TITLE.PM5

(Ann) #1
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.)

Free download pdf