17 at 18:23






                        • 1




                          And in this specific case, the "stuff" that flows is a charge, so the effort per unit stuff is a voltage, the stuff per unit time is a current but the dimension of (effort per unit stuff) × (stuff per unit time) is independent of what "stuff" is.
                          – Alexis
                          Dec 18 at 19:58














                        22












                        22








                        22






                        Here is a simple way to keep this stuff straight.



                        Power is always the product of an effort variable and a flow variable. In hydraulic systems, the effort variable is pressure and the flow variable is the flow rate.



                        For flow in open channels, the effort variable is typically very small (but not zero) and the flow variable is very large. BTW power exchange which occurs at low effort and large flow represents the low-impedance regime.






                        share|cite|improve this answer












                        Here is a simple way to keep this stuff straight.



                        Power is always the product of an effort variable and a flow variable. In hydraulic systems, the effort variable is pressure and the flow variable is the flow rate.



                        For flow in open channels, the effort variable is typically very small (but not zero) and the flow variable is very large. BTW power exchange which occurs at low effort and large flow represents the low-impedance regime.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 16 at 21:47









                        niels nielsen

                        15.5k42650




                        15.5k42650












                        • Very small compared to... flow in pipes? In open channels the effort variable is predominately height rather than pressure, but in any flow it's always the combination (could be evaluated via bernoulli's equation)
                          – Rick
                          Dec 17 at 13:00










                        • flow in pipes, or in hydraulic systems. the driving pressure in open channel flow is so small it is measured in inches of water column rather than pounds per square inch, but it is still pressure.
                          – niels nielsen
                          Dec 17 at 16:49










                        • For a water wheel (as pictured in Cort's answer) the effort from pressure is rather irrelevant compared to the effort from the height difference. If this height effort is included in your driving pressure that wasn't clear to me.
                          – Rick
                          Dec 17 at 17:13






                        • 2




                          Ahh I think I see where we miscommunicated. I think you're saying that the effort variable is measured in units of pressure (which could be calculated as $P_0=rho,g,h+frac12rho,v^2+p$) regardless of which term is dominant in bernoulli's equation. I was just thrown off by your comment that open channels have low effort and thought you were neglecting to consider the height portion of the equation.
                          – Rick
                          Dec 17 at 18:23






                        • 1




                          And in this specific case, the "stuff" that flows is a charge, so the effort per unit stuff is a voltage, the stuff per unit time is a current but the dimension of (effort per unit stuff) × (stuff per unit time) is independent of what "stuff" is.
                          – Alexis
                          Dec 18 at 19:58


















                        • Very small compared to... flow in pipes? In open channels the effort variable is predominately height rather than pressure, but in any flow it's always the combination (could be evaluated via bernoulli's equation)
                          – Rick
                          Dec 17 at 13:00










                        • flow in pipes, or in hydraulic systems. the driving pressure in open channel flow is so small it is measured in inches of water column rather than pounds per square inch, but it is still pressure.
                          – niels nielsen
                          Dec 17 at 16:49










                        • For a water wheel (as pictured in Cort's answer) the effort from pressure is rather irrelevant compared to the effort from the height difference. If this height effort is included in your driving pressure that wasn't clear to me.
                          – Rick
                          Dec 17 at 17:13






                        • 2




                          Ahh I think I see where we miscommunicated. I think you're saying that the effort variable is measured in units of pressure (which could be calculated as $P_0=rho,g,h+frac12rho,v^2+p$) regardless of which term is dominant in bernoulli's equation. I was just thrown off by your comment that open channels have low effort and thought you were neglecting to consider the height portion of the equation.
                          – Rick
                          Dec 17 at 18:23






                        • 1




                          And in this specific case, the "stuff" that flows is a charge, so the effort per unit stuff is a voltage, the stuff per unit time is a current but the dimension of (effort per unit stuff) × (stuff per unit time) is independent of what "stuff" is.
                          – Alexis
                          Dec 18 at 19:58
















                        Very small compared to... flow in pipes? In open channels the effort variable is predominately height rather than pressure, but in any flow it's always the combination (could be evaluated via bernoulli's equation)
                        – Rick
                        Dec 17 at 13:00




                        Very small compared to... flow in pipes? In open channels the effort variable is predominately height rather than pressure, but in any flow it's always the combination (could be evaluated via bernoulli's equation)
                        – Rick
                        Dec 17 at 13:00












                        flow in pipes, or in hydraulic systems. the driving pressure in open channel flow is so small it is measured in inches of water column rather than pounds per square inch, but it is still pressure.
                        – niels nielsen
                        Dec 17 at 16:49




                        flow in pipes, or in hydraulic systems. the driving pressure in open channel flow is so small it is measured in inches of water column rather than pounds per square inch, but it is still pressure.
                        – niels nielsen
                        Dec 17 at 16:49












                        For a water wheel (as pictured in Cort's answer) the effort from pressure is rather irrelevant compared to the effort from the height difference. If this height effort is included in your driving pressure that wasn't clear to me.
                        – Rick
                        Dec 17 at 17:13




                        For a water wheel (as pictured in Cort's answer) the effort from pressure is rather irrelevant compared to the effort from the height difference. If this height effort is included in your driving pressure that wasn't clear to me.
                        – Rick
                        Dec 17 at 17:13




                        2




                        2




                        Ahh I think I see where we miscommunicated. I think you're saying that the effort variable is measured in units of pressure (which could be calculated as $P_0=rho,g,h+frac12rho,v^2+p$) regardless of which term is dominant in bernoulli's equation. I was just thrown off by your comment that open channels have low effort and thought you were neglecting to consider the height portion of the equation.
                        – Rick
                        Dec 17 at 18:23




                        Ahh I think I see where we miscommunicated. I think you're saying that the effort variable is measured in units of pressure (which could be calculated as $P_0=rho,g,h+frac12rho,v^2+p$) regardless of which term is dominant in bernoulli's equation. I was just thrown off by your comment that open channels have low effort and thought you were neglecting to consider the height portion of the equation.
                        – Rick
                        Dec 17 at 18:23




                        1




                        1




                        And in this specific case, the "stuff" that flows is a charge, so the effort per unit stuff is a voltage, the stuff per unit time is a current but the dimension of (effort per unit stuff) × (stuff per unit time) is independent of what "stuff" is.
                        – Alexis
                        Dec 18 at 19:58




                        And in this specific case, the "stuff" that flows is a charge, so the effort per unit stuff is a voltage, the stuff per unit time is a current but the dimension of (effort per unit stuff) × (stuff per unit time) is independent of what "stuff" is.
                        – Alexis
                        Dec 18 at 19:58











                        12














                        In your water example power cannot be equal to current because they have different units (power is an energy per unit time, while current would be something like a number of particles passing through a surface per unit time).




                        ...the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time




                        What you have noticed here through your analogy is that power is proportional to current (as an example, the more force you apply to an object, the larger its acceleration, but this does not mean that force and acceleration are equal). In a circuit element this proportionality is the voltage, since it tells you how much energy is associated with a "unit of current". You would need a similar way to convert your water current to the power generated by that current (although this might be a simplistic model of how power is generated using a water wheel).



                        You also have to keep in mind that it is an analogy, and all analogies have imperfections. With the water analogy, power is generated by water actually pushing on a wheel. In circuits, $P=IV$ is much more general and applies to any charges undergoing a potential difference.






                        share|cite|improve this answer




























                          12














                          In your water example power cannot be equal to current because they have different units (power is an energy per unit time, while current would be something like a number of particles passing through a surface per unit time).




                          ...the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time




                          What you have noticed here through your analogy is that power is proportional to current (as an example, the more force you apply to an object, the larger its acceleration, but this does not mean that force and acceleration are equal). In a circuit element this proportionality is the voltage, since it tells you how much energy is associated with a "unit of current". You would need a similar way to convert your water current to the power generated by that current (although this might be a simplistic model of how power is generated using a water wheel).



                          You also have to keep in mind that it is an analogy, and all analogies have imperfections. With the water analogy, power is generated by water actually pushing on a wheel. In circuits, $P=IV$ is much more general and applies to any charges undergoing a potential difference.






                          share|cite|improve this answer


























                            12












                            12








                            12






                            In your water example power cannot be equal to current because they have different units (power is an energy per unit time, while current would be something like a number of particles passing through a surface per unit time).




                            ...the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time




                            What you have noticed here through your analogy is that power is proportional to current (as an example, the more force you apply to an object, the larger its acceleration, but this does not mean that force and acceleration are equal). In a circuit element this proportionality is the voltage, since it tells you how much energy is associated with a "unit of current". You would need a similar way to convert your water current to the power generated by that current (although this might be a simplistic model of how power is generated using a water wheel).



                            You also have to keep in mind that it is an analogy, and all analogies have imperfections. With the water analogy, power is generated by water actually pushing on a wheel. In circuits, $P=IV$ is much more general and applies to any charges undergoing a potential difference.






                            share|cite|improve this answer














                            In your water example power cannot be equal to current because they have different units (power is an energy per unit time, while current would be something like a number of particles passing through a surface per unit time).




                            ...the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time




                            What you have noticed here through your analogy is that power is proportional to current (as an example, the more force you apply to an object, the larger its acceleration, but this does not mean that force and acceleration are equal). In a circuit element this proportionality is the voltage, since it tells you how much energy is associated with a "unit of current". You would need a similar way to convert your water current to the power generated by that current (although this might be a simplistic model of how power is generated using a water wheel).



                            You also have to keep in mind that it is an analogy, and all analogies have imperfections. With the water analogy, power is generated by water actually pushing on a wheel. In circuits, $P=IV$ is much more general and applies to any charges undergoing a potential difference.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 16 at 19:43

























                            answered Dec 16 at 19:28









                            Aaron Stevens

                            8,79131640




                            8,79131640























                                11














                                One of the challenges with using analogies like the water analogy is making sure you use the correct objects to draw your analogy from. Many answers here argue that hydraulic power depends on volume and pressure. This is true if your water wheel looks something like this:



                                Turbine



                                This is a drawing of what a modern hydroelectric turbine looks like. They are designed to be able to efficiently draw power from large volumes of water with large pressure drops, such as the pressure drop from the bottom of a lake out to atmospheric pressure. In these cases, the water analogy works rather well, as expected.



                                However, when I think of a "water wheel," I have a different image. I imagine something much older:



                                Water wheel



                                These operate differently, and lead to the conclusion you drew, which is that water wheels generate power from current alone. The reason you end up drawing this conclusion is because this sort of water wheel wastes any power from pressure or velocity. The only potential it is effective at generating power from is the gravitational energy of the water from a high altitude to a low altitude. If you were to spray high pressure water at one of these water wheels, most of the energy would be wasted as the water splashes off of the blades. Some energy would indeed be transferred to the wheel, but it would be tremendously wasteful.



                                The water wheel is most efficient at handling cases where the vast majority of the energy of the water is stored up as gravitational potential energy -- energy from being up high. And it's best at converting water which is exactly at the height of the wheel. If you drop water down from high above the wheel it will turn, but most of the energy will be wasted in splashing and sloshing.



                                Thus we would treat this water wheel as a "constant voltage" device, in our electrical analogy. The setup around the wheel has the effect of guaranteeing that most of the potential of the water is its fixed height as it enters the wheel. Any energy above this is wasted as heat. And, indeed, if you look at the math, when your voltage is constant, your power is indeed proportional to the current. It's the special case where this is true.



                                We do indeed have devices that operate this way, but you have to enter the world of semiconductors to do so. Diodes are small semiconductor junctions used to make current flow only one way. Try to flow against it, and they're like a check valve that stops the water flow.



                                Well, almost like a check valve. They operate like a check valve up to a point, called the "breakdown voltage." If you put a voltage higher than this across the diode in the wrong direction, it starts to let current through. It will dissipate any energy that comes from running current across this voltage drop as heat.



                                So what the old fashioned water wheel is most like is a motor with a reverse-biased diode on it. Any potential of the water beyond the gravitational potential energy the wheel can handle is lost to splashing and sloshing. Any potential from higher voltages applied to the diode and motor circuit is lost to heat as the current flows through the diode. The remaining gravitational potential of the water multiplied by the volume of water sent through the wheel tells you how much mechanical power is generated by the water wheel. The voltage across that motor (after the diode limits it) multiplied by the current going through the motor is how much mechanical power the motor generates. The analogy holds, you just need a more complex circuit to model the 6000 year old device!



                                Incidentally, we do actually design circuits like this. In modern circuitry, we often have "zener diodes" which have a carefully tuned breakdown voltage to be a "voltage reference," and we have voltage regulators which are designed to resist the flow of electricity just enough to ensure a specific voltage across the remaining circuit.






                                share|cite|improve this answer























                                • The water wheel actually does benefit from velocity. During the drop the water gets accelerated by gravity. This energy gets absorbed by the wheel's blades as the fluid de-accelerates.
                                  – Muhlemmer
                                  Dec 17 at 16:19










                                • @Muhlemmer Actually, if you think about it, it doesn't really get accelerated. It does have the force of gravity pulling on it, but if it accelerated on the way down, the bottom of the wheel would have to be spinning faster than the top of the wheel. (And it does benefit a little, but its limited) There's also angular acceleration from bending around the wheel, but that's also pretty slight.
                                  – Cort Ammon
                                  Dec 17 at 16:26












                                • +1 for emphasizing the point that while the amount of useful energy a water wheel harvests for a unit quantity of water may be roughly constant, the amount of energy removed per unit quantity is proportional to the pressure difference before and after, with increases in pressure differential leading to increases in waste.
                                  – supercat
                                  Dec 17 at 18:30
















                                11














                                One of the challenges with using analogies like the water analogy is making sure you use the correct objects to draw your analogy from. Many answers here argue that hydraulic power depends on volume and pressure. This is true if your water wheel looks something like this:



                                Turbine



                                This is a drawing of what a modern hydroelectric turbine looks like. They are designed to be able to efficiently draw power from large volumes of water with large pressure drops, such as the pressure drop from the bottom of a lake out to atmospheric pressure. In these cases, the water analogy works rather well, as expected.



                                However, when I think of a "water wheel," I have a different image. I imagine something much older:



                                Water wheel



                                These operate differently, and lead to the conclusion you drew, which is that water wheels generate power from current alone. The reason you end up drawing this conclusion is because this sort of water wheel wastes any power from pressure or velocity. The only potential it is effective at generating power from is the gravitational energy of the water from a high altitude to a low altitude. If you were to spray high pressure water at one of these water wheels, most of the energy would be wasted as the water splashes off of the blades. Some energy would indeed be transferred to the wheel, but it would be tremendously wasteful.



                                The water wheel is most efficient at handling cases where the vast majority of the energy of the water is stored up as gravitational potential energy -- energy from being up high. And it's best at converting water which is exactly at the height of the wheel. If you drop water down from high above the wheel it will turn, but most of the energy will be wasted in splashing and sloshing.



                                Thus we would treat this water wheel as a "constant voltage" device, in our electrical analogy. The setup around the wheel has the effect of guaranteeing that most of the potential of the water is its fixed height as it enters the wheel. Any energy above this is wasted as heat. And, indeed, if you look at the math, when your voltage is constant, your power is indeed proportional to the current. It's the special case where this is true.



                                We do indeed have devices that operate this way, but you have to enter the world of semiconductors to do so. Diodes are small semiconductor junctions used to make current flow only one way. Try to flow against it, and they're like a check valve that stops the water flow.



                                Well, almost like a check valve. They operate like a check valve up to a point, called the "breakdown voltage." If you put a voltage higher than this across the diode in the wrong direction, it starts to let current through. It will dissipate any energy that comes from running current across this voltage drop as heat.



                                So what the old fashioned water wheel is most like is a motor with a reverse-biased diode on it. Any potential of the water beyond the gravitational potential energy the wheel can handle is lost to splashing and sloshing. Any potential from higher voltages applied to the diode and motor circuit is lost to heat as the current flows through the diode. The remaining gravitational potential of the water multiplied by the volume of water sent through the wheel tells you how much mechanical power is generated by the water wheel. The voltage across that motor (after the diode limits it) multiplied by the current going through the motor is how much mechanical power the motor generates. The analogy holds, you just need a more complex circuit to model the 6000 year old device!



                                Incidentally, we do actually design circuits like this. In modern circuitry, we often have "zener diodes" which have a carefully tuned breakdown voltage to be a "voltage reference," and we have voltage regulators which are designed to resist the flow of electricity just enough to ensure a specific voltage across the remaining circuit.






                                share|cite|improve this answer























                                • The water wheel actually does benefit from velocity. During the drop the water gets accelerated by gravity. This energy gets absorbed by the wheel's blades as the fluid de-accelerates.
                                  – Muhlemmer
                                  Dec 17 at 16:19










                                • @Muhlemmer Actually, if you think about it, it doesn't really get accelerated. It does have the force of gravity pulling on it, but if it accelerated on the way down, the bottom of the wheel would have to be spinning faster than the top of the wheel. (And it does benefit a little, but its limited) There's also angular acceleration from bending around the wheel, but that's also pretty slight.
                                  – Cort Ammon
                                  Dec 17 at 16:26












                                • +1 for emphasizing the point that while the amount of useful energy a water wheel harvests for a unit quantity of water may be roughly constant, the amount of energy removed per unit quantity is proportional to the pressure difference before and after, with increases in pressure differential leading to increases in waste.
                                  – supercat
                                  Dec 17 at 18:30














                                11












                                11








                                11






                                One of the challenges with using analogies like the water analogy is making sure you use the correct objects to draw your analogy from. Many answers here argue that hydraulic power depends on volume and pressure. This is true if your water wheel looks something like this:



                                Turbine



                                This is a drawing of what a modern hydroelectric turbine looks like. They are designed to be able to efficiently draw power from large volumes of water with large pressure drops, such as the pressure drop from the bottom of a lake out to atmospheric pressure. In these cases, the water analogy works rather well, as expected.



                                However, when I think of a "water wheel," I have a different image. I imagine something much older:



                                Water wheel



                                These operate differently, and lead to the conclusion you drew, which is that water wheels generate power from current alone. The reason you end up drawing this conclusion is because this sort of water wheel wastes any power from pressure or velocity. The only potential it is effective at generating power from is the gravitational energy of the water from a high altitude to a low altitude. If you were to spray high pressure water at one of these water wheels, most of the energy would be wasted as the water splashes off of the blades. Some energy would indeed be transferred to the wheel, but it would be tremendously wasteful.



                                The water wheel is most efficient at handling cases where the vast majority of the energy of the water is stored up as gravitational potential energy -- energy from being up high. And it's best at converting water which is exactly at the height of the wheel. If you drop water down from high above the wheel it will turn, but most of the energy will be wasted in splashing and sloshing.



                                Thus we would treat this water wheel as a "constant voltage" device, in our electrical analogy. The setup around the wheel has the effect of guaranteeing that most of the potential of the water is its fixed height as it enters the wheel. Any energy above this is wasted as heat. And, indeed, if you look at the math, when your voltage is constant, your power is indeed proportional to the current. It's the special case where this is true.



                                We do indeed have devices that operate this way, but you have to enter the world of semiconductors to do so. Diodes are small semiconductor junctions used to make current flow only one way. Try to flow against it, and they're like a check valve that stops the water flow.



                                Well, almost like a check valve. They operate like a check valve up to a point, called the "breakdown voltage." If you put a voltage higher than this across the diode in the wrong direction, it starts to let current through. It will dissipate any energy that comes from running current across this voltage drop as heat.



                                So what the old fashioned water wheel is most like is a motor with a reverse-biased diode on it. Any potential of the water beyond the gravitational potential energy the wheel can handle is lost to splashing and sloshing. Any potential from higher voltages applied to the diode and motor circuit is lost to heat as the current flows through the diode. The remaining gravitational potential of the water multiplied by the volume of water sent through the wheel tells you how much mechanical power is generated by the water wheel. The voltage across that motor (after the diode limits it) multiplied by the current going through the motor is how much mechanical power the motor generates. The analogy holds, you just need a more complex circuit to model the 6000 year old device!



                                Incidentally, we do actually design circuits like this. In modern circuitry, we often have "zener diodes" which have a carefully tuned breakdown voltage to be a "voltage reference," and we have voltage regulators which are designed to resist the flow of electricity just enough to ensure a specific voltage across the remaining circuit.






                                share|cite|improve this answer














                                One of the challenges with using analogies like the water analogy is making sure you use the correct objects to draw your analogy from. Many answers here argue that hydraulic power depends on volume and pressure. This is true if your water wheel looks something like this:



                                Turbine



                                This is a drawing of what a modern hydroelectric turbine looks like. They are designed to be able to efficiently draw power from large volumes of water with large pressure drops, such as the pressure drop from the bottom of a lake out to atmospheric pressure. In these cases, the water analogy works rather well, as expected.



                                However, when I think of a "water wheel," I have a different image. I imagine something much older:



                                Water wheel



                                These operate differently, and lead to the conclusion you drew, which is that water wheels generate power from current alone. The reason you end up drawing this conclusion is because this sort of water wheel wastes any power from pressure or velocity. The only potential it is effective at generating power from is the gravitational energy of the water from a high altitude to a low altitude. If you were to spray high pressure water at one of these water wheels, most of the energy would be wasted as the water splashes off of the blades. Some energy would indeed be transferred to the wheel, but it would be tremendously wasteful.



                                The water wheel is most efficient at handling cases where the vast majority of the energy of the water is stored up as gravitational potential energy -- energy from being up high. And it's best at converting water which is exactly at the height of the wheel. If you drop water down from high above the wheel it will turn, but most of the energy will be wasted in splashing and sloshing.



                                Thus we would treat this water wheel as a "constant voltage" device, in our electrical analogy. The setup around the wheel has the effect of guaranteeing that most of the potential of the water is its fixed height as it enters the wheel. Any energy above this is wasted as heat. And, indeed, if you look at the math, when your voltage is constant, your power is indeed proportional to the current. It's the special case where this is true.



                                We do indeed have devices that operate this way, but you have to enter the world of semiconductors to do so. Diodes are small semiconductor junctions used to make current flow only one way. Try to flow against it, and they're like a check valve that stops the water flow.



                                Well, almost like a check valve. They operate like a check valve up to a point, called the "breakdown voltage." If you put a voltage higher than this across the diode in the wrong direction, it starts to let current through. It will dissipate any energy that comes from running current across this voltage drop as heat.



                                So what the old fashioned water wheel is most like is a motor with a reverse-biased diode on it. Any potential of the water beyond the gravitational potential energy the wheel can handle is lost to splashing and sloshing. Any potential from higher voltages applied to the diode and motor circuit is lost to heat as the current flows through the diode. The remaining gravitational potential of the water multiplied by the volume of water sent through the wheel tells you how much mechanical power is generated by the water wheel. The voltage across that motor (after the diode limits it) multiplied by the current going through the motor is how much mechanical power the motor generates. The analogy holds, you just need a more complex circuit to model the 6000 year old device!



                                Incidentally, we do actually design circuits like this. In modern circuitry, we often have "zener diodes" which have a carefully tuned breakdown voltage to be a "voltage reference," and we have voltage regulators which are designed to resist the flow of electricity just enough to ensure a specific voltage across the remaining circuit.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Dec 17 at 16:05

























                                answered Dec 17 at 15:44









                                Cort Ammon

                                22.8k34573




                                22.8k34573












                                • The water wheel actually does benefit from velocity. During the drop the water gets accelerated by gravity. This energy gets absorbed by the wheel's blades as the fluid de-accelerates.
                                  – Muhlemmer
                                  Dec 17 at 16:19










                                • @Muhlemmer Actually, if you think about it, it doesn't really get accelerated. It does have the force of gravity pulling on it, but if it accelerated on the way down, the bottom of the wheel would have to be spinning faster than the top of the wheel. (And it does benefit a little, but its limited) There's also angular acceleration from bending around the wheel, but that's also pretty slight.
                                  – Cort Ammon
                                  Dec 17 at 16:26












                                • +1 for emphasizing the point that while the amount of useful energy a water wheel harvests for a unit quantity of water may be roughly constant, the amount of energy removed per unit quantity is proportional to the pressure difference before and after, with increases in pressure differential leading to increases in waste.
                                  – supercat
                                  Dec 17 at 18:30


















                                • The water wheel actually does benefit from velocity. During the drop the water gets accelerated by gravity. This energy gets absorbed by the wheel's blades as the fluid de-accelerates.
                                  – Muhlemmer
                                  Dec 17 at 16:19










                                • @Muhlemmer Actually, if you think about it, it doesn't really get accelerated. It does have the force of gravity pulling on it, but if it accelerated on the way down, the bottom of the wheel would have to be spinning faster than the top of the wheel. (And it does benefit a little, but its limited) There's also angular acceleration from bending around the wheel, but that's also pretty slight.
                                  – Cort Ammon
                                  Dec 17 at 16:26












                                • +1 for emphasizing the point that while the amount of useful energy a water wheel harvests for a unit quantity of water may be roughly constant, the amount of energy removed per unit quantity is proportional to the pressure difference before and after, with increases in pressure differential leading to increases in waste.
                                  – supercat
                                  Dec 17 at 18:30
















                                The water wheel actually does benefit from velocity. During the drop the water gets accelerated by gravity. This energy gets absorbed by the wheel's blades as the fluid de-accelerates.
                                – Muhlemmer
                                Dec 17 at 16:19




                                The water wheel actually does benefit from velocity. During the drop the water gets accelerated by gravity. This energy gets absorbed by the wheel's blades as the fluid de-accelerates.
                                – Muhlemmer
                                Dec 17 at 16:19












                                @Muhlemmer Actually, if you think about it, it doesn't really get accelerated. It does have the force of gravity pulling on it, but if it accelerated on the way down, the bottom of the wheel would have to be spinning faster than the top of the wheel. (And it does benefit a little, but its limited) There's also angular acceleration from bending around the wheel, but that's also pretty slight.
                                – Cort Ammon
                                Dec 17 at 16:26






                                @Muhlemmer Actually, if you think about it, it doesn't really get accelerated. It does have the force of gravity pulling on it, but if it accelerated on the way down, the bottom of the wheel would have to be spinning faster than the top of the wheel. (And it does benefit a little, but its limited) There's also angular acceleration from bending around the wheel, but that's also pretty slight.
                                – Cort Ammon
                                Dec 17 at 16:26














                                +1 for emphasizing the point that while the amount of useful energy a water wheel harvests for a unit quantity of water may be roughly constant, the amount of energy removed per unit quantity is proportional to the pressure difference before and after, with increases in pressure differential leading to increases in waste.
                                – supercat
                                Dec 17 at 18:30




                                +1 for emphasizing the point that while the amount of useful energy a water wheel harvests for a unit quantity of water may be roughly constant, the amount of energy removed per unit quantity is proportional to the pressure difference before and after, with increases in pressure differential leading to increases in waste.
                                – supercat
                                Dec 17 at 18:30











                                2














                                Water and current flow as well as other “mechanical” analogies (pipe resistance vs. electrical resistance, voltage vs. pressure, etc.) are useful for introducing electrical circuit concepts at an elementary level. This is because mechanical concepts are easier to visualize whereas electrical concepts are more abstract. The analogies can only go so far without a deeper understanding.



                                Current does not equal electrical power and neither does water flow equal mechanical power.



                                Electrical current ($frac {Coul}{s}$) times (in phase) voltage ($frac {J}{Coul}$) equals power ($frac {J}{s}$ = watts).



                                Current flow ($frac {m^3}{s}$) times pressure ($frac {N}{m^2}$) equals power ($frac {J}{s}$ = watts).
                                The water pressure for the water wheel can come from water dropping from a height above the wheel (potential energy) or even from a hose directed horizontally, depending on the orientation of the water wheel.



                                To complete the analogy between current and water flow with respect to electrical an mechanical power, you need in addition the analogy between voltage (electrical potential) and pressure (mechanical potential).



                                Hope this helps.






                                share|cite|improve this answer




























                                  2














                                  Water and current flow as well as other “mechanical” analogies (pipe resistance vs. electrical resistance, voltage vs. pressure, etc.) are useful for introducing electrical circuit concepts at an elementary level. This is because mechanical concepts are easier to visualize whereas electrical concepts are more abstract. The analogies can only go so far without a deeper understanding.



                                  Current does not equal electrical power and neither does water flow equal mechanical power.



                                  Electrical current ($frac {Coul}{s}$) times (in phase) voltage ($frac {J}{Coul}$) equals power ($frac {J}{s}$ = watts).



                                  Current flow ($frac {m^3}{s}$) times pressure ($frac {N}{m^2}$) equals power ($frac {J}{s}$ = watts).
                                  The water pressure for the water wheel can come from water dropping from a height above the wheel (potential energy) or even from a hose directed horizontally, depending on the orientation of the water wheel.



                                  To complete the analogy between current and water flow with respect to electrical an mechanical power, you need in addition the analogy between voltage (electrical potential) and pressure (mechanical potential).



                                  Hope this helps.






                                  share|cite|improve this answer


























                                    2












                                    2








                                    2






                                    Water and current flow as well as other “mechanical” analogies (pipe resistance vs. electrical resistance, voltage vs. pressure, etc.) are useful for introducing electrical circuit concepts at an elementary level. This is because mechanical concepts are easier to visualize whereas electrical concepts are more abstract. The analogies can only go so far without a deeper understanding.



                                    Current does not equal electrical power and neither does water flow equal mechanical power.



                                    Electrical current ($frac {Coul}{s}$) times (in phase) voltage ($frac {J}{Coul}$) equals power ($frac {J}{s}$ = watts).



                                    Current flow ($frac {m^3}{s}$) times pressure ($frac {N}{m^2}$) equals power ($frac {J}{s}$ = watts).
                                    The water pressure for the water wheel can come from water dropping from a height above the wheel (potential energy) or even from a hose directed horizontally, depending on the orientation of the water wheel.



                                    To complete the analogy between current and water flow with respect to electrical an mechanical power, you need in addition the analogy between voltage (electrical potential) and pressure (mechanical potential).



                                    Hope this helps.






                                    share|cite|improve this answer














                                    Water and current flow as well as other “mechanical” analogies (pipe resistance vs. electrical resistance, voltage vs. pressure, etc.) are useful for introducing electrical circuit concepts at an elementary level. This is because mechanical concepts are easier to visualize whereas electrical concepts are more abstract. The analogies can only go so far without a deeper understanding.



                                    Current does not equal electrical power and neither does water flow equal mechanical power.



                                    Electrical current ($frac {Coul}{s}$) times (in phase) voltage ($frac {J}{Coul}$) equals power ($frac {J}{s}$ = watts).



                                    Current flow ($frac {m^3}{s}$) times pressure ($frac {N}{m^2}$) equals power ($frac {J}{s}$ = watts).
                                    The water pressure for the water wheel can come from water dropping from a height above the wheel (potential energy) or even from a hose directed horizontally, depending on the orientation of the water wheel.



                                    To complete the analogy between current and water flow with respect to electrical an mechanical power, you need in addition the analogy between voltage (electrical potential) and pressure (mechanical potential).



                                    Hope this helps.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Dec 17 at 14:36

























                                    answered Dec 16 at 22:12









                                    Bob D

                                    2,012211




                                    2,012211























                                        2














                                        The person who said that is wrong. Current doesn't push. Voltage or pressure pushes. Current is how much actually flows.



                                        Useful power is the force of the push (voltage, pressure) multiplied by the current (amps, CFM).



                                        Obviously with no push, you have no power. But with no flow, you also have no power.



                                        In water generation, pressure also equates to height, called "head".



                                        For instance, a PC power supply is a lot like run-of-river hydro -- not much pressure but lots of flow.



                                        A very tall, but low-flow dam like Oroville is a lot like an LED light on a 240V line - lots of pressure, not much flow. Not much flow needed.






                                        share|cite|improve this answer








                                        New contributor




                                        Harper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.























                                          2














                                          The person who said that is wrong. Current doesn't push. Voltage or pressure pushes. Current is how much actually flows.



                                          Useful power is the force of the push (voltage, pressure) multiplied by the current (amps, CFM).



                                          Obviously with no push, you have no power. But with no flow, you also have no power.



                                          In water generation, pressure also equates to height, called "head".



                                          For instance, a PC power supply is a lot like run-of-river hydro -- not much pressure but lots of flow.



                                          A very tall, but low-flow dam like Oroville is a lot like an LED light on a 240V line - lots of pressure, not much flow. Not much flow needed.






                                          share|cite|improve this answer








                                          New contributor




                                          Harper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                          Check out our Code of Conduct.





















                                            2












                                            2








                                            2






                                            The person who said that is wrong. Current doesn't push. Voltage or pressure pushes. Current is how much actually flows.



                                            Useful power is the force of the push (voltage, pressure) multiplied by the current (amps, CFM).



                                            Obviously with no push, you have no power. But with no flow, you also have no power.



                                            In water generation, pressure also equates to height, called "head".



                                            For instance, a PC power supply is a lot like run-of-river hydro -- not much pressure but lots of flow.



                                            A very tall, but low-flow dam like Oroville is a lot like an LED light on a 240V line - lots of pressure, not much flow. Not much flow needed.






                                            share|cite|improve this answer








                                            New contributor




                                            Harper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.









                                            The person who said that is wrong. Current doesn't push. Voltage or pressure pushes. Current is how much actually flows.



                                            Useful power is the force of the push (voltage, pressure) multiplied by the current (amps, CFM).



                                            Obviously with no push, you have no power. But with no flow, you also have no power.



                                            In water generation, pressure also equates to height, called "head".



                                            For instance, a PC power supply is a lot like run-of-river hydro -- not much pressure but lots of flow.



                                            A very tall, but low-flow dam like Oroville is a lot like an LED light on a 240V line - lots of pressure, not much flow. Not much flow needed.







                                            share|cite|improve this answer








                                            New contributor




                                            Harper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.









                                            share|cite|improve this answer



                                            share|cite|improve this answer






                                            New contributor




                                            Harper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.









                                            answered Dec 18 at 1:16









                                            Harper

                                            1211




                                            1211




                                            New contributor




                                            Harper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.





                                            New contributor





                                            Harper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.






                                            Harper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.























                                                1














                                                The analogy with water actually holds really nicely if you consider a water wheel, or other hydroelectric system.



                                                But what you're missing is that the power produced does not only depend on the amount of water going past - it also depends on the speed at which it does so. (this makes sense for the hydro system because kinetic energy depends on velocity and mass)



                                                To make the analogy better, rather than thinking of the speed of the flow, think about how far it has fallen to aquire that speed. At this point you have a volume per second of water - the current - and you have a loss of height, which is literally a potential difference.






                                                share|cite|improve this answer


























                                                  1














                                                  The analogy with water actually holds really nicely if you consider a water wheel, or other hydroelectric system.



                                                  But what you're missing is that the power produced does not only depend on the amount of water going past - it also depends on the speed at which it does so. (this makes sense for the hydro system because kinetic energy depends on velocity and mass)



                                                  To make the analogy better, rather than thinking of the speed of the flow, think about how far it has fallen to aquire that speed. At this point you have a volume per second of water - the current - and you have a loss of height, which is literally a potential difference.






                                                  share|cite|improve this answer
























                                                    1












                                                    1








                                                    1






                                                    The analogy with water actually holds really nicely if you consider a water wheel, or other hydroelectric system.



                                                    But what you're missing is that the power produced does not only depend on the amount of water going past - it also depends on the speed at which it does so. (this makes sense for the hydro system because kinetic energy depends on velocity and mass)



                                                    To make the analogy better, rather than thinking of the speed of the flow, think about how far it has fallen to aquire that speed. At this point you have a volume per second of water - the current - and you have a loss of height, which is literally a potential difference.






                                                    share|cite|improve this answer












                                                    The analogy with water actually holds really nicely if you consider a water wheel, or other hydroelectric system.



                                                    But what you're missing is that the power produced does not only depend on the amount of water going past - it also depends on the speed at which it does so. (this makes sense for the hydro system because kinetic energy depends on velocity and mass)



                                                    To make the analogy better, rather than thinking of the speed of the flow, think about how far it has fallen to aquire that speed. At this point you have a volume per second of water - the current - and you have a loss of height, which is literally a potential difference.







                                                    share|cite|improve this answer












                                                    share|cite|improve this answer



                                                    share|cite|improve this answer










                                                    answered Dec 16 at 22:38









                                                    Flyto

                                                    47845




                                                    47845























                                                        1














                                                        I'd like to put what others said already into equations:



                                                        A river has mass flow rate $dot m$ (kg/s) (the "current"). The water flows with velocity $v$ (m/s). Power is the kinetic energy that is carried per unit time:



                                                        $$
                                                        dot W = frac{dot m v^2}{2}
                                                        $$

                                                        and notice that this has proper units of watt. If we take "voltage" to be $v^2/2$, then we get
                                                        $$
                                                        text{(power)} = text{(current)}timestext{(voltage)}
                                                        $$



                                                        Notice that the mass flow rate is not enough to give you high power. When the river is wide the velocity is low. If you want the water wheel to run fast you should build it at a narrow passage.





                                                        Variation



                                                        The river flows with fixed flow rate $dot m$, the width of the river is constant throughout, but the river is made to go over a waterfall that drops by distance $Delta h$. The velocity at the top of the waterfall is $v_1$ and the bottom $v_2$. By energy balance,
                                                        $$
                                                        frac{v_2^2-v_1^2}{2}=g Delta h
                                                        $$

                                                        The power that is available to us is
                                                        $$
                                                        dot W = dot m frac{v_2^2-v_1^2}{2}
                                                        $$

                                                        We have to subtract $v_1^2$ because the river must continue to flow with its steady velocity after it hits the wheel, or else we will flood. Combining with the energy balance:
                                                        $$
                                                        dot W = dot m g Delta h
                                                        $$

                                                        Here now we have a one-to-one analogy with electric current: $dot m$ is the current and $gDelta h$ is the voltage. Increase the voltage and the same current will produce more power.






                                                        share|cite|improve this answer




























                                                          1














                                                          I'd like to put what others said already into equations:



                                                          A river has mass flow rate $dot m$ (kg/s) (the "current"). The water flows with velocity $v$ (m/s). Power is the kinetic energy that is carried per unit time:



                                                          $$
                                                          dot W = frac{dot m v^2}{2}
                                                          $$

                                                          and notice that this has proper units of watt. If we take "voltage" to be $v^2/2$, then we get
                                                          $$
                                                          text{(power)} = text{(current)}timestext{(voltage)}
                                                          $$



                                                          Notice that the mass flow rate is not enough to give you high power. When the river is wide the velocity is low. If you want the water wheel to run fast you should build it at a narrow passage.





                                                          Variation



                                                          The river flows with fixed flow rate $dot m$, the width of the river is constant throughout, but the river is made to go over a waterfall that drops by distance $Delta h$. The velocity at the top of the waterfall is $v_1$ and the bottom $v_2$. By energy balance,
                                                          $$
                                                          frac{v_2^2-v_1^2}{2}=g Delta h
                                                          $$

                                                          The power that is available to us is
                                                          $$
                                                          dot W = dot m frac{v_2^2-v_1^2}{2}
                                                          $$

                                                          We have to subtract $v_1^2$ because the river must continue to flow with its steady velocity after it hits the wheel, or else we will flood. Combining with the energy balance:
                                                          $$
                                                          dot W = dot m g Delta h
                                                          $$

                                                          Here now we have a one-to-one analogy with electric current: $dot m$ is the current and $gDelta h$ is the voltage. Increase the voltage and the same current will produce more power.






                                                          share|cite|improve this answer


























                                                            1












                                                            1








                                                            1






                                                            I'd like to put what others said already into equations:



                                                            A river has mass flow rate $dot m$ (kg/s) (the "current"). The water flows with velocity $v$ (m/s). Power is the kinetic energy that is carried per unit time:



                                                            $$
                                                            dot W = frac{dot m v^2}{2}
                                                            $$

                                                            and notice that this has proper units of watt. If we take "voltage" to be $v^2/2$, then we get
                                                            $$
                                                            text{(power)} = text{(current)}timestext{(voltage)}
                                                            $$



                                                            Notice that the mass flow rate is not enough to give you high power. When the river is wide the velocity is low. If you want the water wheel to run fast you should build it at a narrow passage.





                                                            Variation



                                                            The river flows with fixed flow rate $dot m$, the width of the river is constant throughout, but the river is made to go over a waterfall that drops by distance $Delta h$. The velocity at the top of the waterfall is $v_1$ and the bottom $v_2$. By energy balance,
                                                            $$
                                                            frac{v_2^2-v_1^2}{2}=g Delta h
                                                            $$

                                                            The power that is available to us is
                                                            $$
                                                            dot W = dot m frac{v_2^2-v_1^2}{2}
                                                            $$

                                                            We have to subtract $v_1^2$ because the river must continue to flow with its steady velocity after it hits the wheel, or else we will flood. Combining with the energy balance:
                                                            $$
                                                            dot W = dot m g Delta h
                                                            $$

                                                            Here now we have a one-to-one analogy with electric current: $dot m$ is the current and $gDelta h$ is the voltage. Increase the voltage and the same current will produce more power.






                                                            share|cite|improve this answer














                                                            I'd like to put what others said already into equations:



                                                            A river has mass flow rate $dot m$ (kg/s) (the "current"). The water flows with velocity $v$ (m/s). Power is the kinetic energy that is carried per unit time:



                                                            $$
                                                            dot W = frac{dot m v^2}{2}
                                                            $$

                                                            and notice that this has proper units of watt. If we take "voltage" to be $v^2/2$, then we get
                                                            $$
                                                            text{(power)} = text{(current)}timestext{(voltage)}
                                                            $$



                                                            Notice that the mass flow rate is not enough to give you high power. When the river is wide the velocity is low. If you want the water wheel to run fast you should build it at a narrow passage.





                                                            Variation



                                                            The river flows with fixed flow rate $dot m$, the width of the river is constant throughout, but the river is made to go over a waterfall that drops by distance $Delta h$. The velocity at the top of the waterfall is $v_1$ and the bottom $v_2$. By energy balance,
                                                            $$
                                                            frac{v_2^2-v_1^2}{2}=g Delta h
                                                            $$

                                                            The power that is available to us is
                                                            $$
                                                            dot W = dot m frac{v_2^2-v_1^2}{2}
                                                            $$

                                                            We have to subtract $v_1^2$ because the river must continue to flow with its steady velocity after it hits the wheel, or else we will flood. Combining with the energy balance:
                                                            $$
                                                            dot W = dot m g Delta h
                                                            $$

                                                            Here now we have a one-to-one analogy with electric current: $dot m$ is the current and $gDelta h$ is the voltage. Increase the voltage and the same current will produce more power.







                                                            share|cite|improve this answer














                                                            share|cite|improve this answer



                                                            share|cite|improve this answer








                                                            edited Dec 19 at 11:52

























                                                            answered Dec 16 at 23:07









                                                            Themis

                                                            2995




                                                            2995























                                                                0














                                                                Power is defined as work done per unit time. So a mass of water moving from one potential to some other lower potetial can do work when it hits the wheel. How much work per unit time? It depends how much mass falls times height times gravitational constant g and all of that divided by time. Water current, on the other hand is just total volume or mass that flows per unit time. Sure, it is connected to the work done or power but is not the same thing. If you define gravitational potential difference as gH and water flow as dm/dt then to have power you have to have curent times this potential difference: gHdm/dt...
                                                                In a conductor, work is done also and energy per unit charge or potential difference is given by U (voltage) and current by dQ/dt so it seems to me that everything is same...






                                                                share|cite|improve this answer


























                                                                  0














                                                                  Power is defined as work done per unit time. So a mass of water moving from one potential to some other lower potetial can do work when it hits the wheel. How much work per unit time? It depends how much mass falls times height times gravitational constant g and all of that divided by time. Water current, on the other hand is just total volume or mass that flows per unit time. Sure, it is connected to the work done or power but is not the same thing. If you define gravitational potential difference as gH and water flow as dm/dt then to have power you have to have curent times this potential difference: gHdm/dt...
                                                                  In a conductor, work is done also and energy per unit charge or potential difference is given by U (voltage) and current by dQ/dt so it seems to me that everything is same...






                                                                  share|cite|improve this answer
























                                                                    0












                                                                    0








                                                                    0






                                                                    Power is defined as work done per unit time. So a mass of water moving from one potential to some other lower potetial can do work when it hits the wheel. How much work per unit time? It depends how much mass falls times height times gravitational constant g and all of that divided by time. Water current, on the other hand is just total volume or mass that flows per unit time. Sure, it is connected to the work done or power but is not the same thing. If you define gravitational potential difference as gH and water flow as dm/dt then to have power you have to have curent times this potential difference: gHdm/dt...
                                                                    In a conductor, work is done also and energy per unit charge or potential difference is given by U (voltage) and current by dQ/dt so it seems to me that everything is same...






                                                                    share|cite|improve this answer












                                                                    Power is defined as work done per unit time. So a mass of water moving from one potential to some other lower potetial can do work when it hits the wheel. How much work per unit time? It depends how much mass falls times height times gravitational constant g and all of that divided by time. Water current, on the other hand is just total volume or mass that flows per unit time. Sure, it is connected to the work done or power but is not the same thing. If you define gravitational potential difference as gH and water flow as dm/dt then to have power you have to have curent times this potential difference: gHdm/dt...
                                                                    In a conductor, work is done also and energy per unit charge or potential difference is given by U (voltage) and current by dQ/dt so it seems to me that everything is same...







                                                                    share|cite|improve this answer












                                                                    share|cite|improve this answer



                                                                    share|cite|improve this answer










                                                                    answered Dec 16 at 19:29









                                                                    Žarko Tomičić

                                                                    853511




                                                                    853511






























                                                                        draft saved

                                                                        draft discarded




















































                                                                        Thanks for contributing an answer to Physics Stack Exchange!


                                                                        • Please be sure to answer the question. Provide details and share your research!

                                                                        But avoid



                                                                        • Asking for help, clarification, or responding to other answers.

                                                                        • Making statements based on opinion; back them up with references or personal experience.


                                                                        Use MathJax to format equations. MathJax reference.


                                                                        To learn more, see our tips on writing great answers.





                                                                        Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                                                        Please pay close attention to the following guidance:


                                                                        • Please be sure to answer the question. Provide details and share your research!

                                                                        But avoid



                                                                        • Asking for help, clarification, or responding to other answers.

                                                                        • Making statements based on opinion; back them up with references or personal experience.


                                                                        To learn more, see our tips on writing great answers.




                                                                        draft saved


                                                                        draft discarded














                                                                        StackExchange.ready(
                                                                        function () {
                                                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f448724%2fthe-water-analogy-seems-to-imply-that-power-current-why-is-this-incorrect%23new-answer', 'question_page');
                                                                        }
                                                                        );

                                                                        Post as a guest















                                                                        Required, but never shown





















































                                                                        Required, but never shown














                                                                        Required, but never shown












                                                                        Required, but never shown







                                                                        Required, but never shown

































                                                                        Required, but never shown














                                                                        Required, but never shown












                                                                        Required, but never shown







                                                                        Required, but never shown







                                                                        Popular posts from this blog

                                                                        數位音樂下載

                                                                        When can things happen in Etherscan, such as the picture below?

                                                                        格利澤436b