Answer:
The value of exit temperature from the nozzle = 719.02 K
Explanation:
Temperature at inlet [tex]T_{1}[/tex] = 450°c = 723 K
Velocity at inlet [tex]V_{1}[/tex] = 55 [tex]\frac{m}{sec}[/tex]
velocity at outlet [tex]V_{2}[/tex] = 390 [tex]\frac{m}{sec}[/tex]
Specific heat at constant pressure for steam [tex]C_{p} = 18723 \frac{J}{kg k}[/tex]
Apply steady flow energy equation for the nozzle
[tex]h_{1} + \frac{V_{1} ^{2} }{2} = h_{2} + \frac{V_{2} ^{2} }{2}[/tex]
[tex]C_{p} T_{1} + \frac{V_{1} ^{2} }{2} = C_{p} T_{2} + \frac{V_{2} ^{2} }{2}[/tex]
Put all the values in the above formula we get,
⇒ 18723 × 723 + [tex]\frac{55^{2} }{2}[/tex] = [tex]C_{p}[/tex] [tex]T_{2}[/tex] + [tex]\frac{390^{2} }{2}[/tex]
⇒ [tex]T_{2}[/tex] = 719.02 K
This is the value of exit temperature from the nozzle.
