# Tube Pressure and Viscosity

1. Nov 14, 2007

### jamesyboy1990

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

where:
l = the length of the tube in cm
r = the radius of the tube in cm
p = the difference in pressure of the two ends of the tube in dynes per cm2
c = the coefficient of Viscosity in poises (dyne-seconds per cm2)
v = volume in cm3 per second
pi = 3.14159..... (i couldn't get the symbol to appear)

I am doing an experiment in which i measure viscosity for different liquids. However, i'm having problems in obtaining value "p". I know that pressure is force times area. In this case, area should be the cross sectional area for the tube at one end, and the cross sectional area of the valve opening at the other end (if i'm not mistaken). However, what will be the force?

2. Relevant equations

where:
l = the length of the tube in cm
r = the radius of the tube in cm
p = the difference in pressure of the two ends of the tube in dynes per cm2
c = the coefficient of Viscosity in poises (dyne-seconds per cm2)
v = volume in cm3 per second
pi = 3.14159..... (i couldn't get the symbol to appear)

3. The attempt at a solution

I know that pressure is force times area. In this case, area should be the cross sectional area for the tube at one end, and the cross sectional area of the valve opening at the other end (if i'm not mistaken). However, what will be the force?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 14, 2007

### Supernats

Isn't pressure defined as force divided by area?

3. Nov 14, 2007

### jamesyboy1990

oooo, typo. your right, i meant force PER UNIT area. but still, any ideas on my problem?

4. Nov 14, 2007

### jamesyboy1990

$$\frac{r^{4}(pi)p}{8cl}$$

5. Nov 14, 2007

### Supernats

No, I was horrible at fluid mechanics. I was just making sure you didn't write it down wrong on your paper and then get confused. Aside from that, I'm useless.

6. Nov 30, 2011

### Lue

Would the force be gravity?