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8-1. A cooling and dehumidifying coil is supplied with 2.4 m^3 /s of air at 29 C dry-bulb and 24 C wet-bulb temperatures, and its cooling capacity is 52 kW. The face velocity is 2.5 m/s. and the coil is of the direct- expansion type provided with refrigerant evaporating at 7 C. The coil has an air-side heat-transfer area of 15 m 2 per square meter of face area per row of tubes. The ratio of the air-side to refrigerant-side area is 14. The values of hr and hc are 2050 and 65 W/m 2 .K, respectively. Calculate (a) the face area, (b) the enthalpy of outlet air, (c) the wetted-surface temperatures at the air inlet, air outlet, and at the point where the enthalpy of air is midway between its entering and leaving conditions, (d) the total surface area, (e) the number of rows of tubes, and (f) the outlet dry-bulb temperature of the air.
Solution: At 29 C dry-bulb and 24 C wet-bulb ha,1=72.5 kJ/kg ga,1 = 0.88 m 3 /kg
(a) Face Area = (2.4 m 3 /s) / (2.5 m/s) Face Area = 0.96 m^2
(b) Enthalpy of outlet air, ha, m = (2.4 m 3 /s) / (0.88 m 3 /kg) = 2.7273 kg/s
m
q h (^) a,2 =ha,1− t
2.7273 kg/s
52 kW ha,2 =72.5kJ/kg−
ha,2 = 53.4 kJ/kg
(c) Wetted Surface Temperature Eq. 8-1.
( a i)
pm
c (^) h h c
hdA dq = −
Eq. 8-2.
dq =hr dAi ( ti−tr)
Eq. 8-3.
pmr i
c a i
i r c h A
hA h h
t t R = −
tr = 7 C A/Ai = 14 hr = 2050 W/m 2 .K hc = 65 W/m^2 .K cpm = 1.02 kJ/kg.K
h h
t t R a i
i r = = −
ha and hi in kJ/kg
Eq. 8-4. hi = 9.3625+1.7861ti+0.01135ti 2 +0.00098855ti 3
Eq. 8-5.
h 9.3625 1.7861t 0.01135t 0.00098855t 0 R
t R
t (^3) i
2 a i i
i (^) − r− + + + + =
At the air inlet, ha,1 = 72.5 kJ/kg
72.5 9.3625 1.7861t 0.01135t 0.00098855t 0
t (^3) i
2 i i
i (^) − − + + + + =
By trial and error: ti = 17.31 C and enthalpy hi = 48.8 kJ/kg at air inlet.
At the air outlet, ha,3 = 53.4 kJ/kg
9.3625 1.7861t 0.01135t 0.00098855t 0
t (^3) i
2 i i
i (^) − − 53. 4 + + + + =
By trial and error: ti = 13.6 C and enthalpy hi = 38.23 kJ/kg at air outlet.
At the midway enthalpy, ha,2 =(1/2)(72.5 kJ/kg + 53.4 kJ/kg) = 62.95 kJ/kg
9.3625 1.7861t 0.01135t 0.00098855t 0
t (^3) i
2 i i
i (^) − − 62. 95 + + + + =
By trial and error: ti = 15.5 C and enthalpy hi = 43.46 kJ/kg at midway enthalpy.
Answer - - - 17.31 C, 15.5 C, and 13.6 C.
(d) Total surface area.
Between 1 and 2.
( ) (mean enthalpydifference)
c
hA q mh h pm
c 12 1 2 =^1 − 2 = −
− −
cpm = 1020 J/kg.K
65 A
2.7273 72.5 62.95^12
A1-2 = 18.93 m^2
Between 2 and 3.
( ) (mean enthalpydifference)
c
hA q mh h pm
c 2 2 =^2 − 3 = −
− −
3 3
cpm = 1020 J/kg.K
65 A
2.7273 62.95^2
A2-3 = 23.59 m^2
Surface Area of Coil = 18.93 m^2 + 23.59 m^2
W W
W W
c
hA G W W^12 i,1 i, pm
c 12 1 2
G = 2.5 kg/s A1-2 = 41.1 m cpm = 1.02 kJ/kg.K = 1020 W/kg.K hc = 55 W/m 2 .K
For W1, psychrometric chart At 30 C dry-bulb and 21 C wet-bulb temperature. W 1 = 0.012 kg/kg Table A-2. ti,1 = 16.28 C, Wi,1 = 0.01163 kg/kg
ti,2 = 12.97 C Wi,2 = 0.00935 kg/kg
Solve for W 2 by substituing to Eq. 8-6.
0.012 W
2.5 0.012 W 2 2
W 2 = 0.0111 kg/kg - - - Ans.
(b) Checking W 2 :
At point 2., ha,2 = 48.66 F, t 2 = 20.56 C
From psychrometric chart, Figure 3-
W 2 = 0.011 kg/kg - - - Ans.
8-3. A direct-expansion coil cools 0.53 kg/s of air from an entering condition of 32 C dry-bulb and 20 C wet-bulb
temperature. The refrigerant temperature is 9 C, hr = 2 kW/m^2 .K, hc = 54 W/m^2 .K, and the ratio of air-side to refrigerant-side areas is 15. Calculate (a) the dry-bulb temperature of the air at which condensation begins and (b) the surface area in square meters of the portion of the coil that is dry.
Solution:
m = 0.53 kg/s At 32 C dry-bulb and 20 C wet-bulb temperatures ha,1 = 57 kJ/kg
(a) Dew-point of entering air = ti,2 = 13.8 C
Equation 8-11.
A
h dAA t 2 −ti,2hcdA=ti,2−tr r i
Then:
A
h A t 2 −ti,2hc=ti,2−tr r i
hc = 54 W/m2.K hr = 2000 W/m2.K tr = 9 C A/Ai = 15
t 2 −13.8 54 =
t 2 = 25.7 C - - - - Ans.
(b)
− = 1 − 2 1 2 r
c i r
pm 1 2 t 2
t t A
Ah
A
h
Gc t t
cpm = 1020 J/kg.K G = 0.53 kg/s
A
A1-2 = 4.47 m2 - - - Ans.
8-4. For a coil whose performance and conditions of entering air are shown in Table 8-1, when the face velocity is 2 m/s and the refrigerant temperature is 4.4 C, calculate (a) the ratio of moisture removal to reduction in dry- bulb temperature in the first two rows of tubes in the direction of air flow in the last two rows and (b) the average cooling capacity of the first two and the last two rows in kilowatts per square meter of face area.
Solution: Use Table 8-1. Face velocity = 2 m/s Refrigerant Temperature = 4.4 C.
(a) First 2-rows: At 30 C dry-bulb, 21.7 C wet-bulb temperature h 1 = 63 kJ/kg W 1 = 0.013 kg/kg γ 1 = 0.08735 m^3 /kg Final DBT = 18.2 C Final WBT = 17.1 C h 2 = 48.5 kJ/kg W 2 = 0.0119 kg/kg
Ratio for the first two rows = (W 1 - W 2 ) / (t 1 - t 2 ) = (0.013 - 0.0119) / (30 - 18.2) = 0.0000932 kg/kg.K - - - Ans.
For the last two rows. Rows of tube = 4 in Table 8-1. Final DBT = 14.3 C Final WBT = 13.8 C
