(adsbygoogle = window.adsbygoogle || []).push({}); Fluid Mechanics Question -- Pipes and Pumps

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

Design a system to deliver [itex]1 m^{3}/s[/itex] 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 [itex]1m^3/s[/itex] 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:

[itex]\frac{V_{1}^{2}}{2g}[/itex] + [itex]\frac{p_{1}}{\gamma}[/itex] + [itex]Z_{1}[/itex] + [itex]h_{p}[/itex] = [itex]\frac{V_{2}^{2}}{2g}[/itex] + [itex]\frac{p_{2}}{\gamma}[/itex] + [itex]Z_{2}[/itex] + [itex]h_{L}[/itex]

V: Velocity

P: Pressure

[itex]\gamma[/itex]: Specific Weight

Z: Height/ Altitude

[itex]h_{p}[/itex]: Pump Head

[itex]h_{L}[/itex]: Head Loss

Head Loss:

[itex]

h_{L} = f \frac{L}{D} \frac{V^{2}}{2g}

[/itex]

f: friction factor

L: Pipe Length

D: Pipe Diameter

Volumetric flow rate to Velocity:

V = Q/A

Q: Volumetric Flow Rate[itex] (m^{3}/s)[/itex]

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[itex]m^{3}/s[/itex]

V = [itex]\frac{Q}{A}[/itex] = 1.053 m/s

f = 0.0171 (from moody diagram)

R[itex]_{e}[/itex] = 1157490

P[itex]_{minimum}[/itex] (absolute) = 101315Pa*0.7 = 70927.5Pa

P[itex]_{minimum}[/itex] (gage) = P_{min,absolute} – P_{atmospheric} = -30397.5Pa

H[itex]_{p}[/itex] = 100m

L ~= 600km = 600,000m

Using Bernoulli’s formula solve for maximum Length before cavitation:

[itex]\frac{V_{1}^{2}}{2g}[/itex] + [itex]\frac{p_{1}}{\gamma}[/itex] + Z[itex]_{1}[/itex] + h[itex]_{p}[/itex] = [itex]\frac{V_{2}^{2}}{2g}[/itex] + [itex]\frac{p_{2}}{\gamma}[/itex] + Z[itex]_{2}[/itex] + [itex]h_{L}[/itex]

P[itex]_{1}[/itex] = 0

P[itex]_{2}[/itex] = -30397.5

V[itex]_{1}[/itex] = V[itex]_{2}[/itex]

Z[itex]_{1}[/itex] = 300m

Z[itex]_{2}[/itex] = 660m

[itex]\gamma[/itex] = 9800

Sub in Head Loss:

[itex]\frac{V_{1}^{2}}{2g}[/itex] + [itex]\frac{p_{1}}{\gamma}[/itex] + Z[itex]_{1}[/itex] + h[itex]_{p}[/itex] = [itex]\frac{V_{2}^{2}}{2g}[/itex] + [itex]\frac{p_{2}}{\gamma}[/itex] + Z[itex]_{2}[/itex] + [itex]f*\frac{L}{D}*\frac{V^{2}}{2g}

[/itex]

Rearrange:

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

Cancelling terms:

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

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

L = -293076m

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Fluid Mechanics Question - Pipes and Pumps

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**