Chapter 15

15. REFRIGERATION 15.1

An ideal single-stage vapor compression refrigeration cycle uses R-22 as the working fluid. The condensing temperature is 110 F (43 C) and the evaporating temperature is 40 F (4.5 C). The system produces 10 tons (35.2 kW) of cooling effect. Determine the (a) coefficient of performance, (b) refrigerating efficiency, (c) hp/ton (kW input per kW of refrigeration), (d) mass flow rate of the displacement, (e) theoretical input to the compressor in hp (kW), and (f) theoretical piston displacement of the compressor in cfm (m3/s).

Solution:

English Units From Chart 4 and Table A-3a At 110 F, P1 = P4 = 241.14 psia At 40 F, P2 = P3 = 83.28 psia i1 = i2 = 42.717 Btu lbm i3 = 108.191 Btu lbm , s3 = s4 = 0.2204 Btu (lbm − F ) , v3 = 0.6561 ft 3 lbm i4 = 119.74 Btu lbm

(a) Coefficient of Performance COP =

i3 − i2 108.191 − 42.717 = = 5.67 i4 − i3 119.74 − 108.191

(b) Refrigerating Efficiency COP COPc Te 40 + 459.67 COPc = = = 7.138 Tc − Te 110 − 40 5.67 ηR = = 0.794 7.138

ηR =

(c) hp/ton

15. REFRIGERATION

Equation 15-3 hp 4.72 4.72 = = = 0.8325 hp ton ton COP 5.67

(d) mass flow rate of refrigerant q&e = m& (i3 − i2 ) q&e 10(12,000 ) m& = = = 1833 lbm hr (i3 − i2 ) 108.191 − 42.717

(e) theoretical input to the compressor in hp hp = (0.8325 hp ton )(10 ton ) = 8.325 hp

(f) theoretical piston displacement Equation 15-10 m& v η v = 3 = 1 .0 PD (1833)(0.6561) = 20.044 cfm PD = m& v3 = 60 SI units From Chart 4 and Table A-3b At 43 C, P1 = P4 = 1.65 MPa At 4.5 C, P2 = P3 = 575.4 kPa i1 = i2 = 253.72 kJ kg i3 = 406.47 kJ kg , s3 = s4 = 1.7436 kJ (kg − K ) , v3 = 0.04091 m3 kg i4 = 432.63 kJ kg

(a) Coefficient of Performance COP =

i3 − i2 406.47 − 253.72 = = 5.84 i4 − i3 432.63 − 406.47

(b) Refrigerating Efficiency COP COPc Te 4.5 + 273.15 COPc = = = 7.212 Tc − Te 43 − 4.5 5.84 ηR = = 0.810 7.212

ηR =

15. REFRIGERATION

(c) hp/ton

Equation 15-4 kW 3.52 3.52 = = = 0.6027 kW ton ton COP 5.84

(d) mass flow rate of refrigerant q&e = m& (i3 − i2 ) q&e 35.2 m& = = = 0.2305 kg s (i3 − i2 ) 406.47 − 253.72

(e) theoretical input to the compressor in hp kW = (0.6027 kW ton )(10 ton ) = 6.027 kW

(f) theoretical piston displacement Equation 15-10 m& v η v = 3 = 1 .0 PD PD = m& v3 = (0.2305)(0.04091) = 0.0094 m 3 s

15.2

A vapor compression refrigeration cycle uses R-22 and follows the theoretical single-stage cycle. The condensing temperature is 48 C, and the evaporating temperature is –18 C. The power input to the cycle is 2.5 kW, and the mass flow rate of refrigerant is 0.05 kg/s. Determine (a) the heat rejected from the condenser, (b) the coefficient of performance, (c) the enthalpy at the compressor exit, and (d) the refrigerating efficiency. Solution:

From Chart 4 and Table A-3b

15. REFRIGERATION

At 48 C, P1 = P4 = 1.8555 MPa At -18 C, P2 = P3 = 264.77 kPa i1 = i2 = 260.51 kJ kg i3 = 397.81 kJ kg , s3 = s4 = 1.7787 kJ (kg − K ) , v3 = 0.08615 m3 kg i4 = 447.81 kJ kg

(a) heat rejected from the condenser m& = 0.05 kg s q&c = m& (i4 − i1 ) = 0.05(447.81 − 260.51) = 9.365 kW

(b) coefficient of performance COP =

i3 − i2 397.81 − 260.51 = = 2.746 i4 − i3 447.81 − 397.81

(c) enthalpy at the compressor exit i4 = 447.81 kJ kg

(d) refrigerating efficiency COP COPc Te − 18 + 273.15 COPc = = = 3.866 Tc − Te 48 − (− 18) 2.746 ηR = = 0.710 3.866

ηR =

15.3

An R-134a system is arranged as shown in Fig. 15-34. Compression is isentropic. Assume frictionless flow. Find the system hp/ton.

Figure 15-34 Schematic for Problem 15-3.

15. REFRIGERATION

Solution:

From Chart 3 and Table A-2a At 100 F, P1 = P4 = 138.83 psia At -10 F, P2 = P3 = 16.626 psia i1′ = 44.943 Btu lbm i3′ = 101.542 Btu lbm At 100 F i1 = i2 = 37.978 Btu lbm i1′ − i1 = i3 − i3′ 44.943 − 37.978 = i3 − 101.542 i3 = 108.507 Btu lbm at 16.626 psia s3 = s4 = 0.2612 Btu (lbm − F ) , v3 = 3.5842 ft 3 lbm i4 = 134.22 Btu lbm

i3 − i2 108.507 − 37.978 = = 2.743 i4 − i3 134.22 − 108.507 Equation 15-3 hp 4.72 4.72 = = = 1.721 hp ton ton COP 2.743 COP =

15.4

Consider a reciprocating compressor operating with R-134a. Refrigerant enters the cylinder at 20 psia (138 kPa) and 20 F (-7 C), but leaves the evaporator saturated at 0.5 F (-18 C). The vapor is discharged from the cylinder at 180 psia (1.24 MPa). Compute the volumetric efficiency for (a) a clearance factor of 0.03; (b) a clearance factor of 0.15; (c) compare the mass flow rates for parts (a) and (b); and (d) compare the power input to the compressor in parts (a) and (b).

15. REFRIGERATION

Solution:

English Units At 3, 0.5 F saturated, Table A-2a, R134a v3 = 2.1362 ft 3 lbm At b, 20 psia, 20 F, Chart 3 vb = 2.413 ft 3 lbm Pb = 20 psia At c, 180 psia Pc = 180 psia Equation 15-8 1   Pc  n  v3 ηv = 1 + C − C      Pb   vb  n = 1.30 for R-134a

(a) C = 0.03 1   180  1.30  2.1362  ηv = 1 + 0.03 − 0.03 = 0.7679   20    2.413 (b) C = 0.15 1   180  1.30  2.1362  ηv = 1 + 0.15 − 0.15 = 0.2983   20    2.413 (c) Equation 15-10 m& v3 PD Assume PD same as Example 15-4

ηv =

15. REFRIGERATION

For Part (a) m& =

η v PD v3

=

(0.7679)(10)(1725) = 3.59 lbm (2.1362)(1728)

min

For Part (b) m& =

η v PD v3

=

(0.2983)(10)(1725) = 1.394 lbm (2.1362)(1728)

min

Mass flow rate of part (a) is larger than part (b) (d) Equation 15-12 and Equation 15-13, η m = 1.0 ( n −1)   n   P n c  & W = m& Pv   − 1  (n − 1) b b  Pb    For Part (a)

 180 (1.30−1)1.30  1 . 30 W& = (3.59) (180)(144)(2.413)  − 1 (1.30 − 1)  20     W& = 642,553 ft − lbf min = 19.47 hp For Part (b)  180 (1.30−1)1.30  1.30 & W = (1.394) (180)(144)(2.413)  − 1 (1.30 − 1)  20     & W = 249,504 ft − lbf min = 7.56 hp Power input flow rate of part (a) is larger than part (b) SI Units At 3, -18 C saturated, Table A-2b, R134a v3 = 0.13597 m3 kg At b, 138 kPaa, -7 C, Chart 3 vb = 0.1503 m 3 kg Pb = 138 kPaa At c, 1.24 MPaa Pc = 1240 kPaa Equation 15-8 1   Pc  n  v3 ηv = 1 + C − C      Pb   vb  n = 1.30 for R-134a

15. REFRIGERATION

(a) C = 0.03 1   1240  1.30  0.13597  ηv = 1 + 0.03 − 0.03 = 0.7849   138    0.1503 (b) C = 0.15 1   1240  1.30  0.13597  ηv = 1 + 0.15 − 0.15 = 0.3057   138    0.1503 (c) Equation 15-10 m& v3 PD Assume PD same as Example 15-4 (10)(1725) = 0.00471 m3 s PD = (3.281)3 (1728)(60)

ηv =

For Part (a) m& =

η v PD v3

=

(0.7849)(0.00471) = 0.0272 kg (0.13597 )

s

(0.3057 )(0.00471) = 0.0106 kg (0.13597 )

s

For Part (b) m& =

η v PD v3

=

Mass flow rate of part (a) is larger than part (b) (d) Equation 15-12 and Equation 15-13, η m = 1.0 ( n −1)   n   P n c W& = m& Pb vb   − 1  (n − 1)  Pb    For Part (a)

 1240 (1.30−1)1.30  1 . 30  (1240)(0.1503) W& = (0.0272) − 1  (1.30 − 1)  138     & W = 14.49 kW For Part (b)  1240 (1.30−1)1.30  1 . 30  W& = (0.0106) − 1 (1240)(0.1503)  (1.30 − 1)  138     W& = 5.65 kW

15. REFRIGERATION

Power input flow rate of part (a) is larger than part (b)

15.5

Consider a four-cylinder, 3 in. bore by 4 in. stroke, 800 rpm, single-acting compressor for use with R-134a. The proposed operating condition for the compressor is 100 F condensing temperature and 40 F evaporating temperature. It is estimated that the refrigerant will enter the expansion valve as a saturated liquid, that vapor will leave the evaporator at a temperature of 45 F, and that vapor will enter the compressor at a temperature of 55 F. Assume a compressor volumetric efficiency of 70 percent and frictionless flow. Calculate the refrigeration capacity in tons.

Solution:  π  (3) (4) PD = 4   (800) = 52.36 ft 3 min 4 1728   ηv = 0.70 2

Entering the compressor at 3’, 55 F P3′ = 54.732 psia at 45 F saturated, Table A-2a Chart 3, v3′ = 0.894 ft 3 lbm m& v3′ PD η v PD (0.70)(52.36) m& = = (60) = 2460 lbm hr v3′ 0.894

Equation 15-10, ηv =

Table A-2a At 1, 100 F, i1 = i2 = 39.538 Btu lbm At 3, 45 F, i3 = 108.6 Btu lbm Refrigeration capacity in tons q&e = m& (i3 − i2 ) = (2460)(108.6 − 39.538) = 169,893 Btu hr = 14.16 tons

15. REFRIGERATION

15.6

Consider the compressor of Fig. 15-7. (a) Construct the pressure-enthapy diagram for a condensing temperature of 130 F (54 C) and an evaporating temperature of 45 F (7 C). (b) What is the heat transfer rate in the evaporator and the power input? (c) Suppose that the load on the evaporator decreases to 110,000 Btu/hr (32 kW), and find the evaporating temperature and power input. Assume that the condensing temperature remains constant.

Solution: (a) Pressure-enthalpy diagram

(b) From Fig. 15-7 Condensing Temperature = 130 F Evaporator Temperature = 45 F Then, Capacity = 145,000 Btu/hr Power Input = 14.7 kW (c) From Fig. 15-7 Condensing Temperature = 130 F Capacity = 110,000 Btu/hr Then, Evaporator Temperature = 32.5 F Power Input = 13.3 kW 15.7

Refer to Problem 15-6a, and sketch the capacity and power curves for 130 F (54 C) from Fig. 15-7. (a) Sketch the capacity curve for the evaporator, assuming its capacity is proportional to the evaporating temperature in the ratio 4000 Btu/(hr-F) (2.1 kW/C). (b) The evaporator load decreases to 130,000 Btu/hr (38 kW) while the condensing temperature decreases to 115 F (46 C). Sketch the new evaporator-capacity curve. (c) Suppose in part (a) that the evaporator operating conditions remain fixed but the condensing temperature increases to 145 F (63 C). What will the capacity and evaporator temperature be?

15. REFRIGERATION

Solution:

(a)

(b)

Use English Units only

15. REFRIGERATION

(c) From Fig. 15-7 Evaporator Temperature = 20 F Condenser Temperature = 145 F Then Capacity = 70,000 Btu/hr Evaporator Temperature = 20 F

15.8

A refrigeration system using the compressor described in Fig. 15-7 is designed to operate with a condensing temperature of 115 F with an evaporating temperature of 50 F when the outdoor ambient is 95 F. After the system is put into operation, the evaporator pressure is measured to be 69 psia,and the power to the compressor is 12 kW. Superheat and subcooling are assumed to be as given in Fig. 15-7, and the ambient is about 95 F. (a) Estimate the condensing and evaporating temperatures, (b) compare the actual and expected performance, and (c) suggest what might be done to obtain the design conditions.

Solution: Figure 15-7 Condensing Temperature = 115 F Evaporating Temperature = 50 F Then Capacity = 175,000 Btu/hr Power Input = 13.7 kW At 69 psia, saturation temperature = 29.5 F, Table A-3a.

15. REFRIGERATION

(a) Evaporating Temperature = 29.5 F Power input = 12 kW Then: Condensing Temperature = 115 F Evaporating Temperature = 29.5 F (b) Actual capacity = 115,000 Btu/hr Expected capacity 175,000 Btu/hr Actual capacity < expected capacity (c) Increase the evaporator pressure to 98.8 psia for evaporator temperature of 50 F. 15.9

Saturated R-22 at 45 F (7 C) enters the compressor of a single-stage system. Discharge pressure is 275 psia (1.90 Mpa). Suction valve pressure drop is 2 psi (13.8 kPa). Discharge valve pressure drop is 4 psi (27.6 kPa). Assume the vapor is superheated 10 F (5.6 C) in the cylinder during the intake stroke. Piston clearance is 5 percent. Determine the (a) volumetric efficiency, (b) pumping capacity in lbm/min (kg/s) for 20 ft3/min (9.44 L/s) piston displacement, and (c) horsepower (kW) requirement if the mechanical efficiency is 80 percent.

Solution:

English Units P4 = 275 psia Pc = Pd = P4 + 4 = 275 + 4 = 279 psia At 45 F, Table A-3a, P3 = 90.791 psia Pb = P4 − 2 = 90.791 − 2 = 88.791 psia Table A-3a, v3 = 0.6029 ft 3 lbm At Point b, tb = 45 + 10 = 55 F

15. REFRIGERATION

Chart 4, Pb = 88.791 psia , tb = 55 F Then, vb = 0.6386 ft 3 lbm n = 1.16 for R22 C = 0.05 (a) Equation 15-8 1   Pc  n  v3 ηv = 1 + C − C      Pb   vb  1   279  1.16  0.6029   ηv = 1 + 0.05 − 0.05   = 0.8646  88.791    0.6386 

(b) Equation 15-10 m& =

η v PD v3

=

(0.8646)(20) = 28.68 lbm 0.6029

min

(c) Equation 15-12 and 15-13 ( n −1)   n     & P m n c  &     W =  Pb vb   − 1   ( ) η n − 1 P  m  b  

 279 (1.16−1)1.16   28.68  1.16  & W = (88.791)(144)(0.6386) − 1    88.791    0.80  (1.16 − 1)   W& = 363,054 ft − lb min = 11.0 hp SI Units P4 = 1.90 MPa = 1900 kPa Pc = Pd = P4 + 27.6 = 1900 + 27.6 = 1927.6 kPa At 7 F, Table A-3b, P3 = 621.94 kPa Pb = P4 − 2 = 621.94 − 13.08 = 608.14 kPa Table A-3b, v3 = 0.03791 m3 kg At Point b, tb = 7 + 5.6 = 12.6 C Chart 4, Pb = 608.14 kPa , tb = 55 F Then, vb = 0.04013 m 3 kg n = 1.16 for R22 C = 0.05 (a) Equation 15-8

15. REFRIGERATION 1   Pc  n  v3 ηv = 1 + C − C      Pb   vb  1   1927.6  1.16  0.03791   ηv = 1 + 0.05 − 0.05   = 0.8642  608.14    0.04013  (b) Equation 15-10

m& =

η v PD

=

(0.8642)(9.44) = 0.2152 kg (0.03791)(1000)

v3 (c) Equation 15-12 and 15-13 ( n −1)   n     & P m n c  &     W =  Pb vb   − 1   ( ) η − n 1 P  m  b  

s

 1927.6 (1.16−1)1.16   0.2152  1.16  & W = (608.14)(0.04013) − 1    608.14    0.80  (1.16 − 1)   W& = 8.2 kW 15.10 Consider the single-stage vapor-compression cycle shown in Fig. 15-35. Design conditions using R-134a are: q& L = 30,000 Btu/hr P3 = 200 psia P1 = 60 psia saturated P3 − P4 = 2 psi P2 = 55 psia C = 0.04 T2 = 60 F η m = 0.90 PD = 9.4 cfm & (a) Determine W , q& H , m& 12 , and sketch the cycle on a P-i diagram. If the load q& L decreases to 24,000 Btu/hr and the system comes to equilibrium with P2 = 50 psia and T2 = 50 F , (b) determine W& , q& H , m& , and locate the cycle on a P-i diagram.

Figure 15-35 Schematic for Problem 15-10.

15. REFRIGERATION

Solution: (a)

q& L = 24,000 Btu hr P2 = 50 psia and T2 = 50 F Equation 15-8 1   Pc  n  v3 ηv = 1 + C − C      Pb   vb  1   P3  n  ηv = 1 + C − C      P2    1   200  1.1  ηv = 1 + 0.04 − 0.04   = 0.899  50    Equation 15-10 η PD m& = v v2 at 50 psia, 50 F, R-134a v2 = 0.975 ft 3 lbm (0.899)(9.4) = 8.6673 lbm min m& = 0.975 ( n −1)   n     & P m n c  & W =   Pb vb   − 1    η m  (n − 1)  Pb    (n −1)    m&  n  P3  n   & W =   P2v2   −1  P2    η m  (n − 1)    200 (1.1−1)1.1   8.6673  1.1 & W = (50)(144)(0.975)  − 1   50    0.90  (1.1 − 1)   & W = 99,882 ft − lb min = 3.03 hp

15. REFRIGERATION P4 = P3 − 2 = 200 − 2 = 180 psia η mW& = m& (i3 − i2 ) J At 50 psia, 50 F i2 = 111.02 Btu lbm (0.90)(99,882) + 111.02 = 124.35 Btu lbm η W& i3 = m + i2 = m& J (8.6673)(778) q& L = 24,000 Btu hr q& L = m& (i2 − i1 ) q& 24,000 i1 = i2 − L = 111.02 − = 64.87 Btu lbm m& (8.6673)(60) i4 = i1 = 64.87 Btu lbm q& H = m& (i3 − i4 ) = (8.6673)(60)(124.35 − 64.87 ) q& H = 30,932 Btu hr

(b)

q& L = 30,000 Btu hr P2 = 55 psia and T2 = 60 F Equation 15-8 1   Pc  n  v3 ηv = 1 + C − C      Pb   vb  1   P3  n  ηv = 1 + C − C      P2    1   200  1.1  ηv = 1 + 0.04 − 0.04   = 0.9107  55    Equation 15-10 η PD m& = v v2

15. REFRIGERATION

at 55 psia, 60 F, R-134a v2 = 0.90165 ft 3 lbm (0.9107 )(9.4) = 9.4944 lbm min m& = 0.90165 ( n −1)   n     & P m n c − 1 W& =   Pb vb    Pb    η m  (n − 1)   (n −1)    m&  n  P3  n   & W =   P2v2   −1  P2    η m  (n − 1)    200 (1.1−1)1.1  9 . 4944 1 . 1   W& =  (55)(144)(0.90165)  − 1   55    0.90  (1.1 − 1)   & W = 103,191 ft − lb min = 3.13 hp P4 = P3 − 2 = 200 − 2 = 180 psia η mW& = m& (i3 − i2 ) J At 55 psia, 60 F i2 = 112.83 Btu lbm (0.90)(103,191) + 112.83 = 125.40 Btu lbm η W& i3 = m + i2 = (9.4944)(778) m& J q& L = 30,000 Btu hr q& L = m& (i2 − i1 ) q& 30,000 i1 = i2 − L = 112.83 − = 60.17 Btu lbm (9.4944)(60) m& i4 = i1 = 60.17 Btu lbm q& H = m& (i3 − i4 ) = (9.4944)(60)(125.40 − 60.17 ) q& H = 37,159 Btu hr 15.11 Consider an ordinary single-stage vapor-compression air-conditioning system. Because of clogged filters the air flow over the evaporator is gradually reduced to a very low level. Explain how the evaporator and compressor will be affected if the system continues to operate. Answer: Air flow over the evaporator when reduced to a very low level will indicate that the compressor is not loaded and also indicates that the evaporator is not transferring the expected quantity of heat to the refrigerant. 15.12 A vapor-compression cycle is subject to short periods of very light load; it is not practical to shut the system down. During these periods of light load, moisture condenses from the air flowing over the evaporator and freezes. Suggest a modification to the system to prevent this condition.

15. REFRIGERATION

Answer: The system can be modified by reducing the suction pressure and evaporator temperature. 15.13 A vapor-compression cycle is subject to occasional overload that leads to the tripping of circuit breakers. Explain how the system can be modified to prevent compressor overload without shutting the system off. Answer: Add an evaporator pressure regulator to maintain a relatively constant minimum pressure in the evaporator. Because most of the evaporator surface is subjected to two-phase refrigerant a constant minimum temperature will also be maintained. Evaporator pressure is sensed internally and balanced by spring loaded diagphragm. When evaporator pressure falls below a set value, the valve will close, restricting the flow of refrigerant so that evaporator pressure will rise, therefore prevent freezing. 15.14 A saturated liquid aqua-ammonia solution at 220 F and 200 psia is throttled to a pressure of 10 psia. Find (a) the temperature after the throttling process, and (b) the relative portions of liquid and vapor in the mixture after throttling. Solution:

(a) Temperature = t 2 = 62.4 F

15. REFRIGERATION

(b) x2 f = 0.3 , x2 g = 0.976 , x2 = 0.368

x2 − x2 f m& v 0.368 − 0.3 lbm vapor = = = 0.1006 m& x2 g − x2 f 0.976 − 0.3 lbm mixture

m& f m& lbm vapor = 1 − v = 1 − 0.1006 = 0.8994 m& m& lbm mixture

15.15 A solution of ammonia and water at 180 F, 100 psia, and with a concentration of 0.25 lbm ammonia per lbm of solution is heated at constant pressure to a temperature of 280 F. The vapor is then separated from the liquid and cooled to a saturated liquid at 100 psia. What are the temperature and concentration of the saturated liquid?

Solution:

15. REFRIGERATION

Temperature = t5 = 134.6 F Concentration = x4 = x5 = 0.437 15.16 It is proposed to use hot water at 180 F (82 C) from a solar collector system to operate a simple absorption cycle. Compute the maximum possible coefficient of performance, assuming an environment temperature of 100 F (38 C) and an air-cooled evaporator with the air temperature at 75 F (24 C). Solution: Equation 15-32 T (T − T ) (COP )max = e g o Tg (To − Te ) English Units: Te = 75 + 459.67 = 534.67 F To = 100 + 459.67 = 559.67 F Tg = 180 + 459.67 = 639.67 F

(COP )max = 534.67(639.67 − 559.67) = 2.675 639.67(559.67 − 534.67 ) SI Units: Te = 24 + 273.15 = 297.15 K To = 38 + 273.15 = 311.15 K Tg = 82 + 459.67 = 355.15 K

15. REFRIGERATION

(COP )max = 297.15(355.15 − 311.15) = 2.63 355.15(311.15 − 297.15) 15.17 Saturated water vapor at 50 F is mixed in a steady flow process with a saturated lithium bromide=water solution having a concentration of 0.60 lbm LiBr per lbm of mixture. The mass of the liquid solution is five times the mass of the water vapor mixed. The mixing process occurs at constant pressure. Find (a) the concentration of the resulting mixture, and (b) the heat that must be removed, in Btu per lbm of the total mixture, if saturated liquid solution is produced. Solution:

(a) m& 1 x1 = m& 6 x6 + m& 10 x10 x6 = 0.60 x10 = 0.00 m& 6 = 5m& 10 m& 1 = m& 6 + m& 10 = 6m& 10 6m& 10 x1 = 5m& 10 x6 5(0.60 ) x1 = = 0.50 6 (b) q& a = m& 10i10 + m& 6i6 − m& 1i1 q& a 1 5 = i10 + i6 − i1 m& 1 6 6 at 50 F saturated water vapor, Table A-1a i10 = 1082.9 Btu lbm , P = 0.178 psia = 9.23 mm Hg Chart 5 At x6 = 0.60 , i6 = −62 Btu lbm At x1 = 0.50 , i6 = −70 Btu lbm q& a 1 5 = (1082.9 ) + (− 62 ) − (− 70 ) = 198.82 Btu lbm m& 1 6 6

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