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CS 2400 - 4800 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 hydraulic system oils and other fluids that are commonly used with shell & tube heat exchangers.

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

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

For example purposes, a hydraulic system has a total input 1200 HP (894Kw) electric motor installed coupled to a pump that produces a flow of 600 GPM @ 3000 PSIG. The temperature differential of the oil entering the pump vs exiting the system is about 6.6°F. Even though the return line pressure operates below 200 psi, calculate the system heat load potential (Q) based upon the prime movers (pump) capability, cooling fluid is water @ 80 o F use one of the following equations to accomplish this: To derive the required heat load (Q) to be removed by the heat exchanger, apply ONE of the following. Note: The calculated heat loads may differ slightly from one formula to the next. This is due to assumptions made when estimating heat removal requirements. The factor (v) represents the percentage of the overall input energy to be rejected by the heat exchanger. The (v) factor is generally about 30% for most hydraulic systems, however it can range from 20%-70% depending upon the installed system components and heat being generated (ie. servo valves, pro- portional valves, etc…will increase the percentage required).

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

E xample a ) Q =600 x 210 x 6.6°F = 831,600 btu / hr b ) Q =[(3000x600)/1714] x .30 x 2545 = 801,808 btu / hr c ) Q =1200 x .30 x 2545 = 916,200 btu / hr d ) Q =894 x .30 x 3415 = 915,909 btu / hr e ) Q =300 x 2545 = 736,500 btu / hr

Constant for a given fluid ( CN )

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

d ) Q = Kw to be removed x 3415 e ) Q = HP to be removed x 2545

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 you may need to 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 ap- plications of water to water or 50 % Ethylene Glycol to water, a 1:1 ratio is common.

E xample ( from step 1,item c )

F ormula

HOT FLUID 

 T = 916,200 BTU/hr 210 CN x 600GPM

=  T Rejected

= 7.37°F

T =

Oil

CN x GPM

COLD FLUID 

BTU / h r

 t = 916,200 BTU/hr 500 CN x 300GPM

= 3.81°F =  t Absorbed

t =

Water

CN x GPM

T in = 117.3 °F T out = 110.0 °F t in = 80.0 °F t out = 86.1 °F

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

110.0°F -80.0°F = 30.0°F = .962

117.3°F -86.1°F = 31.2°F

STEP 3: Calculate Log Mean Temperature Difference (LMTD) To calculate the LMTD please use the following method; L = Larger temperature difference from step 2. M = S/L number ( located in table A ). .962 = .980 LMTD i = L x M To correct the LMTD i for a multipass heat exchangers calculate R & K as follows:

LMTD i = 31.2 x .980 ( from table A ) = 30.6

F ormula

E xample

Locate the correction factor CF B ( from table B ) LMTD c =LMTD i x CF B LMTD c = 30.6 x .997 = 30.5

T in - T out t out - t in

117.3°F - 100°F 86.1°F - 90°F

17.3°F

R =

= {2.82=R}

R =

6.1°F

t out - t in T in - t in

86.1°F - 80°F 117.3°F - 80°F

6.1°F

K =

= {.163=K}

K =

37.3°F

STEP 4: Calculate the area required

Q ( BTU / HR )

916,200

Required Area sq.ft. =

= 300.4 sq.ft.

LMTD c x U ( from table C)

30.5 x 100

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

tel: 434-757-1800 355 American Industrial Drive LaCrosse, VA 23950

email: sales@aihti.com fax: 434-757-1810

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