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Hydro Equations

  1. Jun 8, 2009 #1
    I'm investigating the power available in a free flowing river. I'm attempting to capture this power with a cross-axis (wind turbine style) turbine.

    The equation I have for available power is: P = 0.5*m*A*v^3

    with m=mass, A=area swept by blades, v=horizontal (into the turbine) velocity of the river

    This is the total power available, correct?

    Is the area A the total area swept by the blades (ie pi*r^2) or would it be the area of just the blades themselves?

    Likewise, does anyone have equations relating to the operation and/or efficiency of a turbine in water? I'm looking to determine maximum efficiency, and torque and angular velocity as related to water velocity. Any help would be greatly appreciated.

    I'm an electrical engineer by education, so most of this stuff is new to me...
  2. jcsd
  3. Jun 8, 2009 #2


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    Hello HydroGuy,

    Right away you can tell your equation is not quite right for two reasons:

    1. It doesn't have dimensions of power. Compare:

    A unit of power

    [tex] 1 W = 1 \frac{J}{s} = 1 \frac{N \cdot m}{s} = 1 \frac{kg\cdot m \cdot m}{s \cdot s^2} = 1 \frac{kg\cdot m^2}{s^3} [/tex]

    A unit of whatever quantity is being calculated by your equation:

    [tex] \textrm{mass} \cdot \textrm{area} \cdot \textrm{velocity}^3 [/tex]

    [tex] = \frac {kg \cdot m^2 \cdot m^3}{s^3} = \frac {kg \cdot m^5}{s^3} [/tex]

    2. The presence of "mass" doesn't makes sense in this context.

    "Mass" of what? The entire river? Or the amount of water impinging on the blade per unit time? It is my sneaking suspicion that the m should actually be a rho [itex] \rho [/itex] meaning density. This makes more sense because density is a intrinsic property of the fluid that doesn't depend on the amount you have. The fact that we are off by length to the third power in our dimensional analysis above also greatly strengthens this supposition (since density = mass/volume).

    To give you a sense of why this might be true, I'll give you a sketchy outline/hand-waving derivation that appeals to our mutual knowledge of first year physics, nothing more:

    1. Definition of Power

    The instantaneous work done on the blade is just the total force on it times the instantaneous displacement, s:

    [tex]dW = Fds[/tex]

    Of course, power is just the RATE at which work is done on the blade with time:

    [tex]P = \frac{dW}{dt} = F\frac{ds}{dt} = Fv[/tex]

    where v is the flow rate of the water. This is just the good old "power = force * velocity" relation.

    So what is the force, F, on the blade? Before I answer that question, I want to derive a relation for the mass flow rate, because it will come in handy later. The mass flow rate is just the mass of water arriving at the blade per unit area, per unit time.

    2. Mass Flow Rate

    To derive this, imagine a volume element, i.e. an infinitesimal cylinder of water with cross sectional area dA and length ds. Then obviously this volume element has volume:

    [tex] dV = dAds [/tex]

    Now, if we choose the length so that ds = vdt, then in time interval dt, all of the water in the cylinder will flow out of it. So the rate of flow of water through area dA will just be:

    [tex] dV = dAds = vdAdt [/tex]

    [tex] \frac{dV}{dt} = vdA [/tex]

    This is the volume flow rate (cubic metres per second). To get the mass flow rate, (kilograms per second), we must figure out how much mass is in volume dV. Obviously this is just the density of water multiplied by that volume:

    [tex] dm = \rho dV [/tex]

    And so the mass flowing per unit area per unit time is just given by:

    [tex] \frac{dm}{dtdA} = \rho v [/tex]

    Remember this result for later.

    3. So what is the force???

    Recall that force is the rate of change of momentum (Newton's 2nd).

    [tex] F = \frac{dp}{dt} [/tex]

    Now, let's talk about the momentum "arriving" per unit AREA per unit time. This would be the "flow of momentum" or momentum FLUX. Obviously it would have units of pressure, because it would just be a force per unit area. Since we've already used p, P, AND rho, I'm going to use little f for force per unit area (i.e the fluid pressure):

    [tex] \frac{dp}{dtdA} = f [/tex]

    But what is differential amount of momentum arriving, dp? This amount of momentum, given that the velocity is constant, is just the differential amount of mass, dm, multipied by v:

    [tex] v\frac{dm}{dtdA} = f [/tex]

    But this is just the velocity times the mass flow rate that we derived earlier! Now we make an assumption that the fluid pressure, f, is the the same across the entire blade. N.B. This assumption is WRONG! So we have that the total force on the blade is just equal to f*A, where A is the area of the blade (force = pressure *area):

    [tex] v\frac{dm}{dtdA}A = fA = F [/tex]

    Now substitute the result for the mass flow rate from part 2:

    [tex] v \rho v A = fA = F [/tex]

    Substitute this expression for the force into the expression for the power from part 1:

    [tex]P = Fv = v \rho v A v = \rho A v^3 [/tex]

    So that gives you some idea of where this equation might have come from, assuming the mass was supposed to be, in fact, a density. Now, you might be asking, what about the factor of 0.5? Well, we made an erroneous assumption that certain quantities that are defined at a single point in space were in fact constant everywhere. This assumption was wrong. The fluid pressure, for instance, could have varied in space. That would have necessitated some sort of integration (which is probably how the random factors of 1/2 turn up).

    I KNOW for a fact that in a static situation (water is not flowing), the pressure would vary with depth, h, by the relation:

    [tex] f = f_{\textrm{initial}} + \rho g h [/tex]

    In a dynamical situation, I'm not sure how that changes. It's been a long time since I took fluid mechanics, and I was never very fond of it. I would encourage you to Google fluid statics and fluid dynamics for more details. I hope this helps!
    Last edited: Jun 8, 2009
  4. Jun 8, 2009 #3
    Wow... absolutely excellent reply - thank you. I did in fact mean density instead of mass, which translates to Kg*m2/s3

    Correct me if I am wrong, but that describes the power available in the river, at a certain cross sectional area.

    What I am more interested in, though, is the power transferred to a turbine. Obviously the turbine won't be 100% efficient, so that will invoke some constant. But could I logically say that the power AVAILABLE to the turbine would be the total area swept by the turbine blades (picture wind turbine blades rotating) or would it only be the physical area of the blades looking into the rotor?

    Thanks again for the help.
  5. Jun 8, 2009 #4


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    What is obvious to me is that the area in my equation is the portion of the area of a single blade that is in contact with the water (i.e the part that is being pushed). What I am not sure about is how a hydroelectric turbine is set up, exactly. I was thinking of a "water wheel" type scenario in which one blade enters the water, is pushed along by it, and then leaves just as another is entering, so that our expression for the instantaneous power might actually hold when averaged over longer timescales. (Even though A is a function of t in this scenario) However, any effects due to drag etc would depend on the specific set up. So in short, I'm not really qualified to answer. I imagine there is always some drag, because a blade has to displace water behind it. So maybe the NET force on the blade is less than what you would calculate just considering the momentum flux on the incident side. Now I'm speculating.

    I guess farther down the line there are also efficiency losses in the generator being driven by the turbine due to induction and resistive heating. Maybe you already have a handle on that, since it is electrodynamic, and not hydrodynamic ;-)
  6. Jun 8, 2009 #5
    You need to include the efficiency of the rotor/blade design. There is a theoretical limit of about 59%, and then efficiency of real devices improves as tip to stream speed ratio increases.

    See the attached plot for alpha. This alpha factor will scale your power equation.

    Attached Files:

    • Eff.jpg
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  7. Jun 8, 2009 #6
    I was under the impression that the 59% limit was in specific relation to wind turbines. Are you certain that it applies to water turbines as well? I know that some Pelton wheel turbines can achieve near 100% efficiency, although that is through a conventional dam system, and not an "in-river" turbine like I am talking about.

    I'll try and clarify what I am talking about... it's not a water-wheel style turbine. It is essentially a cross-axis wind-turbine except anchored in the middle of a river to harness river currents instead of wind currents. I'm trying to find equations directly relating to tip speed/water speed ratio, water velocity to turbine angular velocity, etc...

    Thanks again for the help, guys...
  8. Jun 8, 2009 #7
    This is not my field of expertise, so I can't be certain. However, I was told that this relation applies to any fluid by an mechanical engineering professor who understands fluid dynamics and Navier Stokes etc. This consideration of the blade speed effect makes intuitive sense to me and gets back to your original question of whether the area in the formula is the area of the blades or the complete area swept out. The relation uses the complete area swept out, but it makes sense that a slow moving blade would not be able to make full use of the entire area.

    If the relation is not valid, or if there are better relations that apply to the water, I would be very interested in learning about it.
    Last edited: Jun 8, 2009
  9. Jun 9, 2009 #8
    I don't think it could be the area swept out, because picture a turbine with two thin blades, and then a turbine with five meaty blades - intuitively it just makes sense that the five bladed turbine would produce more power although they both "sweep" the same area.

    Unrelated, but does anyone know a good manufacturer/supplier of smaller, variable speed generators (preferably permanent-magnet) for use in wind turbines? I need to begin looking into purchasing a few. Any help is appreciated.
  10. Jun 9, 2009 #9


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    Try posting your lastest query in the electrical engineering subforum as well. Although I get the impression it is not read as much, I do think it is probably frequented by more people who actually work in industry.
  11. Jun 9, 2009 #10
    The power formula is a simple relation derived from the kinetic energy per unit volume (i.e. 0.5*p*v^2) of the flowing fluid stream. The total swept area of the blades helps quantify the available power (A*v is volume of fluid moved through the turbine per unit time) and has nothing to do with the turbine design. The formula you originally quoted (after you correct density in place of mass) only estimates the power in the fluid stream, since energy per unit volume times volume per unit time gives you power. The alpha factor I mentioned is what captures the effect of the turbine design.

    Also, your statement that the five meaty blades is necessarily better than the two thin blades is not correct. Consider the plot that I showed. It can be seen that the high speed propeller with it's thin blades is much better than the Savonius rotor with it's fat blades. Also, the American multiblade and the Dutch 4-arm are not as good at the high speed propeller. The Darius rotor can also be designed with thin wing-like blades, and is very efficient.
    Last edited: Jun 9, 2009
  12. Jun 9, 2009 #11

    Andy Resnick

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    If you could provide a reference for the formula, that would help. As it stands, I found these online sites:

    http://www.awea.org/faq/windpower.html [Broken]
    http://www.windpower.org/EN/tour/wres/pow/index.htm [Broken]

    The second site gives me another clue that your formula is wrong- the first clue is that you have 'mass', when it should be 'mass density' since we are discussing fluids.

    The second site starts with kinetic energy = 1/2 m v^2, or when written for continua, E = 1/2 rho * Volume * v^2. The volume can be written as A*length, where A is the cross-sectional area of the turbine and the length is given by v*t.

    Substituting in everything, and converting energy to power (power = energy * time) gives you the formula P = 0.5 x rho x A x v^3, which is close to what you have.
    Last edited by a moderator: May 4, 2017
  13. Jun 9, 2009 #12
    Thanks for the help guys. I'm still trying to figure out if Betz' limit for efficiency of wind turbines applies to hydro turbines as well.

    Here's my current dilemma: I'm attempting to turn the rotor of a generator that has a moment of inertia J=.066 Kg*m^2. How much torque, or power, will I need the turbine to produce to spin this effectively?
  14. Jun 9, 2009 #13
    This is a tricky thing. First of all, the inertia is not the main concern. The mechanical friction and electrical loading are the real losses and will determine the actual torque that the turbine must provide. The inertia is only a concern while you are trying to accellerate up to speed. At that point, the energy used is stored as kinetic energy and there will be no additional power load to maintain that rotational kinetic energy.

    One issue is deciding what rotational speed (angular velocity) you want to operate at. This will depend on the stream speed and the turbine design. As the plots showed, every turbine design will have an optimum TSR which you want to be near as much as possible. So based on the expected water current velocity and the turbine design, you can establish the approximate rotational speed of the generator (after any gearing). Now, since power is torque times angular velocity, you can calculate the generator torque you need to acheive a desired electrical power (don't forget friction in the calculation). You also need to include generator efficiency in the calculation as well. Once you know generator torque, you can work back through any gearing to know the turbine torque. Then you can make sure the blades are big enough and strong enough to do this.

    Many of the steps here are actually tricky to calculate. Friction is never easy to quantify and strength of materials is not a trivial subject.
    Last edited: Jun 9, 2009
  15. Jun 9, 2009 #14
    Found an excellent article that cleared a few things up here: http://www.carbontrust.co.uk/technology/technologyaccelerator/ME_guide3.htm

    It appears that Betz law DOES apply in the case of hydrokinetic turbines as well - in the end, this does make sense as if you took ALL the energy out of the current, the exiting velocity would be 0 m/s and the blades would not turn.

    It also describes how more blades equals more torque, especially at lower speeds, but lower RPM's. The coefficient of performance, Cp, is what I want to maximize for my turbine. This depends heavily on the tip speed ratio, or TSR. This is the ratio of the linear velocity of the tip of the blade, Vt, to the velocity of the river, Vfs.

    TSR = Vt / Vfs

    And Vt = w * r (w=angular velocity, r=radius)

    A question: can the angular velocity of the turbine tip ever exceed the velocity of the river? I think I heard somewhere that it can't...

    To maximize Cp at varying velocities, they recommend varying the blade pitch, but this would add considerable complexity and consume power, so I'm not sure its very realistic for small or medium size hydrokinetic units.

    Unfortunately, it doesn't give me enough data to construct my own blades at this point, as I don't have any clue how to design them to work at a certain RPM, etc.

    I'm thinking of using wind-turbine blades underwater initially (just for some basic testing) or using a large, high pitch boat prop.

    Any ideas/comments?
  16. Jun 9, 2009 #15
    The velocity of the turbine tip can most certainly exceed the river (stream) velocity. Just look at the plot I posted. TSR is greater than unity for the peak of several designs. In fact, it's hard to have an efficient system unless this constraint is met.
  17. Jun 9, 2009 #16
    Yes, but isn't that the linear velocity, equal to the angular velocity times the radius? Or am I mistaken? I was asking regarding the angular velocity.
  18. Jun 9, 2009 #17
    Yes, I was talking about linear velocity. I don't understand your question if you are asking how angular velocity of the rotor compares with linear velocity of the fluid stream. Each has different units, so you can't really say one is greater or less than the other.
  19. Jun 10, 2009 #18
    Good point, stupid question on my part.

    I'm now trying to find some turbine blades for use in initial testing of a very simplistic setup. I posted a thread on the Mechanical Engineering forum asking this, but I'll ask here as well: does anyone have any advice on where to look for turbine blades that could operate underwater effectively? I've thought of using simply a lightweight boat prop, but I doubt it would be able to produce much power at river speeds between 2 - 6 m/s, and might not even be able to turn the generator. I've also thought of using wind generator blades, but those might break or warp under stress from the water.

    Anyone have any advice on what to do/who to contact to attempt to get a hold of some turbine blades for underwater use?
  20. Jun 18, 2009 #19
    Hi , P= 0.5 x rho x A x v^3

    Angular velocity = 2 x pi x N (rad/sec)

    rotor tip speed = pi x D x N (m/s) , D = rotor diameter
  21. Jun 25, 2009 #20
    Last edited by a moderator: Apr 24, 2017
  22. Jun 26, 2009 #21
    This is a very good paper and is useful to me as well as the OP. However, the constant is not 0.5 for wind turbines. It is about 0.3 at the Betz limit and less than that for real turbines. I've already posted the plot for various rotor designs.

    The factor of 0.5 comes from the kinetic energy formula 0.5*m*v^2, but there is an additional scaling factor alpha which depends on the turbine. Typically 0.6 is the quoted value of maximum obtainable alpha (called the Betz limit). Real wind turbines do not reach this limit (see my previous plot).

    This paper comfirms that the basic kinetic energy formula is basically valid for both wind and water. The interesting thing is that they are talking about the need to include potential energy for hydro as a refinement to the equations because they claim that they see more energy than the theoretical limit found by considering kinetic energy only.

    I can't comment on the validity of the claims of this paper, but it is interesting and I will be looking into this further.

    Anyway, thanks for posting the paper !
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