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Ex 10-1 Refrigerant R-12 enters the compressor of an ideal vapor-compression refrigeration system as saturated vapor at -10oC with a volumetric flow rate of 1 m3/min. The refrigerant leaves the condenser at 35oC, 10 bar. Determine (a) the compressor power, in kW, (b) the refrigerating capacity in tons, and (c) the coefficient of performance. (d) What-if scenario: How would the answers change if the R-12 was replaced with more environmentally friendly R-134a?  |
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Solution Detailed analysis of cycles (steam power, gas power, and refrigeration cycles) executed by a fluid flowing through open devices connected back to back in a loop, is carried out by the open cycle daemons located in the specific branch of the open steady daemons. Launch the refrigeration cycle daemon located at TEST. Daemons. Systems. Open. Steady. Specific. RefriCycles. PC Model page.
Let us set up the vapor compression cycle, shown in the schematic above, as follows : Device-A (compressor): isentropic compression from State-1 to State-2 ; Device-B (condenser): constant pressure heat rejection from State-2 to State-3 ; Device-C (expansion valve): isenthalpic expansion from State-3 to State-4 ; Device-D (evaporator): constant pressure heat addition from State-4 to State-1. Select R-12 as the refrigerant in the state panel and evaluate the principle states one after another, entering the known properties and then pressing the Enter key or the Calculate button. State-1: T1, x1 (=1), and Voldot1 (=1 m3/min) are known. State-2: p1, s2 (=s1), and mdot2 (=mdot1) are known. State-3: p3 (=p2), T3, and mdot3 (=mdot1) are known. State-4: p4 (=p1), h4 (=h3), and mdot4 (=mdot1) are known. After the states are calculated, verify the states on a p-h (as shown below) or a T-s diagram.
On the device panel, analyze each of the four devices. Device-A (Compressor): Select state-1 and 2 as the i1 and e1 states, enter Qdot=0, and Calculate Wdot_ext as -5.91 kW. Device-B (Condenser): Select state-2 and 3 as the i1 and e1 states, enter Wdot_ext=0, and Calculate Qdot as -30.59 kW. Device-C (Valve): Select state-3 and 4 as the i1 and e1 states, enter Qdot=0, and Calculate Wdot_ext as 0 kW. Device-d (Evaporator): Select state-4 and 1 as the i1 and e1 states, enter Wdot_ext=0, and Calculate Qdot as 24.68 kW. Change unit of Qdot by selecting ton from the unit menu, producing the cooling capacity in ton as Qdot=7.02 ton. In the cycle panel, the COP is calculated as 4.17 (see figure below).
For the what-if study, go back to the states panel, select R-134a from the working fluid menu, press Enter and Super-Calculate. The new answers can be found in the cycle panel as shown below.
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Ex 10-2 Consider a two-stage R-12 refrigeration system operating between 0.15 MPa and 1 MPa. The refrigerant leaves the condenser as saturated liquid and is throttled to a flash chamber operating at 0.4 MPa. The vapor from the flash chamber is mixed with the refrigerant leaving the low-pressure compressor and the mixture is compressed by the high-pressure compressor to the condenser pressure. The liquid in the flash chamber is throttled to the evaporator pressure where the cooling load is handled through evaporation. Assume the refrigerant leaves the evaporator and mixer as saturated vapor. (a) Determine the COP. (b) What-if scenario: How would the COP change if the intermediate pressure was increased to 0.6 MPa? |
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Solution Detailed analysis of cycles (steam power, gas power, and refrigeration cycles) executed by a fluid, passing through open devices connected back to back in a loop, is carried out by the open cycle daemons located in the specific branch of the open steady daemons. Launch the refrigeration cycle daemon located
at TEST. Daemons.
Systems. Open. Steady. Specific. RefriCycles. PhaseChange page.
Let us set up the cycle as follows : Device-A: isentropic compression from State-9 to State-4 ; Device-B : constant pressure heat rejection from State-4 to State-5 ; Device-C : isenthalpic expansion from State-5 to State-6 ; Device-D : constant pressure phase separation from State-6 to State-3 ; Device-E : isenthalpic expansion from State-7 to State-8 ; Device-F : constant pressure heat addition from State-8 to State-1 ; Device-G : isentropic compression from State-1 State-2 ; Device-H State-3 into State-9 . The mass flow rates being unknowns, we will take 1 kg/s through the top loop as the basis for this analysis, that is, mdot9=1 kg/s; therefore, mdot3=x6*mdot9 and mdot7=(1-x6)*mdot9 (flash chamber separates saturated vapor from saturated liquid). In evaluating the states, we can follow two approaches. Incorporate known information about devices into state calculations, or rely on device analysis to post the missing state properties for their complete evaluation. For instance, in an isentropic device (say, Device-A) operating between State-1 and -2, we can enter s2 as '=s1'. Alternatively, we could enter Qdot=0 and Sdot_gen=0 (i.e. adiabatic and reversible) in the analysis of Device-A. The Super-Calculate operation, in that case, would solve the entropy-balance equation for Device-A, conclude that s2=s1, substitute s1 for s2 in State-2, and recalculate State-2. For complex cycles such as this one, we will follow the first approach. State-1-8: Enter the known values or relations as described in the
TEST-codes and Calculate the states fully or partially. |
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| Fig. 2.1 Image of Device-D. Solution of the energy equation yields mdot3 which is posted back into State-3. |
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Device-A through H: For each device select a letter, load the anchor states (see the TEST-codes above), enter the known device variable (Qdot or Wdot_ext) and Calculate. Use
Super-Calculate followed by a Super-Iterate to
update all the States and Devices. The COP is calculated as COP=349%. The mass flow rate for the bottom cycle is calculated as
mdot1=0.77 kg/s. |
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For the parametric study, go to the state panel and change p2 to 0.6 MPa. Calculate the State, Super-Calculate and Super-Iterate to update all calculations. The new COP is calculated as 332%. |
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| Fig. 2.2 Image of the cycle panel. Cycle variables are automatically calculated once the cycle is complete. |
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Ex 10-3 In a gas refrigeration system air enters the compressor at 10 o C and 50 kPa and the turbine at 50oC and 250 kPa. The mass flow rate is 0.08 kg/s. Assuming variable specific heat, determine (a) the rate of cooling, (b) the net power input and (c) the COP. (d) What-if scenario: How would the answers change if the working fluid was helium instead? |
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Solution Detailed analysis of cycles (steam power, gas power, and refrigeration cycles) executed by a fluid, passing through open devices connected back to back in a loop, is carried out by the open cycle daemons located in the specific branch of the open steady daemons. Launch the open cycle refrigeration daemon
located at the page TEST. Daemons. Systems. Open. Steady. Specific.RefriCycles.
IdealGas .
Let us set up the cycle as follows: Device-A: compression from State-1 to State-2 ; Device-B: constant pressure heat rejection from State-2 to State-3 ; Device-C : isentropic expansion from State-3 to State-4 ; Device-D : constant pressure heat addition from State-4 to State-1 . State-1-4: Enter known values or relations and Calculate each state fully or partially. |
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In the device panel, work on the four devices.
Device-A through D: As described in the TEST-codes, identify each device by a unique letter, load the anchor states, enter the known device variables (Qdot and Wdot_ext) and Calculate. Use
Super-Calculate to produce the TEST-codes
and the detailed output. In cycle panel no further work is necessary. The COP is calculated as
1.72. Now change the gas to helium in the state panel and
Super-Calculate. The new value for COP is 1.11. |
| Copyright 1998-: Subrata Bhattacharjee |
