TITLE.PM5

FIRST LAW OF THERMODYNAMICS 171

dharm

/M-therm/Th4-4.pm5

Example 4.41. The working fluid, in a steady flow process flows at a rate of 220 kg/min.
The fluid rejects 100 kJ/s passing through the system. The conditions of the fluid at inlet and
outlet are given as : C 1 = 320 m/s, p 1 = 6.0 bar, u 1 = 2000 kJ/kg, v 1 = 0.36 m^3 /kg and C 2 =
140 m/s, p 2 = 1.2 bar, u 2 = 1400 kJ/kg, v 2 = 1.3 m^3 /kg. The suffix 1 indicates the condition at inlet
and 2 indicates at outlet of the system.
Determine the power capacity of the system in MW.
The change in potential energy may be neglected.
Solution. Refer Fig. 4.48.

Fig. 4.48

Conditions of the fluid at point 1 :

Velocity, C 1 = 320 m/s

Pressure, p 1 = 6.0 bar = 6 × 10^5 N/m^2

Internal energy, u 1 = 2000 kJ/kg

Specific volume, v 1 = 0.36 m^3 /kg.

Conditions of the fluid at point 2 :

Velocity, C 2 = 140 m/s

Pressure, p 2 = 1.2 bar = 1.2 × 10^5 N/m^2

Internal energy, u 2 = 1400 kJ/kg

Specific volume, v 2 = 1.3 m^3 /kg

Heat rejected by the fluid, Q = 100 kJ/s (–).

Power capacity of the system :

Applying the energy equation at ‘1’ and ‘2’, we get

mu pv 111 C^1 Zg

2

+++ 2 1

L

N

M

M

O

Q

P

P ± Q =

mu pv 122 C^2 Zg

2

+++ 2 2

L

N

M

M

O

Q

P

P

± W

Taking –ve sign for Q as the system rejects heat and +ve sign for W as the system develops
work.

mu pv

C Zg

1111

2

+++ 2 1

L

N

M

M

O

Q

P

P – Q =

mu pv 222 C^2 Zg

2

+++ 2 2

L

N

M

M

O

Q

P

P

+ W

∴ Wmu u pv pv=−+−+CC− Q

F

HG

I

KJ

L

N

M

M

O

Q

P

P

()(12 11 22)^1 −

2

2

2

2 .[

Q (Z 1 – Z 2 ) g = 0]