Respuesta :
Answer:
The exit velocity air is 77.56 m/s.
Explanation:
Specific heat of air at 60°C is about 1.006 KJ/kg-K (from standard table).
Now from first law for open system
[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]
For ideal gas h=CT
Given that [tex]T_1=30[/tex]°C
[tex]T_2=90[/tex]°C
[tex]V_1=356 m/s[/tex]
By putting the values
[tex]1.006\times (273+30)+\dfrac{356^2}{2000}=1.006\times (273+90)+\dfrac{V_2^2}{2000}+[/tex]
[tex]V_2=77.56 m/s[/tex]
So the exit velocity air is 77.56 m/s.
The velocity at the exit of a diffuser of a jet engine, when the entering velocity is 356 m/s, is 77.56 m/s.
What is first law of thermodynamics?
The first law of thermodynamics, for a process the amount of energy obtained by the system and its surrounding should be equal to the amount of energy declined by the system and its surrounding.
From the first law of thermodynamics, the equation for an open system can be given as,
[tex]Q+h_1+\dfrac{V_1^2}{2}=W+h_2+\dfrac{V_2^2}{2}[/tex]
As the speacif heat of the air is 1.006 kJ/kg-K at 60 degree Celcius. Thus. The enthalpy of the system for the 30°C and 90°C are,
[tex]h_1=1.006(273+30)=304.818\\h_1=1.006(273+90)=365.178\rm kJ/kg[/tex]
The velocity at the enter of a diffuser is 356 m/s. Thus, put the values in the above formula as,
[tex]304.818+\dfrac{356^2}{2\times1000}=365.178+\dfrac{V_2^2}{2\times1000}\\V_1^2=77.56\rm m/s[/tex]
Thus, the velocity at the exit of a diffuser of a jet engine, when the entering velocity is 356 m/s, is 77.56 m/s.
Learn more about the first law of thermodynamics here;
https://brainly.in/question/6861511
