Answer to Question #115775 in Mechanical Engineering for Smith

Question #115775
Air initially at 50 psia and 140 0F undergoes a polytropic process such that the temperature becomes 40 0F. The polytropic exponent for the process is equal to 1.3. Find the final pressure and specific volume.
1
Expert's answer
2020-05-19T09:13:43-0400

"P_1=50 psia, T_1= 140 ^oF,P_2=?, T_2= 40 ^oF" , n= 1.3

we know that for polytropic process as


"\\frac{T_2}{T_1}= (\\frac{P_2}{P_1})^{\\frac{n-1}{n}}"


"\\frac{40}{140}= (\\frac{P_2}{50})^{\\frac{1.3-1}{1.3}}"


0.2857 = "(\\frac{P_2}{50})^{0.231}"


taking log both sides


log(0.2857) = 0.231 log "(\\frac{P_2}{50})"


-2.355= log "(\\frac{P_2}{50})"


"0.00441\\times 50 = P_2"


"P_2=0.2205 psia"





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