TITLE.PM5

FIRST LAW OF THERMODYNAMICS 177

dharm

/M-therm/Th4-5.pm5

Area, A 2 =?

Mass flow rate, m& =?

(i)Velocity at exit of the nozzle, C 2 :

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

h 1 +

C 12

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

C 22

2 + Z^2 g + W

Here Q = 0, W = 0, Z 1 = Z 2

∴ h 1 +

C 12

2 = h^2 +

C 22

2

C 22

2 = (h^1 – h^2 ) +

C 12

2

= (2800 – 2600) × 1000 +

50

2

2

= 201250 N-m

∴ C 22 = 402500

∴ C 2 = 634.4 m/s. (Ans.)

(ii)Mass flow ratem& :

By continuity equation,

m& = AC

v =

AC

v

11

1

=

900 10 50

0

××−^4

.187

kg/s = 24.06 kg/s

∴ Mass flow rate = 24.06 kg/s. (Ans.)

(iii)Area at the exit, A 2 :

Now, m& =

AC

v

22

2

24.06 = A^20498 ×.634 4.

∴ A 2 = 24 06 0498..634 4×. = 0.018887 m^2 = 188.87 cm^2
Hence, area at the exit = 188.87 cm^2. (Ans.)
Example 4.46. In one of the sections of the heating plant in which there are no pumps
enters a steady flow of water at a temperature of 50°C and a pressure of 3 bar (h = 240 kJ/kg). The
water leaves the section at a temperature of 35°C and at a pressure of 2.5 bar (h = 192 kJ/kg). The
exit pipe is 20 m above the entry pipe.
Assuming change in kinetic energy to be negligible, evaluate the heat transfer from the
water per kg of water flowing.
Solution. Refer Fig. 4.53.
Enthalpy at ‘1’, h 1 = 240 kJ/kg
Enthalpy at ‘2’, h 2 = 192 kJ/kg
Z 2 – Z 1 = 20 m
Applying steady flow energy equation,

h 1 +

C 12

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

C 22

2 + Z^2 g + W