Respuesta :
Answer:
a) T₂ = 376.905 K
b) [tex]\eta_n[/tex] = 95.4%
Explanation:
Given:
Initial pressure, P₁ = 1.3 bar
Initial temperature, T₁ = 423 K
Initial velocity of air, v₁ = 40 m/s
Exit pressure, P₂ = 0.85 bar
Exit velocity of the air, v₂ = 307 m/s
gas constant for air, k = 1.4
also,
the specific heat for the air, Cp = 1.005 KJ/kg.K =1.005 × 10³ J/kg.K (standard)
a) applying the energy rate balance, we have
[tex]h_1+\frac{v_1^2}{2}=h_2+\frac{v_2^2}{2}[/tex]
where, h is the enthalpy
on rearranging we get
[tex]\frac{v_2^2-v_1^2}{2}=h_1-h_2[/tex]
also
h = Cp × T
thus,
[tex]\frac{v_2^2-v_1^2}{2}=C_p(T_1-T_2)[/tex]
on substituting the values, we get
[tex]\frac{307^2-40^2}{2}=1.005\times 10^3(423-T_2)[/tex]
or
T₂ = 376.905 K
b) The isentropic nozzle efficiency is given as:
[tex]\eta_n=\frac{V_2^2/2}{V_1^2/2}=\frac{Actual\ expansion}{Ideal\ Expansion}[/tex]
for an isentropic process,we have the isentropic relation as:
[tex]\frac{T_{2s}}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}}[/tex]
on substituting the values and solving, we get
[tex]T_{2s}[/tex] = 374.65 K
by applying the energy equation, we get the ideal exit velocity as:
[tex]V_{2s}=\sqrt{2C_p(T_2-T_1)+V_1^2}[/tex]
on substituting the values, we get
[tex]V_{2s}=\sqrt{2\times1.005\times10^3(423-374.65)+40^2}[/tex]
or
[tex]V_{2s}=314.29\ m/s[/tex]
thus,
substituting in the formula for efficiency, we get
[tex]\eta_n=\frac{307^2/2}{314.29^2/2}[/tex]
or
[tex]\eta_n=0.954[/tex] = 95.4%
