TITLE.PM5

170 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-4.pm5

−−= −+

F

HG

I

KJ −

F

HG

I

KJ

×

R

S

|

|

T

|

|

U

V

|

|

W

|

|

+ −×

L

N

M

M

M

M

M

O

Q

P

P

P

P

P

16000

3600

4500

3600

2300 2800

5600

60

2800

60

2 1000

55 9

1000

22

W () ()1. .5 .81

• 4.44 – W = 1.25 (500 + 3.26 – 0.039) or W = 633.44 kJ/s
  ∴ Power output of the turbine = 633.44 kW. (Ans.)
  Example 4.40. Steam at a 6.87 bar, 205°C, enters in an insulated nozzle with a velocity of
  50 m/s. It leaves at a pressure of 1.37 bar and a velocity of 500 m/s.
  Determine the final enthalpy of steam.
  Solution. Pressure of steam at the entrance, p 1 = 6.87 bar
  The velocity with which steam enters the nozzle, C 1 = 50 m/s
  Pressure of steam at the exit, p 2 = 1.37 bar
  Velocity of steam at the exit, C 2 = 500 m/s.

p = 6.87 bar,

205°C

C = 50 m/s

1

1

p = 1.37 bar

C = 500 m/s

2

2

1

2

Fig. 4.47

The steady flow energy equation is given by

h 1 + C^1

2

2 + Z^1 g + Q = h^2 +

C 22

2

+ Z 2 g + W ...(i)

Considering the nozzle as an open system, it is evident that :

— there is no work transfer across the boundary of the system (i.e., W = 0)

— there is no heat transfer because the nozzle is insulated (i.e., Q = 0).

— the change in potential energy is negligible since there is no significant difference in

elevation between the entrance and exit of the nozzle [i.e. (Z 2 – Z 1 ) g = 0].

Thus eqn. (i) reduces to

h 1 +

C 12

2 = h 2 +

C 22

2

∴ (h 2 – h 1 ) +

CC 22 12

2

−

= 0

From steam table corresponding to 6.87 bar, h 1 = 2850 kJ/kg

∴ (h 2 – 2850) + ()()5002 1000^50

(^22) −
× = 0
or h 2 – 2850 + 123.75 = 0 or h 2 = 2726.25 kJ
Hence final enthalpy of steam = 2726.25 kJ. (Ans.)