What is the temperature at point 1? What is the volume at point 3? What then is
ID: 2216697 • Letter: W
Question
What is the temperature at point 1?What is the volume at point 3?
What then is the temperature at point 3?
What is the volume at point 2
What is the heat transfer from steps 2 -> 3?
What is the heat Transfer from steps 4->1?
Because these two were the only transfer of heat they give the total heat flow. What then was the total work input required for this refrigerator?
What is the refrigerators COP?
If the power input to operate this refrigerator is 60 cycles per second. How much power do I need to run it? his quiz is all about this Brayton cycle The little black arrows show the direction of a refrigerator While the numbering of the points are for an engine. Sorry about the confussion, just remember a refrigerator is the opposite direction of an engine. Also notice that only the position 1 and 4 have a known volume you will solve for the other volumes, and temperatures What are the number of moles for this problem? What is the temperature at point 1? What is the volume at point 3? What then is the temperature at point 3? What is the volume at point 2 What is the heat transfer from steps 2 - > 3? What is the heat Transfer from steps 4- >1? Because these two were the only transfer of heat they give the total heat flow. What then was the total work input required for this refrigerator? What is the refrigerators COP? If the power input to operate this refrigerator is 60 cycles per second. How much power do I need to run it?
Explanation / Answer
consider point 4 we have
PV= nRT
150 kPa*100cm3= n 8.31*250
n=0.0072 moles
1-->4 it is isobaric process
V4/T4 =V1/T1
T1=200k
4-->3
P4 (V4)^()= P3(V3)^()
considering mono atomic we get V3 =38.15 cm^(3) [according to figure it slightly differ i dont know why but i am sure about the procedur that i carried out]
at 3 we have PV = nRT
T3=475.1 K
1-->2 adibatic process we have
P1 V1^ = P2 V2^
V2=30.52 cm3,T2=382.57K
Qh=n Cp (T2-T3)= 2.5R*(382.57-475.1)*n=-13.84
Qc= nCp (T4-T1)=7.479
total input work in full cycle w = -Q=6.361
cop = mod (Qh)/w=13.84/6.361=2.19
work in one cycle =6.31
power = w*no of cycles per sec =378.6 W
may be there are some mistakes in sign or value may crept in but the procedure is same
so plz rate and encourage
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