Answer: 436.5k, 330.8 kpa
Explanation:
(A) Temperature
500K = 503.02k h1 ( from table)
P1 = 600kpa
Solution
h2 = h1 -( v2^2 -v1^2) / 2
= 503.02 - ( 380^2 - 120^2) /2 (1kj/kg/1000mls)
= 436.5k
(B) pressure
P2 = (A1 T2 V1) / (A2 T1 V2) X P1
= 2( 436.5k x 120m/s) / 1( 500k x 380m/s) x 600kpa
P2 = 330.8kpa
a) The outlet temperature of the air is 436.832 K.
b) The outlet pressure of the air is 331.076 kilopascals.
Let suppose that the air flowing in the adiabatic nozzle behaves ideally, by mass balance and the equation of state for ideal gases we have an expression for the outlet pressure of the air ([tex]P_{out}[/tex]), in kilopascals:
[tex]P_{out} = \left(\frac{T_{out}}{T_{in}} \right)\cdot \left(\frac{A_{in}}{A_{out}} \right)\cdot \left(\frac{v_{in}}{v_{out}} \right)\cdot P_{in}[/tex] (1)
Where:
And by energy balance and definition of enthalpy change for ideal gases we construct the following formula for the outlet temperature ([tex]T_{out}[/tex]), in Kelvin:
[tex]T_{out} = T_{in} - \frac{v_{out}^{2}-v_{in}^{2}}{2\cdot \bar{c}_{p}}[/tex] (2)
If we know that [tex]T_{in} = 500\,K[/tex], [tex]v_{out} = 380\,\frac{m}{s}[/tex], [tex]v_{in} = 120\,\frac{m}{s}[/tex], [tex]\bar{c}_{p} = 1029\,\frac{J}{kg\cdot K}[/tex], [tex]\frac{A_{in}}{A_{out}} = 2[/tex] and [tex]P_{in} = 600\,kPa[/tex], then we have the outlet temperature and pressure of the adiabatic nozzle are:
a) Outlet temperature
[tex]T_{out} = 500\,K - \frac{\left(380\,\frac{m}{s} \right)^{2}-\left(120\,\frac{m}{s} \right)^{2}}{2\cdot \left(1029\,\frac{J}{kg\cdot K} \right)}[/tex]
[tex]T_{out} = 436.832\,K[/tex]
The outlet temperature of the air is 436.832 K.
b) Outlet pressure
[tex]P_{out} = \left(\frac{436.832\,K}{500\,K} \right)\cdot (2)\cdot \left(\frac{120\,\frac{m}{s} }{380\,\frac{m}{s} } \right)[/tex]
[tex]P_{out} = 331.076\,kPa[/tex]
The outlet pressure of the air is 331.076 kilopascals.
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