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Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat per unit mass (in joules per kilogram) necessary to choke the flow at the exit of the duct, as well as the pressure and temperature at the duct exit, for an inlet Mach number of

(a) M1 = 2.0
(b) M1 = 0.2.

Respuesta :

saqib097

Answer:

The solution for the given problem is done below.

Explanation:

M1 = 2.0

[tex]\frac{p1}{p*}[/tex] = 0.3636

[tex]\frac{T1}{T*}[/tex] = 0.5289

[tex]\frac{T01}{T0*}[/tex] = 0.7934

Isentropic Flow Chart:  M1 = 2.0 , [tex]\frac{T01}{T1}[/tex] = 1.8

T1 = [tex]\frac{1}{0.7934}[/tex] (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= [tex]\frac{T02}{T1}T1[/tex] = (1.8)(288K) = 518.4 K.

Q= Cp(T02-T01) = [tex]\frac{1.4(287 J / (Kg.K)}{1.4-1}(653.4-518.4)K[/tex] = 135.7*[tex]10^{3}[/tex] J/Kg.

shadrachadamu

Answer:

q=135.7×10^3 J/kg

T_0^*= 653.4K

Explanation:

q=135.7×10^3 J/kg

T_0^*= 653.4K

M_1=2.0 => p_1/p^* =0.3636 T_1/T^* =0.5289 T_01/T_0 =0.7934

Isentropic flow properties chart:

M_1=2.0 => T_01/T_1 =1.8

T_0^*= ‖(T_O^*)/T_01 ‖‖T_01/T_1 ‖ T_1=‖1/0.7934‖(1.8)(288K)= 653.4K

In order to choke the flow at the exit (M_2=1),the above T_0^*= T_02; p_2=p^*

T_0^* must be the stagnation temperature at the exit.

At the inlet

T_02=T_02/T_1 T_1=(1.8)(288K)=518.4K

q=c_p (T_02-T_01 )=γ^R/(γ-1) (T_02-T_01 )=1.4(287J/(kg.K))/(1.4-1) (653.4-518.4)K=135.7×10^3 J/kg

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