# Homework Help: Flow Rate Problem -

1. Apr 7, 2010

### mparsons06

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

SAE no. 10 oil has a viscosity of 0.2 Pa· s. How long (in sec) would it take to pour 6 liters of oil through a funnel with a neck 12 cm long and 2.6 cm in diameter. Assume that it is poured in such a way that the oil level is kept just above the top of the tube.

Hint: The specific gravity (= ratio of its density to that of water) of the oil is 0.70.

2. Relevant equations

The pressure difference is given by
Dp = rgL

1000 liter = 1 m^3 = 10^6 cm^3

The flow rate is given by Poiseuille's law:

Q = p * r^4 * Dp / 8*h*L

3. The attempt at a solution

You can calculate the pressure difference:
Dp = rgL
Dp = (700 kg/m^3)(9.81 m/s^2)(0.0012 m)
Dp = 82.4 Pa

And the time it takes for the oil to flow through is the volume of the oil divided by the flow rate:

t = V / Q = 8 * V * h * L / (p * r^4 * Dp)
t = 8 (0.006 m^3)*(0.2)*(0.012 m) / p * (0.5)*(0.013 m)^4 * (82.4 Pa)
t = 8.97 X 10^-9

But I know that can't be right. Can someone please help me with where I went wrong? And try to explain it to me?

Last edited: Apr 7, 2010
2. Apr 8, 2010

### Redbelly98

Staff Emeritus

I found two errors.
12 cm is not 0.0012 m.

I agree with the next-to-last line here, except for the pressure which you must recalculate based on my earlier comment. Even so, something went really wrong in calculating that final value for t here. Everything to the right of the "/" here is in the denominator, i.e. you must divide by all of those values. Technically, those terms should be enclosed in a pair of parantheses, so that you have
t = 8 (0.006 m^3)*(0.2)*(0.012 m) / (p * (0.5)*(0.013 m)^4 * (82.4 Pa))
It's encouraging that you thought about your answer and realized something is wrong

3. Apr 11, 2010

### huybinhs

"You can calculate the pressure difference:
Dp = rgL
Dp = (700 kg/m^3)(9.81 m/s^2)(0.0012 m)
Dp = 82.4 Pa"

=> r = 700 kg/m^3 ???? I'm confused ???

4. Apr 11, 2010

### Redbelly98

Staff Emeritus

The density of the oil is 0.70 times that of water, which is 1000 kg/m^3.

5. Apr 11, 2010

### mparsons06

I don't know what "p" is. Help?

6. Apr 12, 2010

### Redbelly98

Staff Emeritus

p is π here, 3.14159...

7. Apr 12, 2010

### mparsons06

So I calculated the equation using pi:

t = V / Q = 8 * V * h * L / (p * r^4 * Dp)
t = 8 (0.006 m^3)*(0.2)*(0.012 m) / [ pi * (0.5)*(0.013 m)^4 * (82.4 Pa) ]
t = 1.9492 sec

But it is still incorrect... Any ideas?

8. Apr 12, 2010

### Redbelly98

Staff Emeritus

Dp is not 82.4 Pa. See post #2.

9. Apr 12, 2010

### mparsons06

You can calculate the pressure difference:
Dp = rgL
Dp = (700 kg/m^3)(9.81 m/s^2)(0.12 m)
Dp = 824 Pa

And the time it takes for the oil to flow through is the volume of the oil divided by the flow rate:

t = V / Q = 8 * V * h * L / (pi * r^4 * Dp)
t = 8 (0.006 m^3)*(0.2)*(0.012 m) / (pi * (0.5)*(0.013 m)^4 * (824 Pa))
t = 0.1949 sec

It is still incorrect.

10. Apr 12, 2010

### Redbelly98

Staff Emeritus

Hmm, now that I look more carefully I am seeing several problems with the calculation.

1. The numbers you have do not give 0.1949 s
2. 12 cm is not equal to 0.012 m.
3. Where does the (0.5) come from?

11. Apr 12, 2010

### mparsons06

So if I recalculate, fixing my errors and taking out 0.5:

t = V / Q = 8 * V * h * L / (pi * r^4 * Dp)
t = 8 (0.006 m^3)*(0.2)*(0.12 m) / (pi * (0.026 m)^4 * (824 Pa))
t = 0.974 sec

Still wrong.

Last edited: Apr 12, 2010
12. Apr 13, 2010

### Redbelly98

Staff Emeritus

1 more try: 0.026 m is the diameter; the radius is 0.013 m. Use the radius here.

Last edited: Apr 13, 2010
13. Apr 13, 2010

### mparsons06

You were right! Thank you for helping me out!

14. Apr 13, 2010

### Redbelly98

Staff Emeritus