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Question #196190

•   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

1
2021-05-24T05:03:02-0400

"\\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 \np_c-\\log 989"

"\\ln log x=\\ln 4.0512"

"P_C=11251kP_a"

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