SP#3. A cylinder fitted with a frictionless piston contains 10 kg of superheated
ID: 700855 • Letter: S
Question
SP#3. A cylinder fitted with a frictionless piston contains 10 kg of superheated refrigerant R-134a vapor at 150°C and a pressure of 3000 +/- 23 kPa. The cylinder is now cooled at constant pressure until the volume of the R-134a in the cylinder is 0.08 m3. Show a schematic for the process and estimate the temperature of the R-134a after cooling. Write the properties for each state of the process that you need to determine the change in internal energy and work associated with the cooling Calculate the work during the process and indicate if it is "into" or "out of" the system for this process? What is the heat transfer for this process? a. b. c. d.Explanation / Answer
a) We will make use of R-134a property table to calculate all the answers
mass = 10 Kg
State point 1
T = 150 oC
P = 3000 KPa 3 MPa
from property table
V1 = 0.0094 m3/Kg
U1 =342.9 KJ/Kg
H1 = 371.1 KJ/Kg
Now for state 2
Volume = 0.08/10 = 0.008 m3/Kg
the process as specified is a constant process
P = 3 MPa
now we will look up in property table and find
T and other properties at given P = 3 MPa, v = 0.008 m3/Kg
From property table
T = 120 oC
U2 = 309.4 KJ/Kg
H2 =333.2 KJ/Kg
For showing in a diagram one may simply plot the state point in a PT or PV diagram of the initial and the final state point
b)
Hence once we find H and U from property table we can easily find the value of internal table enthalpy or flow work
U1 =342.9 KJ/Kg
H1 = 371.1 KJ/Kg
U2 = 309.4 KJ/Kg
H2 =333.2 KJ/Kg
C)We know that
H = U + PV
Flow work is PV work
Flow work = ( H2-U2) - (H1-U1) = (333.2 - 309.4)- ( 371.1- 342.9) =-4.4 KJ/Kg = -4.4*10 = -44 KJ
Negative sign indicated here that the energy is flowing "out of" of the system
d) Heat transfer = (H2 m*(Hout-Hin) = 10*(333.2-371.1) =-379 KJ
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