# Evaporator

Topics: Thermodynamics, Volumetric flow rate, Volume Pages: 3 (265 words) Published: December 9, 2012
* 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 ft