Pipe Design Problem (Diameter and Pump Determination)

[kedwardr]

You are the Chief Engineer for a prominent hydraulics consulting company. Your client wants to pay you lots of money to design a system to deliver 1 m3/s from reservoir A at elevation 300 m to reservoir B at 500 m. The distance from A to B is 1000 km and the elevation between the two reservoirs is given by z (m) = 300 + 1.2x – 0.001x2, where x is the distance from A in km.

The system is to consist of a pipeline and a series of pumps. Commercial steel pipe with a roughness e = 0.05 cm and a pressure rating of 150 m head is available. The total cost of purchasing pipe and constructing a pipeline is listed below for various pipe sizes. Pumps that can deliver 1 m3/s at 100 m head cost $10,000,000 each (installed).

Determine the most economical design (i.e. pipe size and number and location of pumps). Hint: Vary your pipe diameter and consider pumps in series.


Pipe diameter (m) Total Cost ($/km)
0.9 80,000
1.0 100,000
1.1 120,000
1.2 150,000
1.4 200,000


Note 1: Theoretically, a pipeline may be designed to allow pressure to fall to the vapour pressure. In practice, however, water usually contains dissolved gasses that will vapourize well before the vapour pressure point is reached. Such gasses dissolve very slowly. They can move with the water in the form of large bubbles that disrupt the flow. Therefore, negative pressures shouldn’t be allowed to exceed about 70% of atmospheric pressure – use this figure in your design calculations.

Note 2: The pressure rating of a pipe is the maximum pressure that it can withstand.





I have calculated the length of the pipe. However, I am not sure how to proceed because we need pumps and we do not know what the diameter is. Moreover, We have to make sure cavitation does not occue (-7.0m pressure head) and the pressure in the pipe is no greater than 150 m head.

Please help me!

Thanks
K

 

[RTW69]

You need to calculate the arc length of the pipe, which it sounds like you have done. The pumps need to be close enough the reservoir A so as not to cause cavitation. You will do 5 calculations, one for each pipe diameter. Select the pipe diameter and number of pumps combination that gives you the lowest cost. Smaller diameter pipe means lower pipe cost but larger pump cost, so it is a trade off.

 

[kedwardr]

Hello,

Thanks for the reply. Can you explain why the pumps has to be close to Reservoir A as not to cause cavitation? Or do you mean to say we have to place pumps along the pipe until the maximum point of the pipe?

We are given that each pump can deliver a total head of 100m. However, the question is how many pumps should I use? I am stuck with this question all day long. Should we do this graphically and place a pump each time pressure head is 0m? Please advice me on how can i determine the numbers and locations of the pump.

Thanks
Kevin

 

[RTW69]

P1,P2 and V1 is probably 0 since you are pumping to large reservoirs. Use the energy equation to determine pump energy in meters. Pump energy= Delta Z + V2^2/2g+HL.

Since you selected a pipe diameter, you can find pipe velocity. You can now have everything needed to find HL/meter of pipe. You now have the pump energy in meters so you can determine the number of pumps. You will need two pumps just to pump the 200 meters in elevation. The rest is to over come the friction in the piping. The pumps will be spread out to roughly every 100 m of head loss. You may want to create an energy and hydraulic grade line to visualize pump locations. Calculate pump and pipe costs. Do the whole process over for the next pipe diameter. Select the cheapest cost option.

 

[kedwardr]

Yes. I have calculated the total length of the pipe(1162 km), and the respective major losses due to friction.

However on that particular equation of Pump energy= Delta Z + V2^2/2g+HL. Isn't V2 = 0?? Since the velocity on both of the reservoir is negligible (i.e. very large areas). Thus the energy equation can be simplified down to:

Pump energy= Delta Z + +HL. ?

Where Head Loss is just friction loss (we were told to neglect minor losses). Thus it is denoted by f(L/D) V2/2g, where V = Q/A of the pipe with a given diameter.

Is there any tips on drawing EGL and HGL? Cause It is pretty tedious to be drawn by hand.
And how do you know the rate of decrease of the EGL/HGL?? I know that the EGL and HGL will decrease in time unless there are pumps, but to determine the location of the pump correctly, we need to have an exact slope too.

Thanks

 

[nkonjeni]

P1,P2 and V1 is probably 0 since you are pumping to large reservoirs. Use the energy equation to determine pump energy in meters. Pump energy= Delta Z + V2^2/2g+HL.

Since you selected a pipe diameter, you can find pipe velocity. You can now have everything needed to find HL/meter of pipe. You now have the pump energy in meters so you can determine the number of pumps. You will need two pumps just to pump the 200 meters in elevation. The rest is to over come the friction in the piping. The pumps will be spread out to roughly every 100 m of head loss. You may want to create an energy and hydraulic grade line to visualize pump locations. Calculate pump and pipe costs. Do the whole process over for the next pipe diameter. Select the cheapest cost option.

is the approach you mentioned above assuming that you'd place the pumps at the water surface? is that why p1 and p2 are 0?

Im having a lot of trouble with pump placement? I am trying to ensure that all the points that i pick are on the parabola but the process is so arduous I am convinced I am using the wrong approach...