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DEVICE IS REFRIGERATION 12 23 At 3 4 AT 2 AT 3 : AT 4: MASS ADIABATIC COMPRESSION COOL AT P1 = P2 = ADIABATIC EXPANSION HEATING AT CONSTANT PRESSURE A BAR P2 P3 P1 = P4 ASK: A) B) T2 c) HEAT 3.2 BAR M = 116g CONSTANT C SPECIFIC HEAT RATIO OF WORKING Y = 1.5 WORKI HEAT CAPACITE ф ти UNDERGOES REVERSE BRAYTON CYCLE PRESSURE T1 = -5.15e EXCHANGE S T5 = 17.85°C SUBSTANCE is IDEALLY SUBSTANCE cu & Cp IN CELCIUS IN J/g Q25 $ Q 41

DEVICE IS REFRIGERATION 12 23 At 3 4 AT 2 AT 3 : AT 4: MASS ADIABATIC COMPRESSION COOL AT P1 = P2 = ADIABATIC EXPANSION HEATING AT CONSTANT PRESSURE A BAR P2 P3 P1 = P4 ASK: A) B) T2 c) HEAT 3.2 BAR M = 116g CONSTANT C SPECIFIC HEAT RATIO OF WORKING Y = 1.5 WORKI HEAT CAPACITE ф ти UNDERGOES REVERSE BRAYTON CYCLE PRESSURE T1 = -5.15e EXCHANGE S T5 = 17.85°C SUBSTANCE is IDEALLY SUBSTANCE cu & Cp IN CELCIUS IN J/g Q25 $ Q 41

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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DEVICE is REFRIGERATION & 12 23 3 4 4 → 1 At 1. AT 2: AT 3 AT 4 : : ADIABATIC COMPRESSION COOL AT CONSTANT PRESSURE P1 = ADIABATIC EXPANSION HEATING AT CONSTANT PRESSURE P2 = P1 = A BAR P2 P3 3.2 BAR ASK: A) B) T2 c) HEAT 74 T1 HEAT CAPACITIE C 4 тн E SPECIFIC HEAT RATIO OF WORKING Y = 1.5 T3 = EXCHANGE S UNDERGOES REVERSE BRAYTON CYCLE -5.15e MABS OF WORKI G SUBSTANCE M = 116g 17.85°C SUBSTANCE cu & Cp iN CELCIUS Q23 IN s/g & Q41 is IDEA
Transcribed Image Text:DEVICE is REFRIGERATION & 12 23 3 4 4 → 1 At 1. AT 2: AT 3 AT 4 : : ADIABATIC COMPRESSION COOL AT CONSTANT PRESSURE P1 = ADIABATIC EXPANSION HEATING AT CONSTANT PRESSURE P2 = P1 = A BAR P2 P3 3.2 BAR ASK: A) B) T2 c) HEAT 74 T1 HEAT CAPACITIE C 4 тн E SPECIFIC HEAT RATIO OF WORKING Y = 1.5 T3 = EXCHANGE S UNDERGOES REVERSE BRAYTON CYCLE -5.15e MABS OF WORKI G SUBSTANCE M = 116g 17.85°C SUBSTANCE cu & Cp iN CELCIUS Q23 IN s/g & Q41 is IDEA
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