How to Solve Hydraulic Jump Problems with Conservation of Momentum and Mass?

[member 428835]

Homework Statement

The question is stated here, though I'm happy to repost but they include a picture. I should say this is not homework, I'm doing problems for practice.
http://web.mit.edu/2.25/www/5_pdf/5_04.pdf

Homework Equations

Conservation of momentum/mass

The Attempt at a Solution

Conservation of mass between stations 1 and 2 is $h_1V_1=h_2V_2$. I'm unsure how to approach momentum since I would typically make my CV the fluid between stations 1 and 2, but there is viscous dissipation in the jump, as illustrated. Any ideas?

 

[haruspex]

Homework Statement

The question is stated here, though I'm happy to repost but they include a picture. I should say this is not homework, I'm doing problems for practice.
http://web.mit.edu/2.25/www/5_pdf/5_04.pdf

Homework Equations

Conservation of momentum/mass

The Attempt at a Solution

Conservation of mass between stations 1 and 2 is $h_1V_1=h_2V_2$. I'm unsure how to approach momentum since I would typically make my CV the fluid between stations 1 and 2, but there is viscous dissipation in the jump, as illustrated. Any ideas?

Viscous dissipation involves internal forces, so no net affect on momentum.
You are told to ignore horizontal stress from streambed, so no affect from that either.

 

[member 428835]

Thanks! Then the momentum equation on the CV of fluid from station 1 to station 2 (assuming the hydraulic jump is stationary) is $$\partial_t \iiint \rho \vec{V} \, dV + \iint_{\partial V} \rho \vec{V} (\vec{V_{rel}}\cdot \hat{n})\, dS = \sum F\implies\\
\iint_{\partial V} \rho \vec{V} (\vec{V_{rel}}\cdot \hat{n})\, dS = -\iint_{\partial V} P \hat{n}\, dS\implies\\
\int_0^b \int_0^{h_1} \rho V_1 \hat{i} (V_1 \hat{i} \cdot(-\hat{i}))\, dz\,dy+\int_0^b \int_0^{h_2} \rho V_2 \hat{i} (V_2 \hat{i} \cdot(\hat{i}))\, dz\,dy = -\int_0^b \int_0^{h_1}P_1 (-\hat{i}) \, dz\,dy -\int_0^b \int_0^{h_2}P_2 \hat{i} \, dz\,dy\implies\\
\rho (V_2^2 h_2-V_1^2 h_1) = P_1h_1-P_2h_2$$ Hydrostatic pressure is $P=\rho g z$, so assuming this holds then average pressure along $A_1$ is $$P_1 = \frac{1}{b h_1}\int_0^b \int_0^{h_1} \rho g z \, dz \, dy = \frac{\rho g h_1}{2}$$ where I assume atmospheric pressure is 0. Then the final result for the momentum balance is
$$V_2^2 h_2-V_1^2 h_1 = \frac{g h_1^2}{2}-\frac{g h_2^2}{2}$$ This along with continuity gives us the solution in terms of asked quantities. Is this correct?

Viscous dissipation involves internal forces, so no net affect on momentum.

So viscous dissipation does not change the time derivative of momentum? Could you briefly explain why?

You are told to ignore horizontal stress from streambed, so no affect from that either.

So how would this balance change if we were not to ignore the horizontal streambed stress? I think the additional force would be $$-\iint_{\partial V} \mu \vec{V} \otimes \vec{V} \cdot \hat{n} \, dS$$ Would we have to take this to Navier Stokes?

 

[Nidum]

Solutions for this problem and the previous nozzle problem can be easily found in undergraduate fluid mechanics textbooks and also online .

 

[member 428835]

Solutions for this problem and the previous nozzle problem can be easily found in undergraduate fluid mechanics textbooks and also online .

Could you list the books?

 

[Chestermiller]

Thanks! Then the momentum equation on the CV of fluid from station 1 to station 2 (assuming the hydraulic jump is stationary) is $$\partial_t \iiint \rho \vec{V} \, dV + \iint_{\partial V} \rho \vec{V} (\vec{V_{rel}}\cdot \hat{n})\, dS = \sum F\implies\\
\iint_{\partial V} \rho \vec{V} (\vec{V_{rel}}\cdot \hat{n})\, dS = -\iint_{\partial V} P \hat{n}\, dS\implies\\
\int_0^b \int_0^{h_1} \rho V_1 \hat{i} (V_1 \hat{i} \cdot(-\hat{i}))\, dz\,dy+\int_0^b \int_0^{h_2} \rho V_2 \hat{i} (V_2 \hat{i} \cdot(\hat{i}))\, dz\,dy = -\int_0^b \int_0^{h_1}P_1 (-\hat{i}) \, dz\,dy -\int_0^b \int_0^{h_2}P_2 \hat{i} \, dz\,dy\implies\\
\rho (V_2^2 h_2-V_1^2 h_1) = P_1h_1-P_2h_2$$ Hydrostatic pressure is $P=\rho g z$, so assuming this holds then average pressure along $A_1$ is $$P_1 = \frac{1}{b h_1}\int_0^b \int_0^{h_1} \rho g z \, dz \, dy = \frac{\rho g h_1}{2}$$ where I assume atmospheric pressure is 0. Then the final result for the momentum balance is
$$V_2^2 h_2-V_1^2 h_1 = \frac{g h_1^2}{2}-\frac{g h_2^2}{2}$$ This along with continuity gives us the solution in terms of asked quantities. Is this correct?

This is what I got.

So viscous dissipation does not change the time derivative of momentum? Could you briefly explain why?

The system is at steady state.

So how would this balance change if we were not to ignore the horizontal streambed stress? I think the additional force would be $$-\iint_{\partial V} \mu \vec{V} \otimes \vec{V} \cdot \hat{n} \, dS$$ Would we have to take this to Navier Stokes?

If you wanted to include the horizontal streambed stress, you would have to get the viscous shear stress at the wall. But to do that, you would have to solve the turbulent NS equations (say using a CFD code) inside the control volume.

 

[Chestermiller]

Thanks! Then the momentum equation on the CV of fluid from station 1 to station 2 (assuming the hydraulic jump is stationary) is $$\partial_t \iiint \rho \vec{V} \, dV + \iint_{\partial V} \rho \vec{V} (\vec{V_{rel}}\cdot \hat{n})\, dS = \sum F\implies\\
\iint_{\partial V} \rho \vec{V} (\vec{V_{rel}}\cdot \hat{n})\, dS = -\iint_{\partial V} P \hat{n}\, dS\implies\\
\int_0^b \int_0^{h_1} \rho V_1 \hat{i} (V_1 \hat{i} \cdot(-\hat{i}))\, dz\,dy+\int_0^b \int_0^{h_2} \rho V_2 \hat{i} (V_2 \hat{i} \cdot(\hat{i}))\, dz\,dy = -\int_0^b \int_0^{h_1}P_1 (-\hat{i}) \, dz\,dy -\int_0^b \int_0^{h_2}P_2 \hat{i} \, dz\,dy\implies\\
\rho (V_2^2 h_2-V_1^2 h_1) = P_1h_1-P_2h_2$$ Hydrostatic pressure is $P=\rho g z$, so assuming this holds then average pressure along $A_1$ is $$P_1 = \frac{1}{b h_1}\int_0^b \int_0^{h_1} \rho g z \, dz \, dy = \frac{\rho g h_1}{2}$$ where I assume atmospheric pressure is 0. Then the final result for the momentum balance is
$$V_2^2 h_2-V_1^2 h_1 = \frac{g h_1^2}{2}-\frac{g h_2^2}{2}$$ This along with continuity gives us the solution in terms of asked quantities. Is this correct?

This is what I got.

So viscous dissipation does not change the time derivative of momentum? Could you briefly explain why?

The system is at steady state.

So how would this balance change if we were not to ignore the horizontal streambed stress? I think the additional force would be $$-\iint_{\partial V} \mu \vec{V} \otimes \vec{V} \cdot \hat{n} \, dS$$ Would we have to take this to Navier Stokes?

If you wanted to include the horizontal streambed stress, you would have to get the viscous shear stress at the wall. But to do that, you would have to solve the turbulent NS equations (say using a CFD code) inside the control volume.

 

[Nidum]

Could you list the books?

The one I actually have is Massey - Mechanics of Fluids 2nd Edition . There is good coverage in it of hydraulic jumps and related things like broad weirs and under water obstructions

 

[Nidum]

http://www.slideshare.net/NiteshSingh36/massey-mechanicsoffluids-1

 

[member 428835]

Thank you all very much!