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To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.
By definition the entropy change would be defined as
Using the Boyle equation we have
Where,
= Specific heat at constant pressure
= Initial temperature of gas
= Final temprature of gas
R = Universal gas constant
= Initial specific Volume of gas
= Final specific volume of gas
According to the statement, it is an isothermal process and the tank is therefore rigid
The equation would turn out as
<em>Therefore the entropy change of the ideal gas is 0</em>
Into the surroundings we have that
Where,
Q = Heat Exchange
T = Temperature in the surrounding
Replacing with our values we have that
<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>
Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 , = 1.8
T1 = (1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02= = (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) = = 135.7*
J/Kg.