TITLE.PM5

204 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-6.pm5

Process of expansion : Reversible adiabatic
As the process is reversible adiabatic, it will be represented by a vertical line on T-s diagram
by 1-2 as it is also constant entropy process.
The condition at point ‘2’ can be calculated by equating the entropy at point ‘1’ and point ‘2’,
i.e., s 1 = s 2 .....per kg of steam

7.102 = sxs sf 222 +− 2 ()g f

= 1.3609 + x 2 (7.2984 – 1.3609)

∴ x 2 =

7 3609

7 3609

.102 1.

.2984 1.

−

− = 0.967

h 2 = hxhf 22 + 2 fg = 439.4 + 0.967 × 2244.1 = 2609.44 kJ/kg

h 1 (at 15 bar and 350°C) = 3147.5 kJ/kg

Applying the first law energy equation for steady flow process,

h 1 + C^1

2

2 = h^2 +

C 22

2 + W

i.e., W = (h 1 – h 2 ) +

CC 12 22

2

F −

HG

I

KJ

= 3147.5 – 2609.44 +

60 180

210

22

3

−

×

F

HG

I
KJ
= 3147.5 – 2609.44 – 14.4 = 523.66 kJ/kg.
Hence work done per kg of steam = 523.66 kJ/kg. (Ans.)
Example 4.60. Steam at 10 bar and 200°C enters a convergent divergent nozzle with a
velocity of 60 m/s and leaves at 1.5 bar and with a velocity of 650 m/s. Assuming that there is no
heat loss, determine the quality of the steam leaving the nozzle.
Solution. Initial pressure of steam, p 1 = 10 bar
Initial temperature of steam, t 1 = tsup = 200°C
Initial velocity, C 1 = 60 m/s
Final velocity, C 2 = 650 m/s
Final pressure, p 2 = 1.5 bar
Heat loss = nil
Quality of steam at the outlet :
It is a steady-state non-work developing system. Applying the steady flow energy equation
to the process, we get

h 1 + C^1

2

2

= h 2 + C^2

2

2

(Q Q = 0, W = 0)

∴ h 2 = h 1 +

CC 12 22

2

F −

HG

I

KJ

At 10 bar, 250ºC : h 1 = 2827.9 kJ/kg (from steam tables)

∴ h 2 = 2827.9 +

60 650

210

22

3

−

×

L

NM

O

QP = 2618.45 kJ/kg