Question
I have the answer but don't know how to get to it for part B. I need to see the work done and the answer in the correct units for part B. I will provide answer in parentheses after the question.
Commercial aircraft fly near 200mb where typically the outside temperature is -60C
(a) Calculate the temperature of air if compressed adiabatically to the cabin pressure of 1000mb. (337.5K) - I DON'T NEED THIS ONE SOLVED...JUST FOR F.Y.I.
(b)How much specific heat must be added or removed (isobarically) to maintain the cabin at 25C? (ANSWER: 9.5cal/g has to be removed)
Explanation / Answer
Given initial temperature of aircraft T1 = -60 oC = 273 - 60 = 213 K initial pressure P 1 = 200 mb final pressure P 2 = 1000 mb here the air is compresed adiabatiacally the relation between pressure and temperature is P1 1- T 1 = P2 1- T 2 ...........(1) for air is 1.4 from equation we can written as ( P 1 / P 2 ) 1- = ( T 2 / T1 ) ( 200 m b / 1000 mb ) 1-1.4 = ( T 2 / T1 ) 1.4 0.2-0.4 = ( T 2 / T1 ) 1.4 1.9036 = ( T 2 / T1 ) 1.4 taking log on both sides
0.6437 = 1.4 ln ( T2 / T1 ) 0.4598 = ln ( T2/ T1 ) 1.5838 = T2 / T1 T2 = 213*
1.5838196087665790445526433827527 = 337.35 K b) initial volume of gas V1 = n R T1 / P1 = 8.31 * 210 / 200 mb 1 bar = 10 5 Pa 1 m b = 10 2 Pa there fore V 1 = 0.0872 m 3 final volume V2 = n R T2 / P2 = 8.31 * 337.35 / 1000 mb = 0.028 m 3 . heat loss Q = n C T ( 1/ -1 ) ( P2 V2 - P1 V1 ) = n C T ( 1/ 1.4- 1 ) ( 0.028 * 105 - 0.0872 *2*104 ) = n c T here T = 25 degrees = 1.4 c = .............. J / kg K 1 Joule = 0.239 cal 0.4598 = ln ( T2/ T1 ) 1.5838 = T2 / T1 T2 = 213*
1.5838196087665790445526433827527 = 337.35 K b) initial volume of gas V1 = n R T1 / P1 = 8.31 * 210 / 200 mb 1 bar = 10 5 Pa 1 m b = 10 2 Pa there fore V 1 = 0.0872 m 3 final volume V2 = n R T2 / P2 = 8.31 * 337.35 / 1000 mb = 0.028 m 3 . heat loss Q = n C T ( 1/ -1 ) ( P2 V2 - P1 V1 ) = n C T ( 1/ 1.4- 1 ) ( 0.028 * 105 - 0.0872 *2*104 ) = n c T here T = 25 degrees = 1.4 c = .............. J / kg K 1 Joule = 0.239 cal