Can't seen to get the right solution-A sluice gate dams water up

1. Apr 8, 2006

JSBeckton

Can't seen to get the right solution-

A sluice gate dams water up 5m. A 1-cm dia hole inthe bottom of the gate allows liquid water at 20C to come out. Neglest any changen in internal energy. Find exit velocity and mass flow rate.

governing eqn- hi=he+1/2v^2

I found
P1=(1000)(9.81)(5)=49050Pa
=49050+101.325
=150.375kPa

P2=101.325

he=175.5 (at 20C, 101.375kPa)
hi=217.8 (at 20C, 150.375kPa)

when I sub and solve for V i get 9.18 m/s
It should be 9.9 m/s

Where am I going wrong?
Thanks

Last edited: Apr 8, 2006
2. Apr 8, 2006

Cyrus

This is the second time you posted in the wrong place. Please put your questions in the Homework help section from now on.

The exit velocity is a function of the height,

$$V= \sqrt {2gh}$$

the mass flow rate is:

$$\dot{m}=\rho \sqrt {2gh}A_{jet}$$

You are also using an incompressible liquid, so the specific density is constant.

Your question is poorly written, find the exit velocity at what height?

Last edited: Apr 8, 2006
3. Apr 8, 2006

Staff: Mentor

As Cyrus indicated the OP belongs in the Introductory Physics or Engineering Homework section.

Using the formula for V posted by Cyrus will give 9.9 m/s.

What are the he and hi values given.

The formula used, hi=he+1/2v^2, would indicate h is specific enthalpy.

Last edited: Apr 8, 2006
4. Apr 8, 2006

JSBeckton

Thanks, I mistakenly thought that this was a ME course specific help board, i will not post HW questions here again. As far as the porly worded question, you will have to talk to Mr. Sonntag about that because other than the fat fingered mispelling of a couple of words the question was worded verbatium. I believe the height is specified in the problem, at 5m below the surface.

5. Apr 8, 2006

JSBeckton

No values are given, those are the values I found in the tables at the paramaters in ().

Now that I see what Cyrus was saying, it seems like an introductory physics problem, it there a way to do it using the specific enthalpy and the first law of thermodynamics?

Last edited: Apr 8, 2006
6. Apr 9, 2006

Cyrus

It is not possible, JS. Think about it. You will have to say, in rate form,

$$\dot {E_{in}} - \dot { E_{out}} = \frac {dm}{dt}|_{cv}$$

which will give you:

$$m(z)gz- m(z)_{out} h_{out} - m(z)_{out} \frac{V^2}{2} = m_2u_2 - m_1u_1$$

So, what's going on here? If this is a big dam, then the change in mass at state 1 and state 2 of the control volume is esentially the same, thus the difference is equal to zero (Internal energy is a function of temperature, mainly) , and:

$$\dot{m}(z)gz - \dot{m}(z)_{out} h_{out} - \dot{m}(z)_{out} \frac{V^2}{2} = 0$$

And now we have a problem, becuase will give you a complex root for the velocity. Therefore, we conclude that it is not possible to forumlate a solution using this method, and we fall back on the first method.

Edit:

Here is a better reason to see why this is not possible, you will be in violation of the conservation of mass:

$$\dot{m_{in}} - \dot{m_{out}} = \frac {dm}{dt} |_{cv}$$

Therefore, this method has invalid assumptions and cannot be solved this way.

$$- \dot{m_{out}} = 0$$

Is not true.

Last edited: Apr 9, 2006
7. Apr 9, 2006

JSBeckton

I see, thanks.