Answer to Question #196190 in Chemical Engineering for Lebohang Radebe

Question

•   Question 2 of 4  

Question assume ideal and optimal operating conditions for the cycle and the re-heater. Air enters the cycle at 300K and 100kPa and it leaves the combustor at 975K and 1089kPa.

 

Calculate the (a) pressure (kPa) of the air leaving the re-heater

Answer 1

$\frac{T_b}{T_a}=(\frac{P_b}{P_a})^{\frac{\gamma-1}{\gamma}}$

$\frac{975}{300}=(\frac{1089}{100})^{\frac{\gamma-1}{\gamma}}$

finding gamma

$\frac{975}{300}\cdot \:300=\left(\frac{1089}{100}\right)^{\frac{\gamma-1}{\gamma}}\cdot \:300$

$\gamma=\frac{\ln \left(\frac{1089}{100}\right)}{\ln \left(\frac{1089}{325}\right)}=1.97474$

$\frac{975}{300}=(\frac{P_C}{1089-100})^{\frac{\gamma-1}{\gamma}}$

$\frac{975}{300}=(\frac{P_C}{1089-100})^{\frac{1.97474-1}{1.97474}}$

$\log\frac{975}{300}=(\frac{0.97474}{1.97474})\log\frac{P_c}{989}$

$1.056=\log p_c-\log 989$

$\ln log x=\ln 4.0512$

$P_C=11251kP_a$