This document discusses problems involving pumps. It provides data on the pressure, vacuum gauge reading, correction head, discharge, specific weight of water, input power, and output power for two different pumps. For the first pump, the output power is calculated to be 1.3 kW when the discharge is 0.85 x 10-2 m3/sec. The efficiency is calculated to be 5.3351% when the input power is 0.3267 HP. For the second pump, the discharge is calculated to be 0.0722 m3/sec given the output power of 10.13 HP and other data. The efficiency is calculated to be 13.3% when the input power is 0.7591 HP.

1. Problems based on pumps
- DN VAISNAVI
17BBT057

2. A Centrifugal pump has a pressure of 9 𝑘𝑔𝑓/𝑚2
and
shows vacuum gauge reading of 440 mm Hg . The
correction head of pump is 0.45m . Take specific
weight of water as 9.81 𝐾𝑁/𝑚3
.
1.Find the output power of the pump when the
discharge is 0.85 x 10−2
𝑚3
/𝑠𝑒𝑐.
2.Also find the efficiency if the input power is 0.3267
HP.

3. Given data :
pressure P = 9 𝑘𝑔𝑓/𝑚2
vacuum gauge reading = 440 mm of Hg
= 0.44 x 13.6 m of water
= 5.98 m of water.
Correction head h(c) = 0.45m
discharge Q = 0.85 x 10−2 𝑚3/𝑠𝑒𝑐
specific weight of water = 9.81 𝐾𝑁/𝑚3
.

4. Solution :
output power p(o) = w x Q x H
to find H ( total head ) ,
H = P + vacuum gauge reading + correction head h(o)
= 9 + 5.98 + 0.45
H = 15.43
output power p(o) = w x Q x H
= 9.81 x 0.85 x 10−2 x 15.43
= 1.3 KW

5. To find efficiency
Efficiency in % = Output power / Input power
= 1.743 / 0.3267
= 5.3351
1 kilowatt = 1.341 HP
1.3 kW = 1.743 HP

6. The output power of gear oil pump is 10.13HP . Find
the discharge of pump if the pressure of pump is
12 𝑘𝑔𝑓/𝑚2
and vacuum gauge reading of 170 mm
of mercury . The correction head of the pump is
negligible . Take specific weight of water as
9.81 𝐾𝑁/𝑚3
.
1.To find the discharge of pump
2.Find the efficiency if input power is 0.7591HP.

7. Given data :
output power p(o) = 10.13HP
W = 9.81 𝐾𝑁/𝑚3
p = 12 𝑘𝑔𝑓/𝑚2
vacuum gauge reading = 170 mm of Hg
= 0.17 x 13.6 m of water
= 2.31 m of water .
Input power , P (i) = 0.7591HP

8. Solution :
Total head = p + vacuum gauge reading
= 12 + 2.3
= 14.3
Output power = Q X H X W
Q = output power / H X W
= 10.13 / 14.3 x 9.81
= 0.0722 𝑚3/𝑠𝑒𝑐
efficiency in % = output power / input power
= 10.13/0.7591 = 13.3

9. Characteristic curve: