TITLE.PM5

176 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

Increase in enthalpy of circulating water, Qwater = – 92 kJ/kg

Amount of heat transferred to atmosphere, Qatm. =?

Applying steady-flow 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

or Q = (h 2 – h 1 ) + CC^2

2

1

2

2

F −

HG

I

KJ

+ (Z 2 – Z 1 )g + W

Assuming change in P.E. and K.E. to be negligible.

∴ Q = (h 2 – h 1 ) + W = 70 + (– 175) = – 105 kJ

But Q = Qatm + Qwater or –105 = Qatm + (– 92)

∴ Qatm = – 13 kJ/kg.

Thus heat transferred to atmosphere = 13 kJ/kg. (Ans.)

+Example 4.45. At the inlet to a certain nozzle the enthalpy of fluid passing is 2800 kJ/kg,

and the velocity is 50 m/s. At the discharge end the enthalpy is 2600 kJ/kg. The nozzle is horizon-

tal and there is negligible heat loss from it.

(i)Find the velocity at exit of the nozzle.

(ii)If the inlet area is 900 cm^2 and the specific volume at inlet is 0.187 m^3 /kg, find the mass

flow rate.

(iii)If the specific volume at the nozzle exit is 0.498 m^3 /kg, find the exit area of nozzle.

Solution. Refer Fig. 4.52.

1

2

h = 2800 kJ/kg

C = 50 m/s

A = 900 cm

1

1

1

2

h = 2600 kJ/kg

C=?

A=?

2

2

2

Fluid in Fluid out

Fig. 4.52

Conditions of fluid at inlet (1) :

Enthalpy, h 1 = 2800 kJ/kg

Velocity, C 1 = 50 m/s

Area, A 1 = 900 cm^2 = 900 × 10–4 m^2

Specific volume, v 1 = 0.187 m^3 /kg

Conditions of fluid at exit (2) :

Enthalpy, h 2 = 2600 kJ/kg

Specific volume, v 2 = 0.498 m^3 /kJ