Evaporator
* For Refrigeration Effect RE
* Enthalpy Properties
h1 = 348 h2 = 400 h3 = 260 h4 = 260
* For mass in the system
Qs = m (h1 – h4)
65 kW = m (348 – 260)
m=0.739 kg/s
EVAPORATOR
* Qe = 65 kw * Temperature at the evaporator is -31.56 0C * Specific volume is 0.528 m3/kg (from ph diagram) * Consider steel material for coil: Uo=90-120 Btuhrft2 0F ; use 105Btuhrft2 0F * Allowable velocity at the liquid line is 125 – 450 ft/min; use 287.5 ft/min
* For the volumetric flow rate in the coil:
V’ = mv
V’= (0.739 kg/s) (0.528 m3/kg)
V’= 0.391 m3/s = 0.664 ft3/min
* For the area of the coil:
Acoil = V'v
=0.665287.5
Acoil = 2.32 x 10-3 ft2 = 0.335 in2
Dcoil = 0.653 in
Steel Pipe Specification | Nominal Size | 1.25 in | Outside Diameter | 1.375 in | Inside Diameter | 1.245 in | Source: http://www.engineeringtoolbox.com/astm-copper-tubes-d_779.html |
* LMTD:
Inside tube: Tin = To = -31.56
Outside tube:
Consider the temperature before and after the operation of the refrigeration system.
Ti = -30 0C
Tf = -26.11 0C LMTD = -30--31.56-[-26.11-31.56]ln-30--31.56[-26.11-31.56]
LMTD = 3.11 0C
* For the surface area S of the evaporator:
QTotal = SUoLMTD
S = Q UoLMTD
S = Q UoLMTD
= 221 789.20628Btuhr (18.48)105 Btuhrft2 0F(37.598 0F)
S= 1038.22 ft2
* For length of the evaporator coil: S=πDL
L = SπD
L = 1038.22ft2π(1.37512)ft
L= 1427.64
* Enthalpy Properties
h1 = 348 h2 = 400 h3 = 260 h4 = 260
* For mass in the system
Qs = m (h1 – h4)
65 kW = m (348 – 260)
m=0.739 kg/s
EVAPORATOR
* Qe = 65 kw * Temperature at the evaporator is -31.56 0C * Specific volume is 0.528 m3/kg (from ph diagram) * Consider steel material for coil: Uo=90-120 Btuhrft2 0F ; use 105Btuhrft2 0F * Allowable velocity at the liquid line is 125 – 450 ft/min; use 287.5 ft/min
* For the volumetric flow rate in the coil:
V’ = mv
V’= (0.739 kg/s) (0.528 m3/kg)
V’= 0.391 m3/s = 0.664 ft3/min
* For the area of the coil:
Acoil = V'v
=0.665287.5
Acoil = 2.32 x 10-3 ft2 = 0.335 in2
Dcoil = 0.653 in
Steel Pipe Specification | Nominal Size | 1.25 in | Outside Diameter | 1.375 in | Inside Diameter | 1.245 in | Source: http://www.engineeringtoolbox.com/astm-copper-tubes-d_779.html |
* LMTD:
Inside tube: Tin = To = -31.56
Outside tube:
Consider the temperature before and after the operation of the refrigeration system.
Ti = -30 0C
Tf = -26.11 0C LMTD = -30--31.56-[-26.11-31.56]ln-30--31.56[-26.11-31.56]
LMTD = 3.11 0C
* For the surface area S of the evaporator:
QTotal = SUoLMTD
S = Q UoLMTD
S = Q UoLMTD
= 221 789.20628Btuhr (18.48)105 Btuhrft2 0F(37.598 0F)
S= 1038.22 ft2
* For length of the evaporator coil: S=πDL
L = SπD
L = 1038.22ft2π(1.37512)ft
L= 1427.64