5 - Prob 4.12 PROBLEM 4.12 PROBLEM STATEMENT In an air...

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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 = 1 2 m & & 2. Energy balance, = & & & & 2 2 OUT IN IN IN IN IN OUT OUT OUT OUT c c c c g g 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 + m h + 2g v IN c g + z g & OUT - W & 2 OUT OUT OUT c m h + 2g v OUT c g + z g = & & 1 1 2 2 0 m h = m h Applying the mass balance equation, h 1 = h 2 . From Table 14s: 1 L 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: 2 L,2 2 LV,2 h -h 106.19 - 53.31 x = = h 195.3 = 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 = 80 kg/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 kg/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 c c g + z - Q m h + g 2g & & v = OUT c g + z g 0 & QUANTITATIVE SOLUTION: Conservation of mass: & & & & & & & & & & & IN OUT 1 3 2 4 1 2 W 3 4 STM m = m m +m = m +m m = m = m , m = m = m Conservation of energy: { } { } & & & & & & & & IN IN OUT OUT 1 1 3 3 2 2 4 4 W 2 1 STM 3 4 m h = m h m h +m h = m h +m h 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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