Respuesta :
Answer:
a. The inlet temperature is approximately 305.232 K
b. The inlet pressure is approximately 452.0108 kPa
c. The area ratio between the inlet and exit is approximately 2.2509
Explanation:
a. From the energy equation related to the question, we have;
[tex]C_p \cdot (T_i - T_e) = \dfrac{1}{2} \cdot \left (v_e^2 - v_1^2 \right)[/tex]
Where;
[tex]C_p[/tex] = The specific heat capacity for helium = 5.913 kJ/(kg·K)
[tex]T_i[/tex] = The inlet temperature
[tex]T_e[/tex] = The exit temperature = 300 K
[tex]v_i[/tex] = The inlet velocity = 25 m/s
[tex]v_e[/tex] = The exit velocity = 250 m/s
Therefore, we have;
[tex]T_i= \dfrac{ \dfrac{1}{2} \cdot \left (v_e^2 - v_1^2 \right)}{C_p} + T_e = \dfrac{ \dfrac{1}{2} \times \left (250^2 - 25^2 \right)}{5.913 \times 1000} + 300 \approx 305.232[/tex]
The inlet temperature = [tex]T_i[/tex] ≈ 305.232 K
b. From the following equation for the critical pressure, for helium, we have;
[tex]\dfrac{P_c}{P_i} = \left (\dfrac{2}{n + 1} \right ) ^{\dfrac{n}{n - 1} } = 0.487[/tex]
Where;
[tex]P_c[/tex] = The critical pressure = 2.26 atm for helim
[tex]P_i[/tex] = The inlet pressure
n = The polytropic constant
We have;
[tex]\dfrac{2.26 \ atm}{P_i} = 0.487[/tex]
[tex]\therefore P_i = \dfrac{2.26 \ atm}{0.487} \approx 4.641 \ atm[/tex]
The inlet pressure, [tex]P_i[/tex] ≈ 4.641 atm ≈ 452.0108 kPa
c. The inlet to exit pressure ratio is given as follows;
[tex]P_e = \dfrac{A_i \times T_e \times v_i}{A_e \times T_i \times v_e} \times P_i[/tex]
Therefore, we have;
[tex]\dfrac{A_i}{A_e} = \dfrac{P_e \times T_i \times v_e}{ T_e \times v_i \times P_i} = \dfrac{100 \times 305.232 \times 250}{ 300 \times 25 \times 452.0108} = 2.2509[/tex]
The area ratio between the inlet and exit, [tex]A_i/A_e[/tex] ≈ 2.2509.
