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UCS & URCS Series selection

STEP 1: Calculate the heat load The heat load in BTU/HR or (Q) can be derived by using several methods. To simplify things, we will consider general specifications for hy - draulic system oils and other fluids that are commonly used with shell & tube heat exchangers.

Kw = Kilowatt (watts x 1000) T in = Hot fluid entering temperature in °F T out = Hot fluid exiting temperature in °F t in = Cold fluid temperature entering in °F t out = Cold fluid temperature exiting in °F Q = BTU / HR

Terms GPM = Gallons Per Minute CN = Constant Number for a given fluid  T = Temperature differential across the potential PSI = Pounds per Square Inch (pressure) of the operating side of the system MHP = Horsepower of the electric motor driving the hydraulic pump

For example purposes, a 2,000 HP gear box lubrication system is provided with a flow of 80 GPM. The temperature differential of the oil entering the pump @ 200°F vs exiting the system @ 230°F is about 30.0°F. Though our return line pressure is below 100 psi, calculate the system heat load potential (Q) based upon the measured  T and the flow rate or by using the overall efficiency in our case 90%. To derive the required heat load (Q) to be removed by the heat exchanger, apply the following. Note: The calculated heat load may differ slightly from one formula to the next. This is due to assumptions made when estimating heat removal requirements.

F ormula a ) Q = GPM x CN x actual  T b ) Q = [ (PSI x GPM) / 1714 ] x ( v ) x 2545 c ) Q = MHP x ( v ) x 2545 d ) Q = Kw to be removed x 3415 e ) Q = HP x (1 - % efficiency) x 2545

E xample a ) Q = 80 x 210 x 30°F = 504,000 btu / hr

Constant for a given fluid ( CN )

1) Oil .............................. CN = 210 2) Water.......................... CN = 500 3) 50% E. Glycol............ CN = 450

d ) Q = 1,490 x (1 - 0.9) x 3415 = 508,835 btu / hr e ) Q = 2,000 x (1 - 0.9) x 2545 = 509,000 btu / hr

STEP 2: Calculate the Mean Temperature Difference

When calculating the MTD you will be required to choose a liquid flow rate to derive the Cold Side  T. If the water flow is unknown, assume a number based on what is available. As a normal rule of thumb, for oil to water cooling a 2:1 oil to water ratio is used. For applications of water to water or 50 % Ethylene Glycol to water, a 1:1 ratio is common.

E xample  T = 504,000 BTU/hr ( from step 1,item b ) = 30°F =  T Rejected 210 CN x 80GPM

F ormula

HOT FLUID  T =

Oil

CN x GPM

COLD FLUID  t =

BTU / hr

 t = 504,000 BTU/hr

= 25.2°F =  T Absorbed

Water

500 CN x 40GPM ( for a 2:1 ratio )

CN x GPM

T in = Ho t Fluid entering temperature in degrees F T out = Hot Fluid exiting temperature in degrees F t in = Cold Fluid entering temperature in degrees F t out = Cold Fluid exiting temperature in degrees F T out - t in S [smaller temperature difference] S = L [larger temperature difference] = L T in - t out ( )

T in = 125.3 °F T out = 120.0 °F t in = 70.0 °F t out = 74.5 °F

S ( ) = L

230°F - 40°F = 190°F 230°F - 65.2°F = 164.8°F

164.8°F = .867

190°F

STEP 3: Calculate Log Mean Temperature Difference (LMTD)

To calculate the LMTD please use the following method;

L = 190 o F M = .933

L = Larger temperature difference from step 2 M = S/L number ( located in table A ).

LMTD i = L x M

LMTD i = 190 x .933 ( from table A ) = 177.3

To correct the LMTD i for a multipass heat exchangers calculate R & K as follows:

F ormula

E xample

T in - T out t out - t in

230°F - 200°F 65.2°F - 40°F

30°F

R =

= {1.19=R}

R =

Locate the correction factor CF B ( from table B ) LMTD c =LMTD i x CF B LMTD c = 177.3 x 1 = 177.3

25.2°F

t out - t in T in - t in

65.2°F - 40°F 230°F - 40°F

25.2°F 190°F

K =

= {0.132=K}

K =

note: AIHTI reserves the right to make reasonable design changes without notice.

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355 American Industrial Drive LaCrosse, VA 23950

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fax: 434-757-1810

email: sales@aihti.com

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