Chapter 13

CHAPTER 13 - EXPANSION DEVICES

13-1.

Using the method described in Sec. 13-5 and entering conditions given in Table 13-1 for example 13-1 at position 4, compute the length of tube needed to drop the temperature to 36 C. Use property values from Refrigerant 22 tables when possible.

Solution: At Table 13-1, position 4 Temperature = 37 C. p4 = 1425.8 kPa x4 = 0.023 3

ν4 = 0.001230 m /kg

h4 = 249.84 kJ/kg V4 = 5.895 m/s At Position 5, t = 36 C Eq. 13-15 2418.4  p  ln  = 15.06 − t + 273.15  1000 

 p 5  2418.4  = 15.06 − 36 + 273.15  1000 

ln

p5 = 1390.3 kPa Eq. 13-16. ν f5 = 0.777 + 0.002062t + 0.00001608t 2 2

ν f5 =

0.777 + 0.002062(36) + 0.00001608(36 ) 1000 3

nf5 = 0.000872 m /kg Eq. 13-17. −4.26 + 94050 (t + 273.15) p ν g5 = 1000 −4.26 + 94050 (36 + 273.15) (1390300) ν g5 = 1000 3

nf5 = 0.01665 m /kg Eq. 13-18. h f5 = 200.0 + 1.172t + 0.001854t 2 2

h f5 = 200.0 + 1.172(36) + 0.001854(36 ) hf5 = 244.6 kJ/kg Eq. 13-19 h g5 = 405.5 + 0.3636t − 0.002273t 2

2

h g5 = 405.5 + 0.3636(36) − 0.002273(36 ) hg5 = 415.64 kJ/kg

Eq. 13-20 µ f5 = 0.0002367 − 1.715 × 10 −6 t + 8.869 × 10 −9 t 2 2

µ f5 = 0.0002367 − 1.715 × 10 −6 (36) + 8.869 × 10 −9 (36 ) µf5 = 0.0001865 Pa.s

Eq. 13-21. 6

9

9 2

µ g5 = 11.945 × 10 − + 50.06 × 10 − t + 0.2560 × 10 − t

Page 1 of 14


µ g5 = 11.945 × 10

−6

+ 50.06 × 10

−9

(36) + 0.2560 × 10 −9 (36)2

µg5 = 0.00001408 Pa.s

Eq. 13-14. x=

− b ± b 2 − 4ac

2a 2  w 

( g5 − ν f5 )  a = ν

2

1

  A  2 2

w/A = 4792.2 kg/s.m from Ex. 13-1. 2

2

a = (0.01665 − 0.000872) (4792.2) b = 1000(h g5 − h f5 ) + ν f5 ( ν g5

1 = 2858.54 2

 w  − ν f5 )   A 

2

2

b = 1000(415.64 − 244.6) + 0.00872(0.01665 − 0.000872)(4792.2 ) b = 171,356 2

 w  1 2 V c = 1000(h f5 − h 4 ) +   ν f5 − 4 2  A  2 2

c = 1000(244.6 − 249.84) + (4792.2)

2

2

(5.895) 1 (0.000872)2 − 2 2

c = -5,248.65 2

x=

− b ± b − 4ac

2a 2

x=

− 171,356 ± 171,356 − 4(2854.54)(- 5248.65)

2(2858.54)

= 0.031

Then: h5 = hf5 + x(hg5 - hf5) h5 = 244.6 + 0.031 (415.64 - 244.6) h5 = 249.9 kJ/kg ν5 = νf5 + x( νg5 - νf5) ν5 = 0.000873 + 0.031 (0.01665 - 0.000872) 3

ν5 = 0.001361 m /kg µ 5 = µ f5 + x µ g5 − µ f5 µ 5 = 0.0001865 + 0.031(0.00001408 − 0.0001865) µ5 = 0.0001812 Pa.s

w ν 5 = (4792.2)(0.001361) A V5 = 6.522 m/s V5 =

VD D  w  =   µν µ  A  D = 1.63 mm = 0.00163 m At 4: Re =

µ 4 = µ f4 = 0.0002367 − 1.715 × 10 −6 t + 8.869 × 10 −9 t 2

t4 = 37 C

Page 2 of 14


2

µ 4 = 0.0002367 − 1.715 × 10 −6 (37) + 8.869 × 10 −9 (37 ) µ 4 = 0.0001854

Re =

(0.00163) (0.0001812)

(4792.2) = 43,109

Eq. 13-9. 0.33 f = Re 0.25 0.33 = 0.02303 f 4 = (42,132)0.25 f 5 =

0.33

(43,109)0.25

= 0.02290

0.02303 + 0.02290 = 0.022965 2 5.895 + 6.522 = 6.2085 m/s Vm = 2 Eq. 13-4 f m =

 ∆L V 2  (p 4 - p 5 ) - f  A = w (V5 − V4 ) D 2 ν   Eq. 13-7 f

∆L V 2

∆L V w

D 2 A ∆L V w w = (V5 − V4 ) (p 4 - p 5 ) - f D 2 A A ∆L V w = 4792.2(6.522 − 5.895) 1000(1425.8 - 1390.3) - f D 2 A ∆L V w = 32,495.3 f D 2 A ∆L (6.2085) (0.022965) (4792.2) = 32,495.3 (0.00163) 2 ∆L4-5 = 0.155 m - - - Ans . 13-2.

D 2 ν

= f

A capillary tube is to be selected to throttle 0.011 kg/s of refrigerant 12 from a condensing pressure of 960 kPa and a temperature of 35 C to an evaporator operating at -20 C. (a) Using Figs. 13-7 and 13-8, select the bore and length of a capillary tube for this assignment. (b) If the evaporating temperature had been 5 C rather than -20 C, would the selection of part (a) be suitable? Discuss assumptions that have been made.

Solution: Table A-5, p = 960 kPa, tsat = 40 C, Subcooling = 40 C - 35 C = 5 C (a) Use bore diameter D = 1.63 mm Fig. 13-7, 960 kPa inlet pressure, saturated. Flow rate = 0.0089 kg/s Fig. 13-8. Flow correction factor = (0.011 kg/s)/(0.0089 kg/s) Flow correction factor = 1.24 Then Length = 1,230 mm = 1.23 m L - - - Ans . (b) Use positions from 35 C to -20 C at 5 C increment. Table A-5, 35 C, sat. p = 847.72 kPa.

Page 3 of 14


At position 1, h1 = 233.50 kJ/kg 3

ν1 = 0.78556 L/kg = 0.000786 m /kg

Table 15-5, µ1 = 0.000202 Pa.s p1 = 960 kPa w 0.011 = = 5271.4 kg/s.m 2 2 A π(0.00163) 4 w ν 1 = (5271.4)(0.000786) A V1 = 4.143 m/s V1 =

Re 1 = Re 1 =

V1D µ 1 ν 1

D   w     A  µ 1 

=

(5271.4)(0.00163)

= 42,537 0.000202 0.33 0.33 = = 0.02298 f 1 = 0.25 Re 1 (42537)0.25

At position 2, 30 C p2 = 744.90 kPa hf2 = 228.54 kJ/kg hg2 = 363.57 kJ/kg 3

νf2 = 0.77386 L/kg = 0.000774 m /kg 3

νg2 = 23.5082 L/kg = 0.02351 m /kg µf2 = 0.0002095 Pa.s µg2 = 0.00001305 Pa.s

x=

− b ± b 2 − 4ac

2a 2  w 

( g2 − ν f2 )  a = ν

2

1

  A  2

2

2

a = (0.02351 − 0.000774) (5271.4 ) b = 1000(h g2 − h f2 ) + ν f2 ( ν g2

1 = 7182.1 2

 w  − ν f2 )   A 

2

2

b = 1000(363.57 − 228.54) + 0.000774(0.02351 − 0.000774)(5271.4 ) b = 135,519 2

2

 w  1 2 V1  ν f2 − 2  A  2

c = 1000(h f2 − h1 ) + 

2

c = 1000(228.54 − 233.50) + (5271.4)

1 2

(0.000774 )2 −

c = -4,960.3 2

x=

− b ± b − 4ac

2a

Page 4 of 14

(4.143)2 2


x=

− 135,519 ± 135,519 2 − 4(7182.1)(- 4960.3)

2(7182.1)

= 0.0365

Then: h2 = hf2 + x(hg2 - hf2) h2 = 233.47 kJ/kg ν2 = νf2 + x( νg2 - νf2) 3

ν2 = 0.001604 m /kg µ 2 = µ f2 + x µ g2 − µ f2 µ2 = 0.0002023 Pa.s

w ν 2 = (5271.4)(0.001604) A V2 = 8.455 m/s V2 =

Re 2 = Re 2 =

VD D  w  =   µν µ 2  A 

(5271.4)(0.00163) 0.0002023

= 42,474

0.33

f 2 =

Re 0.25 0.33

f 2 =

= 0.02299

(42,474)0.25

0.02298 + 0.02299 = 0.022985 2 4.142 + 8.455 = 6.299 m/s Vm = 2 ∆L V w w = (V2 − V1 ) (p 1 - p 2 ) - f D 2 A A f m =

1000(960 - 744.9) - (0.022985 )

∆L

(6.299)

(0.00163 )

2

(5271.4) = 5271.4(8.455 − 4.143)

∆L1-2 = 0.8217 m


νf2 = 0.76286 L/kg = 0.000763 m /kg 3


x=

− b ± b 2 − 4ac

2a 2  w 

( g3 − ν f3 )  a = ν

2

1

  A  2

Page 5 of 14


2

2

a = (0.026854 − 0.000763) (5271.4 )

1 = 9458.1 2

 w    A 

b = 1000(h g3 − h f3 ) + ν f3 ( ν g3 − ν f3 )

2

2

b = 1000(361.68 − 223.65) + 0.000763(0.026854 − 0.000763)(5271.4 ) b = 138,583 2

 w  1 2 V c = 1000(h f3 − h 2 ) +   ν f3 − 2 2  A  2 2

c = 1000(223.65 − 233.47) + (5271.4)

2

1 (8.455) (0.000763)2 − 2 2

2

c = -9,847.7 x= x=

− b ± b 2 − 4ac

2a − 138,583 ± 138,583 2 − 4(9458.1)(- 9847.7)

2(9458.1)

= 0.0707



w ν 3 = (5271.4)(0.002608) A V3 = 13.748 m/s V3 =

Re 3 = Re 3 = f 3 = f 3 =

VD D  w  =   µν µ 3  A 

(5271.4)(0.00163) 0.0002026

= 42,411

0.33 Re 0.25 0.33

(42,411)0.25

= 0.0230


1000(744.9 - 651.62 ) - (0.0230 )

∆L

(11.102)

(0.00163 )

2

(5271.4 ) = 5271.4(13.748 − 8.455) ∆L2-3 = 0.1584 m

Page 6 of 14



νf4 = 0.75246 L/kg = 0.00075246 m /kg 3


x=

− b ± b 2 − 4ac

2a 2  w 

( g4 − ν f4 )  a = ν

2

1

  A  2

2

2

a = (0.0307802 − 0.00075246) (5271.4 )

( g4 b = 1000(h g4 − h f4 ) + ν f4 ν

 w  − ν f4 )   A 

1 = 12,528 2 2

2

b = 1000(359.73 − 218.82) + 0.00075246(0.0307802 − 0.00075246)(5271.4 ) b = 141,538 2

 w  1 2 V3  ν f4 − 2  A  2

2

c = 1000(h f4 − h 3 ) + 

2

c = 1000(218.82 − 233.41) + (5271.4)

2

(13.748) 1 (0.00075246)2 − 2 2

c = -14,677 2

x=

− b ± b − 4ac

2a 2

x=

− 141,538 ± 141,538 − 4(12528)(- 14677)

2(12528)

= 0.1028



w

ν 4 = (5271.4)(0.003839) A V4 = 20.237 m/s

V4 =

Re 4 = Re 4 =

VD D  w  =   µν µ 4  A 

(5271.4)(0.00163) 0.0002032

= 42,285

Page 7 of 14


0.33

f 4 =

Re 0.25 0.33

f 4 =

(42,285)0.25

= 0.0230


1000(651.62 - 567.29) - (0.0230)

∆L

(16.993)

(0.00163)

2

(5271.4) = 5271.4(20.237 − 13.748 ) ∆L3-4 = 0.0793 m


νf5 = 0.74262 L/kg = 0.00074262 m /kg 3


x=

− b ± b 2 − 4ac

2a 2  w 

( g5 − ν f5 )  a = ν

2

1

  A  2

2

1 = 16,701 2

2

a = (0.0354133 − 0.00074262) (5271.4 )

 w    A 

b = 1000(h g5 − h f5 ) + ν f5 ( ν g5 − ν f5 )

2

2

b = 1000(357.73 − 214.05) + 0.00074262(0.0354133 − 0.00074262)(5271.4 ) b = 144,396 2

 w  1 2 V c = 1000(h f5 − h 4 ) +   ν f5 − 4 2  A  2 2

c = 1000(214.05 − 233.31) + (5271.4)

2

1 2

(0.00074262)2 −

(20.237)2

c = -19,457 2

x=

− b ± b − 4ac

2a 2

x=

− 144,396 ± 144,396 − 4(16,701)(- 19,457)

2(16,701)

= 0.1327

Then: h5 = hf5 + x(hg5 - hf5)

Page 8 of 14

2


h5 = 233.12 kJ/kg ν5 = νf5 + x( νg5 - νf5) 3


w ν 5 = (5271.4)(0.005343) A V5 = 28.165 m/s V5 =

Re 5 = Re 5 =

VD D  w  =   µν µ 5  A 

(5271.4)(0.00163) 0.0002059

= 41,731

0.33

f 5 =

Re 0.25 0.33

f 5 =

= 0.02309

(41,731)0.25

0.0230 + 0.02309 = 0.02305 2 20.237 + 28.165 = 24.201 m/s Vm = 2 f m =

(p 4 - p5 ) - f

∆L V w

D 2 A

=

1000(567.29 - 491.37) - (0.02305)

w (V5 − V4 ) A ∆L (24.201)

(0.00163)

2

(5271.4) = 5271.4(28.165 − 20.237) ∆L4-5 = 0.0378 m


νf6 = 0.73326 L/kg = 0.00073326 m /kg 3


x=

− b ± b 2 − 4ac

2a 2  w 

( g6 − ν f6 )  a = ν

2

1

  A  2

2

1 = 22,431 2

2

a = (0.0409137 − 0.00073326) (5271.4 )

 w    A 

b = 1000(h g6 − h f6 ) + ν f6 ( ν g6 − ν f6 )

2

Page 9 of 14


2

b = 1000(355.69 − 209.32) + 0.00073326(0.0409137 − 0.00073326)(5271.4 ) b = 147,189 2

 w  1 2 V c = 1000(h f6 − h 5 ) +   ν f6 − 5 2  A  2 2

c = 1000(209.32 − 233.12) + (5271.4)

2

1 2

(0.00073326)2 −

(28.165)2 2

c = -24,189 x= x=

− b ± b 2 − 4ac

2a − 147,189 ± 147,189 2 − 4(22,431)(- 24,189)

2(22,431)

= 0.1604



w ν 6 = (5271.4)(0.007178) A V6 = 37.838 m/s V6 =

Re 6 = Re 6 = f 6 = f 6 =

VD D  w  =   µν µ 6  A 

(5271.4)(0.00163) 0.0002085

= 41,211

0.33 Re 0.25 0.33

(41,211)0.25

= 0.02316

0.02309 + 0.02316 = 0.02313 2 28.165 + 37.838 = 33 m/s Vm = 2 ∆L V w w = (V6 − V5 ) (p 5 - p 6 ) - f D 2 A A f m =

1000(491.37 - 423.30 ) - (0.02313 )

∆L

(33.0 )

(0.00163 ) 2

(5271.4 ) = 5271.4(37.838 − 28.165 ) ∆L5-6 = 0.0138 m

At position 7, 5 C p7 = 363.55 kPa hf7 = 204.64 kJ/kg hg7 = 353.60 kJ/kg

Page 10 of 14


3

νf7 = 0.72438 L/kg = 0.00072438 m /kg 3


x=

− b ± b 2 − 4ac

2a 2  w 

( g7 − ν f7 )  a = ν

2

1

  A  2

2

1 = 30,380 2

2

a = (0.0474853 − 0.00072438) (5271.4 )

b = 1000(h g7 − h f7 ) + ν f7 ( ν g7

 w  − ν f7 )   A 

2

2

b = 1000(353.60 − 204.64) + 0.00072438(0.0474853 − 0.00072438)(5271.4 ) b = 149,901 2

 w  1 2 V c = 1000(h f7 − h 6 ) +   ν f7 − 6 2  A  2 2

c = 1000(204.64 − 232.80) + (5271.4)

2

(37.838) 1 (0.00072438)2 − 2 2

c = -28,869 x=

− b ± b 2 − 4ac

2a 2

x=

− 149,901 ± 149,901 − 4(30,380)(- 28,869)

2(30,380)

= 0.1856



w ν 7 = (5271.4)(0.009403) A V6 = 49.567 m/s V7 =

Re 7 = Re 7 = f 7 = f 7 =

VD D  w  =   µν µ 7  A 

(5271.4)(0.00163) 0.0002111

= 40,703

0.33 Re 0.25 0.33

(40,703)0.25

= 0.02323

Page 11 of 14

2



1000(423.3 - 363.55) - (0.02320)

∆L

(43.703)

(0.00163)

2

(5271.4) = 5271.4(49.567 − 37.838) ∆L6-7 = -0.0013 m ~ 0.000 m

Assume choked flow is at approximately 5 C. L = ∆L1-2 +∆L2-3 +∆L3-4 +∆L4-5 +∆L5-6 +∆L6-7 L = 0.8217 m + 0.1584 m + 0.0793 m + 0.0378 m + 0.0138 m + 0 m L = 1.111 m

Ans.

By assuming choked flow length the same , choked flow is at 5 C. 5 C is still suitable for the selection of part (a) as it is the choked flow temperature.

13-3.

A refrigerant 22 refrigerating system operates with a condensing temperature of 35 C and an evaporating temperature of -10 C. If the vapor leaves the evaporator saturated and is compressed isentropically, what is the COP of the cycle (a) if saturated liquid enters the expansion device and (b) if the refrigerant entering the expansion device is 10 percent vapor as in Fig. 13-3?

Solution: Table A-6. At 1, -10 C, h1 = 401.555 kJ/kg s1 = 1.76713 kJ/kg At 2, 35 C, constant entropy, Table A-7 h2 = 435.212 kJ/kg (a)

At 35 C saturated. h3 = hf = 243.114 kJ/kg h4 = h3 = 243.114 kJ/kg h1 − h 4

401.555 − 243.114 h 2 − h1 435.212 − 401.555 COP = 4.71 - - - Ans . COP =

(b)

=

h3 = hf + x (hg - hf ) hf = 243.114 kJ/kg hg = 415.627 kJ/kg x = 0.10 h3= 243.114 + (0.10)(415.627 - 243.114) h3 = 260.365 kJ/kg h4 = h3 = 260.365 kJ/kg h1 − h 4

401.555 − 260.365 h 2 − h1 435.212 − 401.555 COP = 4.20 - - - Ans . COP =

=

Page 12 of 14


13-4.

Refrigerant 22 at a pressure of 1500 kPa leaves the condenser and rises vertically 10 m to the expansion valve. The pressure drop due to friction in the liquid line is 20 kPa. In order to have no vapor in the refrigerant entering the expansion valve, what is the maximum allowable temperature at that point?

Solution: say ν = ν1 p2 = p1 - gH/ ν1 - ∆p H = 10 m g = 9.81 m/s ∆p = 20 kPa p1 = 1500 kPa Table A-6. 3

ν1=0.8808 L/kg = 0.0008808 m /kg

p2 = (1500)(1000) - (9.81)(10) / (0.0008808) - (20)(1000) p2 = 1,368.62 kPa Table A-6. t2 = 35.4 C - - - Ans . 13-5.

A superheat-controlled expansion valve in a refrigerant 22 system is not equipped with an external equalizer. The valve supplies refrigerant to an evaporator coil and comes from the factory with a setting that requires 5K superheat in order to open the valve at an evaporator temperature of 0 C. (a) What difference in pressure on opposite sides of the diaphragm is required to open the valve? (b) When the pressure at the entrance of the evaporator is 600 kPa, how much superheat is required to open the valve if the pressure drop of the refrigerant through the coil is 55 kPa?

Solution: Using Fig. 13-15 and deriving equation by assuming parabolic curve. Let y - pressure, kPa and x = temperature , C. 2

2

y2 - y1 = A (x2 - x1 ) + B(x2 - x1) At 5 C evaporator temperature, 5 K superheat 100 kPa pressure differential x1 = 5 C, x2 = 5 C + 5 = 10 C y2 - y1 = 100 kPa 2

2

100 = A (10 - 5 ) + B (10 - 5) 100 = 75A + 5B - - Eq. 1 At -30 C evaporator temperature 12 C superheat 100 kPa pressure differential x1 = -30 C, x2 = -30 C + 12 = -18 C y2 - y1 = 100 kPa 2

2

100 = A ((-18) -(-30) ) + B ((-18) -(-30)) 100 = -576A + 12 B - - Eq. 2 But 5B = 100 - 75A Then 100 = -576A + 12 (20 - 15A) A = 0.185185 B = 17.222222

Page 13 of 14


Therefore: 2

2

y2 - y1 = 0.185185 (x2 - x1 ) + 17.222222(x2 - x1) (a)

At 0 C evaporator temperature, 5 K superheat x1 = 0 C x2 = 0 C + 5 = 5 C 2

2

y2 - y1 = 0.185185 (5 -0 ) + 17.222222(5 -0) y2 - y1 = 90.74 kPa - - - Ans.

(b)

At 0 C evaporator temperature, p = 497.59 kPa ∆p = 600 kPa + 55 kPa - 497.59 kPa = 157.41 kPa. x1 = 0 C Then 2

2

157.41 = 0.185185 (x2 -0 ) + 17.222222(x2 -0) x2 = 8.4 C x2 - x1 = 8.4 K - - - Ans.

13-6.

The catalog of an expansion valve manufacturer specifies a refrigerating capacity of 45 kW for a certain valve when the pressure difference across the valve is 500 kPa. The catalog ratings apply when vapor-free liquid at 37.8 C enters the expansion valve and the evaporator temperature is 4.4 C. What is the expected rating of the valve when the pressure difference across it is 1200 kPa?

Solution: Eq. 13-22 Velocity = C 2(pressuredifference) m/s With all other data as constant except for pressure difference and refrigerating capacity. Refrigerating Capacity α 2(pressuredifference) m/s Then: New Refrigerating Capacity = (45 kW )

1200 kPa 500 kPa

= 69.7 kW - - - - Ans .

-000-

Page 14 of 14

Chapter 13

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