TITLE.PM5

174 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

Fig. 4.50

Specific volume, v 2 = 0.14 m^3 /kg

Increase in enthalpy of air passing through the compressor,

(h 2 – h 1 ) = 150 kJ/kg

Heat lost to the surroundings,

Q = – 700 kJ/min = – 11.67 kJ/s.

(i)Motor power required to drive the compressor :

Applying energy equation to the system,

mh

C Zg

1 1

2

++ 2 1

F

HG

I

KJ + Q =

mh 2 C^2 Zg

2

++ 2 2

F

HG

I

KJ

+ W

Now Z 1 = Z 2 (given)

∴ mh

C

1 1

2

2

+

F

HG

I

KJ

+ Q = mh

C

2 2

2

2

+

F

HG

I

KJ

+ W

Wmh h=−+CC−

L

N

M

M

O

Q

P

P

() 12 1

2

2

2

2 + Q

=−+−

×

L

N

M

M

O

Q

P

P

0 2 150 12 90

2 1000

22

. + (– 11.67)

= – 42.46 kJ/s = – 42.46 kW

∴ Motor power required (or work done on the air) = 42.46 kW. (Ans.)