# Viscous drag of water

• I

Hi
i want to understand effects of viscious drag of water in fixed cross section open channel flow.
My understanding is water viscosity and channel surface create drag opposing movement of water, changing kinetic energy of water to heat.
Result is warmer water and slower water flow.
If the water flow in channel should maintain constant Q per second, lost kinetic energy has to be replenished by water head , potential energy,gravity, slope of open channel...etc. Is this explanation correct so far?

Yes, But say what you wish to say. Please don't play "twenty questions"

I want to know how it applies to this scenario
u have long channel without slope H1 = H2
according to above, water closer to free fall should be slower and warmer,
slower water would mean lower Q2, but that's not the case in reality since Q1=Q2
what am i missing here ?

You are missing that the water is essentially incompressible so Q2 is not slower. Viscosity causes the pressure to vary along the pipe (not flow rate), with pressure highest at the left end.

vanhees71 and hutchphd
I don't understand what pressure are you talking about, its open channel flow not flow in the pipe

I don't understand what pressure are you talking about, its open channel flow not flow in the pipe

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In that case, the viscous drag balances the along-channel component of the gravitational force on the fluid.

Unless my intuition is mistaken, there will have to be some slope to the open channel so that the gravitational head will match the pressure head that would otherwise exist in an unpitched, closed channel with a similar flow rate.

In the absence of such a slope, the water surface itself will slope downward toward the exit -- a shallower, faster moving stream toward the exit in order to maintain an equilibrium.

Lnewqban and Chestermiller
What you're missing is that your premise is nonphysical. You need a slope to maintain that flow, and as jbriggs said, if your channel isn't sloped, the water surface still will be.

ok ,if the channel is 100km long, H1 is 1 meter H2 is zero, than slope angle along whole channel is really negligible as you can see in calculation
for 1km part of the channel, it is 100x smaller angle
so how can this molecular level slope drive the water forward
Q on 100km long channel is not 100x smaller than on 1 km channel
Q on 100km long channel is not 1000x smaller than on 0.1 km channel

1 meter drop in 100km? Your flow rate will be negligibly small. You'd effectively have a stagnant channel.

jbriggs444
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Lnewqban
Hi
i want to understand effects of viscious drag of water in fixed cross section open channel flow.
My understanding is water viscosity and channel surface create drag opposing movement of water, changing kinetic energy of water to heat.
Result is warmer water and slower water flow.
If the water flow in channel should maintain constant Q per second, lost kinetic energy has to be replenished by water head , potential energy,gravity, slope of open channel...etc. Is this explanation correct so far?
Perhaps the problem with that description is that it happens backwards in real life.
Without a driving force, the molecules of water will not move horizontally.
The natural balance that is reached between that driving force and the resistive force determines the unique terminal velocity of the whole mass-flow of water.
The heating also reaches a natural balance with the cooling effect of surfaces temperature and evaporation.