# Homework Help: Fluid Mechanics Question - Pipes and Pumps

1. Apr 4, 2012

### eyraincg

Fluid Mechanics Question -- Pipes and Pumps

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

Design a system to deliver $1 m^{3}/s$ of water from reservoir A at elevation 300m to reservoir B at 500m. The distance from A to B is 1000km (1000,000m) and the elevation between the two reservoirs can be approximated as parabolic with a maximum elevation of 660m reached 600km from A.

Negative pressures shouldn’t be allowed to exceed about 70% of atmospheric pressure, use this figure to prevent cavitation.

The steel pipe with a roughness of e=0.5cm and diameter of 1.1m and a pressure rating of 150m head is available. The pumps can deliver $1m^3/s$ at 100m head.

*Since the horizontal distance is significantly greater than the vertical distance. I approximated the parabola with straight lines from A to the maximum point of 600,000m from A and elevation 660m and then from that point to B.
I am then doing calculations separately for the 2 parts

*I am trying to calculate amount of pumps that would make this system work and prevent cavitation

*Any help would be appreciated, thanks!

2. Relevant equations

Extended bernoulli’s equation:
$\frac{V_{1}^{2}}{2g}$ + $\frac{p_{1}}{\gamma}$ + $Z_{1}$ + $h_{p}$ = $\frac{V_{2}^{2}}{2g}$ + $\frac{p_{2}}{\gamma}$ + $Z_{2}$ + $h_{L}$
V: Velocity
P: Pressure
$\gamma$: Specific Weight
Z: Height/ Altitude
$h_{p}$: Pump Head
$h_{L}$: Head Loss

$h_{L} = f \frac{L}{D} \frac{V^{2}}{2g}$
f: friction factor
L: Pipe Length
D: Pipe Diameter

Volumetric flow rate to Velocity:

V = Q/A

Q: Volumetric Flow Rate$(m^{3}/s)$
A: Area
V = Velocity

3. The attempt at a solution

For part 1 [0km to 600km] only:
D = 1.1m
A = 0.95 m^{2}
Q = 1$m^{3}/s$
V = $\frac{Q}{A}$ = 1.053 m/s

f = 0.0171 (from moody diagram)
R$_{e}$ = 1157490

P$_{minimum}$ (absolute) = 101315Pa*0.7 = 70927.5Pa
P$_{minimum}$ (gage) = P_{min,absolute} – P_{atmospheric} = -30397.5Pa

H$_{p}$ = 100m
L ~= 600km = 600,000m
Using Bernoulli’s formula solve for maximum Length before cavitation:

$\frac{V_{1}^{2}}{2g}$ + $\frac{p_{1}}{\gamma}$ + Z$_{1}$ + h$_{p}$ = $\frac{V_{2}^{2}}{2g}$ + $\frac{p_{2}}{\gamma}$ + Z$_{2}$ + $h_{L}$

P$_{1}$ = 0
P$_{2}$ = -30397.5
V$_{1}$ = V$_{2}$
Z$_{1}$ = 300m
Z$_{2}$ = 660m
$\gamma$ = 9800

$\frac{V_{1}^{2}}{2g}$ + $\frac{p_{1}}{\gamma}$ + Z$_{1}$ + h$_{p}$ = $\frac{V_{2}^{2}}{2g}$ + $\frac{p_{2}}{\gamma}$ + Z$_{2}$ + $f*\frac{L}{D}*\frac{V^{2}}{2g}$
Rearrange:

L = $(\frac{V_{1}^{2}}{2g}$ + $\frac{p_{1}}{\gamma}$ + Z$_{1}$ – Z$_{2}$ + h$_{p}$ - $\frac{V_{2}^{2}}{2g}$ - $\frac{p_{2}}{\gamma}$ ) $*\frac{2D*g}{fV^{2}}$
Cancelling terms:

L = (Z$_{1}$ – Z$_{2}$ +h$_{p}$ - $\frac{p_{2}}{\gamma}$) $*\frac{2D*g}{fV^{2}}$

L = (300m – 660m + 100m +3.0986) *1144.77
L = -293076m

I don’t understand why this is the answer, what am I doing wrong?