# Homework Help: Bernoulli's Equation and A Water Tank w/ Small Outlet

1. Aug 3, 2008

### yellowgators

1. The problem statement, all variables and given/known data
A large storage tank (diameter 10m) is filled 10m deep with water (density= 1000 kg/m^3, viscosity= 1.0x10^-3 Pa*s). The outlet at the bottom consists of 2 pipes, as shown. The large pipe has a diameter d2= 0.1m, and the smaller have a diameter d3= 0.05m and length of 20m. A pressure and flow gauge at point A indicates an absolute pressure of 1.95x10^5 Pa, and a velocity of 6 m/s for the water inside the large pipe. You may assume points A,B, and C are all at the same height, and that the height is 0.5m above the bottom of the tank.

Link below takes you to the diagram....
http://farm4.static.flickr.com/3073/...339014.jpg?v=0 [Broken]

a) Determine the atmospheric pressure. (Warning: it will not turn out to be exactly 1 atm.)

b) If the outlet of the small pipe is atmospheric pressure, how many cubic meters of water leave the tank each second?

c) How long will it take to empty the tank? (Assume the outward flw is a constant for this calculation!)

d) Find the water velocity and pressure at point B, inside the smaller pipe.

e) At point C, the storage tank has a square hole, 5cm x 5cm. A little dutch boy is holding a patch in place over the hole. How much force does he have to exert (horizontally) to keep the patch in place? Explain the reasoning.

2. Relevant equations
P_absolute=P_gauge + P_atm
Bernoulli's equation (rho)gy_1+1/2(rho)v_1^2=(rho)gy_2+1/2(rho)v_2^2

3. The attempt at a solution
For part A:
P_absolute=P_gauge + P_atm
P_atm=P_absolute-(rho)gh=1.95x10^5Pa-1000kg/m^3(9.8N/kg)(9.5m)=101900 Pa

For part B:
P_atm=P_outlet
Bernoulli's equation (rho)gy_1+1/2(rho)v_1^2=(rho)gy_2+1/2(rho)v_2^2
The speed of the emerging water is v_2. v_1 is the speed of the water at the surface, which is moving slowly because the tank is draining. The continuity equation requires that
v_1A_1=v_2A_2
Since the cross-sectional area of the outlet A_2 is much smaller than the area of the top of the tank A_1, the speed of the water at the surface v_1 is negligibly small compared with v_2. Setting v_1=0 and canceling out (rho), Bernoulli's equation becomes:
gy_1=gy_2+1/2v_2^2 or g(y_1-y_2)=1/2(v_2^2)
Therefore v_2=sqrt[2g(y_1-y_2)]=sqrt(2*9.8N/kg*9.5m)=13.6m/s
So the velocity of water coming out of the outlet pipe 13.6m/s.

How can I use this value to solve for cubic meters leaving the pipe?
Should I have taken viscosity of water into account anywhere here?
Any help would be greatly appreciated.

Last edited by a moderator: May 3, 2017
2. Aug 3, 2008

### yellowgators

Looking over Part B again I realize that Poiseuille's law will give me the units I want and includes the viscosity of the water.
(delta-V)/(delta-t)=[(pi)/8]*[((delta-P)/L)/(eta)]*r^4
All values are given except the change in pressure, delta-P.
The pressure at one end of the pipe is 1.95x10^5 Pa and at the other end is atmospheric pressure, 101900 Pa; delta-P is 93100 Pa.
So,
(delta-V)/(delta-t)=[(pi)/8]*[(93100Pa/2m)/1.0x10^-3Pa s]*(0.05m)^4=114 m^3/s

Now I'm on the right track... right?

3. Aug 3, 2008

### yellowgators

Part C:
Volume-tank=h(pi)r^2=10m(pi)(5m)^2=785 m^3
Volume-tank/(delta-V/delta-t)=t t= 6.89s

Part D:
To find v_2: continuity equation v_1(A_1)=v_2(A_2)
Therefore v_2=(v_1*A_1)/A_2= [6.0m/s(pi)(0.05m)^2]/[pi*(0.025m)^2]=24 m/s
Bernoulli's equation adjusted to a constant height:
P_1+1/2(rho)v_1^2=P_2+1/2(rho)v_2^2
P_2=P_1+1/2(rho)v_1^2-1/2(rho)v_2^2=1.95x10^5 Pa + 1/2(1000kg/m^3)(6.0m/s)^2 - 1/2(1000kg/m^3)(24m/s)^2=-75000Pa

I didn't think negative Pascals was possible.
Where am I going wrong?