# Pressure as a function of time.

1. Jun 21, 2010

### wwshr87

1. The problem statement, all variables and given/known data

Find pressure as a function of time. At time t = 0, the water level is 2 feet.
Cross-sectional area is 8 ft2 for the top and 1 ft2 for the bottom.
Please see the attachment for further information.

2. Relevant equations

p=(ro)gh

3. The attempt at a solution

Since I want to find the pressure as a function of time; I will need to find h as a function of time, then multiply by ro and g. How can I do this?

Thanks

2. Jun 21, 2010

There is no attachment. I assume water is flowing out somewhere?

3. Jun 21, 2010

### wwshr87

Water in flowing in at a rate of 3-t, and flowing out at a rate of t^2. Initial water level is 2 ft.

I'm sorry I am new to this. I tried attaching the file again. Thanks.

#### Attached Files:

• ###### waterpressure.JPG
File size:
12.7 KB
Views:
69
4. Jun 22, 2010

### hikaru1221

Which pressure do you want to find? Pressure on the pipe, pressure of the water flow at the lower pipe, etc?

5. Jun 22, 2010

### wwshr87

I want to find the pressure at the bottom of the tank.
Thanks

6. Jun 22, 2010

### hikaru1221

I'm not sure what you meant by "pressure at the bottom of the tank". Forget it for a while. Back to your main problem, finding h. From the two water flowing rates, you can calculate the net rate at which water gets in/out of the top pipe, can't you? And from the relation: $$\frac{dM_{net}}{dt}=\rho S_{top}\frac{dh}{dt}$$, it's easy to find dh/dt and h. What is the dimension of $$\dot {M}$$ by the way?

The main problem I want to talk about is the pressure. From your equation $$p=\rho gh$$, I can guess that you want to find the pressure exerting on the bottom, correct? It is not easy at all. If you look at it more carefully, the pressure on the bottom near the edge is different from the pressure at the places farther. The flow's speed is not homogeneous in the lower pipe. Moreover the pressure due to the water flowing in must be taken into account.

Last edited: Jun 22, 2010
7. Jun 22, 2010

### wwshr87

I have attached the solution I have for this problem. The part I do not understand is how they calculate dv/dh.

#### Attached Files:

• ###### pressuresolution.JPG
File size:
19 KB
Views:
103
8. Jun 22, 2010

### hikaru1221

Sorry, I don't get it either.