5 - Prob 4.12 9/18/04 PROBLEM 4.12 PROBLEM STATEMENT: In an...

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Prob 4.12 9/18/04 PROBLEM 4.12 PROBLEM STATEMENT: In an air conditioning system, saturated liquid refrigerant R-134a at T=40°C flows through an insulated expansion valve, and reaches a final pressure of 320 kPa. What is the quality of the refrigerant after the expansion? DIAGRAM DEFINING SYSTEM AND PROCESS: 2 1 Expansion valve State 1: T 1 = 40°C x 1 = 0 State 2: P 2 = 320 kPa GIVEN: R-134a through an expansion valve T 1 = 40°C, x 1 = 0, P 2 = 320 kPa FIND: Exit quality, x 2 . ASSUMPTIONS: Adiabatic process, NKEPE, SFSS GOVERNING RELATIONS: 1. Mass balance, m= 12 m && 2. Energy balance,   =     ∑∑ & & 2 2 OUT IN IN IN IN IN OUT OUT OUT OUT cc c c gg Q+ m h+ + z -W m h + + z 0 2g g 2g g v v QUANTITATIVE SOLUTION: From the energy balance equation: & IN Q & 2 IN IN IN c +mh+ 2g v IN c g +z g     & OUT -W & 2 OUT OUT OUT c mh+ 2g v OUT c g g = 11 2 2 0 m h = m h Applying the mass balance equation, h 1 = h 2 . From Table 14s: 1L 2 h = h (T = 40°C) =106.19 kJ/kg =h From Table 15s, at State 2, the refrigerant exists as a saturated mixture. The quality x 2 can be calculated as follows: 2L , 2 2 LV,2 h-h 106.19 -53.31 x= = h1 9 5 . 3 =0.27=27% 5 DISCUSSION OF RESULTS: The process of dropping the pressure of a fluid through a restriction (valve, orifice, capillary tube) is called “throttling”. This problem shows that adiabatic throttling causes a saturated liquid to “flash” into a liquid-vapor mixture, an important part of refrigeration cycles.
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Prob 4.12 9/18/04 PROBLEM 4.14 PROBLEM STATEMENT: An adiabatic condenser operates as shown below. If the process is steady-state steady flow, determine the mass flow rate of cooling water required (kg/s). DIAGRAM DEFINING SYSTEM AND PROCESS: Q & & STM m= 8 0 k g / s 4 3 2 1 Cooling water Steam P 4 = P 3 T 4 = 60°C P 3 = 60 kPa x 3 = 92% P 2 = P 1 T 2 = 85°C P 1 = 0.1013 MPa T 1 = 50°C GIVEN: Water, m , adiabatic and SSSF & STM = 80 kg/s State 1: P 1 = 0.1013 MPa, T 1 = 50°C State 2: P 2 = P 1 , T 2 = 85°C State 3: P 3 = 60 kPa, x 3 = 92% State 4: P 4 = P 3, T 4 = 60°C FIND: m ( & W k g / s ) . ASSUMPTIONS: Externally adiabatic, no work, NKEPE, SSSF, treat liquid water as incompressible liquid. GOVERNING RELATIONS: 3. Mass balance, in out m= m && 4. Energy balance, 2 IN IN IN IN c Q+ m h+ 2g & & v     2 OUT IN OUT OUT OUT cc g +z- Q m h+ g2 g & & v = OUT c g +z g 0 & QUANTITATIVE SOLUTION: Conservation of mass: ∑∑ & & & &&&&&& IN OUT 1 3 2 4 12W34S T M m= m m+m=m+m m = m = m , m = m = m Conservation of energy: { } { } && & & IN IN OUT OUT 11 3 3 2 2 4 4 W2 1 S T M3 4 mh= m h mh +mh =mh +mh m( h-h)=m ( h-h)
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Prob 4.12 9/18/04 The specific enthalpy values at each state can be determined from the tables for water.
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This note was uploaded on 03/19/2008 for the course ME 326 taught by Professor Schmidt during the Spring '07 term at University of Texas.

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5 - Prob 4.12 9/18/04 PROBLEM 4.12 PROBLEM STATEMENT: In an...

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