This example is for the hydraulic machin

1. EXAMPLE 7.2 CENTRIFUGAL PUMP
•A centrifugal pump lifts water under a static head of 36 m of water of which 4 m
is suction lift. Suction and delivery pipes are both 150 mm in diameter. The head
loss in suction pipe is 1.8 m and in delivery pipe 7 m. The impeller is 380 mm in
diameter and 25 mm wide at mouth and revolves at 1200 r.p.m. Its exit blade
angle is 35°. If the manometric efficiency of the pump is 82 percent determine:
• (i) The discharge through the pump, and
• (ii) The pressure at the suction and delivery
• branches of the pump
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5. EXAMPLE 7.3 SPECIFIC SPEED AND RATIONAL SPEED
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6. EXAMPLE 7.4: HEAD AND POWER BY A PUMP
• Water is being pumped from the lower reservoir at elevation of +100m to the
upper reservoir at elevation of +150m with a flow rate of 2m3/s. And, the pipe has
a length of 900m and diameter of 60cm. If the efficiency of the pump is 75%,
a) what is the shaft power required to drive the pump?
•Take C =120 (New welded steel pipe material) Haz. W eqn. Ans: 2,859.314KW
b) What is the cost of electric energy to fill a reservoir volume of 8000m3 if motor
efficiency =90%?
• Take cost of energy =1.5Birr/kwhr Ans: 5,289.73Birr
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7. Example 7.5 Power and Energy from turbine
A Hydropower dam has the following main features:
• Head water level = +500m, Tail water level = +464m , Total head loss=1.5m,
• Two Francis Turbines of capacities 14,000KW have been installed,
• Hydraulic efficiency = 86%, Conveyance efficiency = 92%,
• Average number of days in a month =30days , Service hour per day=10hr
a) Estimate the flow rate available in the penstocks?
b) Determine the maximum energy generation per month?
c) Determine the specific speed of the turbine if the rotational speed is 150rpm?

8. Solution
a) Total installed Capacity = 2*14,000 = 28,000KW
• Gross Head, HG= HWL-TWL =500-464 = 46m
• Net Head , Hnet =HG- hf = 46-1.5 = 44.5m
• Over all efficiency =Hydraulic efficiency*Conveyane efficiency =H *C =
0.86*0.92 = 0.791
• Total installed Capacity , P= o*w*Q*Hnet , 28,000= 0.791*9.81* Q*44.50
• The flow rate , Q = 81.087 m3/se Discharge per penstock=Q/4=20.27m3/s
• b) Energy, E=P*t = 28000KW*30days* 10hr/day = 8,400,000KWhr
• C) 𝑵𝒔 =
𝑵 𝑷
𝑯𝟓/𝟒 =
𝟏𝟓𝟎∗ 𝟏𝟒𝟎𝟎𝟎
𝟒𝟒.𝟓𝟓/𝟒 = 𝟏𝟓𝟒. 𝟒𝟐