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A water pump has been operating at a pump station at speed 1000 rpm. The pipeline...

Question:

A water pump has been operating at a pump station at speed 1000 rpm. The pipeline system characteristics at the station are a

A water pump has been operating at a pump station at speed 1000 rpm. The pipeline system characteristics at the station are as follows. Head loss in pipeline: 80 1 Discharge (LS) 0 20 40 60 Head (m) 4 10 20 The static head is 20 m. The pump characteristics at speed 1000 rpm are provided by the manufacturer and given as follows. Discharge (IS) 0 20 40 50 60 Head (m) 50 45 14 Efficiency (%) 60 69 60 40 33 25 2 a). Determine the duty point of the present pump-pipeline system, and the power consumption of the pump. It is proposed to add another pump, identical to the present one. The two pumps will be connected in parallel and operating at a speed of 1000 rpm. b). Determine the maximum discharge rate of the proposed pump-pipeline system. c). Determine the total power consumption of the proposed system corresponding to the maximum discharge rate, and comment on the suitability of the proposed system.

Answers

Ans a) We know, System curve equation can be given as :

Hp = Z + H Q^2    

where, Z = Static head = 20 m

H = head loss

=> Hp = 20 + H Q^2

At flow rate = 20 L/s or 0.02 m^3 /s.

Hp = 20 + 1(0.022) = 20 m

At flow rate = 40 L/s or 0.04 m^3 /s.

Hp = 20 + 4(0.4-) = 20 m

Similarly we can calcultate head for all flow rates. Final results areshown in table below:

Q (Lit/s) Hp (m)
0 20
20 20
40 20
60 20.036
80 20.13

Plot the given system curve and performance curve as shown below :

80 70 60 50 Hp (m) 40 Performance curve 30 System curve Efficiency 20 10 0 0 20 40 60 80 100 Q(Lit/s)

The point of intersection of performance curve and system curve is operating point or duty point.Hence,

Operating flow rate corresponding to duty point = 55 L/s or 0.055 m^3 s

Operating efficiency corresponding to opearting point (\eta) = 50%

We know,

Power consumption (P) = \rho g Q Hp / \eta

where, Hp = Operating head corresponding to operating point = 20 m

Putting values,

=> P = 1000 x 9.81 x 0.055 x 20 / 0.50

=> P = 21.58 kW

Ans b) We know, when two similar pumps are conneted in series, the resulting performance head will be twice of head of individual pumps but flow rate remain same. Hence, performance curve for pumps in series is as follows :

Q (Lit/s) Hp (m)
0 50 + 50 = 100
20 45 + 45 = 90
40 33 + 33 = 66
50 25 + 25 = 50
60 14 + 14 = 28

Plot the given system curve and new performance curve as shown below

120 100 80 Hp (m) 60 -Performance curve -System curve 40 20 0 0 20 40 60 80 100 olit/s)

The point of intersection of performance curve and system curve is operating point or duty point.Hence,

Operating flow rate corresponding to duty point = 63 L/s

Hence, Maximum flow rate of proposed system is 63 L/s   

Ans c) Now, for proposed system,

Q = 63 L/s or 0.063 m^3 /s

Hp = 20 m

Efficiency (\eta) corresponding to flow rate 63 L/s is approximate to 36%

Power consumption (P) = \rho g Q Hp / \eta

=> P = 1000 x 9.81 x 0.063 x 20 / 0.36

=> P = 34.34 kW

Since, power consumption of system increases significantly (from 21.58 kW to 34.34 kW) but the increase in flow rate is small (from 55 L/s to 63 L/s), the proposed system is not suitable as the efficiency of system is very low.

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