Question

Steam enters an adiabatic nozzle at 2.5 MPa and 450oC with a velocity of 55 m/s and exits at 1 MPa and 390 m/s. If the nozzle has an inlet area of 6 cm2 , determine (a) the exit temperature. (b) the rate of entropy generation for this process.

Answer 1

Answer:

(a). The value of temperature at turbine exit = 687.64 K

(b). The rate of entropy generation for this process =  $S_{gen}$ = 0.3157 $\frac{KJ}{Kg k}$

Explanation:

Pressure at inlet $P_{1}$ = 2.5 M pa

Temperature $T_{1}$ = 450 °c = 723 K

Velocity at inlet $V_{1}$ = 55 $\frac{m}{sec}$

Pressure at exit $P_{2}$ = 1 M pa

Velocity at exit $V_{2}$ = 390 $\frac{m}{sec}$

Value of specific heat for steam = 2.108 $\frac{KJ}{Kg k}$

(a). Apply steady flow energy equation for adiabatic nozzle

⇒ $h_{1}$ + $\frac{V_{1} ^{2} }{2000}$ = $h_{2}$ + $\frac{V_{2} ^{2} }{2000}$  ----------- ( 1 )

⇒ $h_{2}$ - $h_{1}$ = $\frac{V_{1} ^{2} }{2000}$ - $\frac{V_{2} ^{2} }{2000}$

⇒ $C_{p}$ ( $T_{2}$ - $T_{1}$ ) =  $\frac{V_{1} ^{2} }{2000}$ - $\frac{V_{2} ^{2} }{2000}$

⇒ 2.108 ( $T_{2}$ - 723 ) = $\frac{55 ^{2} }{2000}$ - $\frac{390 ^{2} }{2000}$

⇒ 2.108 ( $T_{2}$ - 723 ) = - 74.53

⇒ ( $T_{2}$ - 723 ) = - 35.36

⇒ $T_{2}$ = 687.64 K

This is the value of temperature at turbine exit.

(b). Rate of entropy generation for this process is given by

$S_{gen}$ = $C_{p}$ $log_{e}$ $\frac{T_{2} }{T_{1} }$ - R $log_{e}$ $\frac{P_{2} }{P_{1} }$

Put all the values in above equation we get

⇒ $S_{gen}$ = 2.108 $log_{e}$ $\frac{687.64}{723}$ - 0.46 $log_{e}$ $\frac{1}{2.5}$

⇒ $S_{gen}$ = - 0.1057 + 0.4214

⇒ $S_{gen}$ = 0.3157 $\frac{KJ}{Kg k}$

This is the value of entropy generation for this process.