Consider a two-stage cascade refrigeration system operating between the pressure limits of 1.2 MPa and 200 kPa with refrigerant-134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where the pressure in the upper and lower cycles are 0.4 and 0.5 MPa, respectively. In both cycles, the refrigerant is a saturated liquid at the condenser exit and a saturated vapor at the compressor inlet, and the isentropic efficiency of the compressor is 80 percent. If the mass flow rate of the refrigerant through the lower cycle is 0.15 kg/s, determine...
a.) The mass flow rate of the refrigerant through the upper cycle
b.) The rate of heat removal from the refrigerated space
c.) The COP of this refrigerator
9 comments:
Im missing some little peice of info to find mdot2. I have the equation that ballances the two flow rates and the qc upper and qh lower. problem is i have 2 unknowns- m2 and h5 which is part of the qc2 equation. how do i get h5? or is there another equation with these same unknowns?
im getting lost at how to use the isentropic efficiency to find the enthalpy values after the stream exits the compressor.
My notes about a cascade refrigeration cycle state that the mass flow rates for the two cycles must be the same! Why isn't this so?
Graham 4:36 PM
You are correct, apply the 1st Law to the evaporator to determine Qc=Q12 (using the diagram from the TCD Notebook).
The upper cycle does not exchange heat with the cold reservoir. So, you are correct to not use any terms from it to calculate Qc.
Good work.
AdamAnt 5:17
The key to determining the mass flow rate in the upper cycle is the fact that all of the heat that is rejected by the lower cycle in the heat exchanger enters the upper cycle. So, once you have all the H^ values, you can apply the 1st Law to the heat exchanger and the only unknown will be Mdot for the upper cycle.
If you numbered the streams the way I did in the TCD Notebook, then stream 5 leaves the expansion valve in the upper cycle. You can show that the expansion valve is isenthalpic, so H5 = H8.
You should be good to go now.
Anon 11:25 PM
It is like you have 2 different compressors that you are looking at. One is the actual one and one is isentropic. Both are adiabatic. Began by analyzing the adiabatic compressor to determine the H^ at outlet, H^3s, where the "s" in the subscript reminds you that it is for the isentropic compressor. S3s = S2.
The isentropic efficiency is the ratio of the work required by the isentropic compressor to the work required by the actual compressor. So, you can use the isentropic efficiency to calculate the work required by the actual compressor. Then, you can use the 1st Law to determinine the H^ of the actual compressor effluent, H^3.
Anon 1:12 AM
The mass flow rates in cascade refrigeration cycles do not need to be the same.
Q out of the lower cycle must be equal to Q into the upper cycle. But this does not mean that the mass flow rates must be the same. The heat of vaporization or heat capacity of the fluids in the different cycles may be different...even if it is the same working fluid. Both heat of vaporization and Cp are different at different temperatures.
Im having a hard time finding the net work for this problem. Do I take into account the valves? Im just using the work of the compressors now and it is not working, I tried using vdP for them but that doesn't seem to help either. Any suggestions?
Claire 10:35 PM
The only places where work crosses the system boundary is at the compressors. So, you should apply the 1st Law to each compressor and add up the shaft work. I cannot tell what is going wrong for you. If you can give me some more info, I may be able to help.
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