# Change in gauge pressure using viscous flow through a pipe.

Tags:
1. Jul 26, 2015

### KAC

1. The problem statement, all variables and given/known data
Water flows at 0.25 L/s through a 9.0-m-long garden hose 2.0 cm in diameter that is lying flat on the ground. The temperature of the water is 20 ∘C. What is the gauge pressure of the water where it enters the hose?

Side question: does the velocity of the water flow need to be converted to m/s rather than L/s? Upon conversion this becomes m^3/s, and I am not sure if this is suitable units for the velocity.

R = .01m
n = 1.0*10^-3 Pa*s
L = 9m
v= 0.25L/s

2. Relevant equations
Q = (pi*R^4*p)/8*n*L (p = pressure change; n = coefficient of viscosity, in this case 1.0*10^-3 due to the temperature of the water; L= length of the tube, 9m; R = .01m)
Q = v(avg)*A (A = cross sectional area, of a tube is pi*r^2)
v(avg) = (R^2/8nL)/p.

3. The attempt at a solution

I am unsure of where to begin. The first equation I would think to use is v(avg) = (R^2/8nL)*p so that I could plug this into Q = v(avg)A but the first equation requires p and that is what I am solving the problem for. To start, would it be suitable to use the velocity given and change the units to m^3/s? If I take this route, I get:

v = 0.25L/s
1000L = 1m^3
0.25L/s * 1m^3/1000L = 0.00025 m^3/s = v

Q = v*A; A = pi*R^2
A = pi* 0.01^2; A = 0.000314 m^2
Q = 0.00025m^3/s * 0.000314 m^2; Q = 7.85 * 10^-8

Q = (pi*R^4*p)/8nL
p= [(8nL)*Q]/pi*R^4
p = [(8 * (1.0*10^-3 Pa*s)) * 9m]/pi * 0.01^4
p = (0.072)/3.14*10^-8
p = 2292993 or 2.293 * 10 ^6.

This answer is incorrect and I am weary of the steps I took to get there. Can someone please point me in the correct direction? The first concern I have is converting L/s for velocity. Should I be leaving velocity in terms of L/s or changing them into m^3/s as I have done here? Also, is this velocity able to be used at all? The formula calls for average velocity, but with the formula for average velocity, I am unable to solve because it calls for the use of p, which I do not have.

Any help is appreciated. Thank you in advance.

2. Jul 26, 2015

### Staff: Mentor

What do you think the units of velocity are?

Chet

3. Jul 26, 2015

### KAC

Chet,

I think that the units of velocity should be m/s. Since it is a fluid and not an object in this case, I am assuming that m^3/s is suitable to use?

4. Jul 26, 2015

### SteamKing

Staff Emeritus
Are m3/s the same as m/s?

You're mixing rate of flow with flow velocity, which are two different concepts.

If you had carried the units in the calculation of Q = A * v as you should have, this would also tell you that rate of flow cannot be substituted for flow velocity.

5. Jul 26, 2015

### Staff: Mentor

So, out of the whole list of the symbols you have shown, which one does the 0.25 l/ s correspond to?

6. Jul 26, 2015

### KAC

If I convert L to m3, it corresponds to Q, the flow rate of the fluid. So in this case, I don't actually need to solve for Q because this is already given?
After conversions Q = 2.5*10-4; if I plug this into my equation:

p= [(8nL)*Q]/pi*R^4
p=[(8*(1.0*10-3 Pa*s)*9m)*2.5*10-4 m3/s]/pi*0.01m4
p= 1.8*10-5/3.14*10-8
p = 573.2 Pa

Does my form appear to be correct here?

7. Jul 26, 2015

### Staff: Mentor

I haven't checked your arithmetic, but this now looks right. You should check the reynolds number to confirm that the flow is laminar. If it is turbulent, then a different equation prevails, and pressure drop is larger.

Chet