TITLE.PM5

FIRST LAW OF THERMODYNAMICS 179

dharm

/M-therm/Th4-5.pm5

C 22

2 = (h^1 – h^2 ) +

C 12

2 = (915 – 800) × 1000 +

300

2

2

∴ C 22 = 320000 or C 2 = 565.7 m/s.
Hence relative velocity of gas leaving the jet pipe = 565.7 m/s. (Ans.)
Example 4.48. A centrifugal pump delivers 50 kg of water per second. The inlet and outlet
pressures are 1 bar and 4.2 bar respectively. The suction is 2.2 m below the centre of the pump
and delivery is 8.5 m above the centre of the pump. The suction and delivery pipe diameters are
20 cm and 10 cm respectively.
Determine the capacity of the electric motor to run the pump.
Solution. Refer Fig. 4.54.

Electric

motor

W

Water out

2

Boundary

Centrifugal

pump

(^1) Water in
2.2 m
8.5 m
Fig. 4.54
Quantity of water delivered by the pump, mw = 50 kg/s
Inlet pressure, p 1 = 1 bar = 1 × 10^5 N/m^2
Outlet pressure, p 2 = 4.2 bar = 4.2 × 10^5 N/m^2
Suction-below the centre of the pump = 2.2 m
Delivery-above the centre of the pump = 8.5 m
Diameter of suction pipe, d 1 = 20 cm = 0.2 m
Diameter of delivery pipe, d 2 = 10 cm = 0.1 m