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FBF 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.

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 hydraulic system has a 125 HP (93Kw) electric motor installed coupled to a pump that produces a flow of 80 GPM @ 2500 PSIG. The temperature differential of the oil entering the pump vs exiting the system is about 5.3°F. Even though our return line pressure operates below 100 psi, we must calculate the system heat load potential (Q) based upon the prime movers (pump) capability. We can 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 dif- fer 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, proportional 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

E xample a ) Q =80 x 210 x 5.3°F = 89,040 btu / hr b ) Q =[(2500x80)/1714] x .30 x 2545 = 89,090 btu / hr c ) Q =125 x .30 x 2545 = 95,347 btu / hr

Constant for a given fluid ( CN )

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

c ) Q = MHP x ( v ) x 2545 d ) Q = Kw to be removed x 3415 e ) Q = HP to be removed x 2545

d ) Q =28 x 3415 = 95,620 btu / hr e ) Q =37.5 x 2545 = 95,437 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 your 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 applications of water to water or 50 % Ethylene Glycol to water, a 1:1 ratio is common.

E xample  T = 89,090 BTU/hr ( from step 1,item b ) = 5.3°F =  T Rejected 210 CN x 80GPM

F ormula

HOT FLUID  T =

Oil

CN x GPM

COLD FLUID  t =

BTU / h r

 t = 89,090 BTU/hr

= 4.45°F =  T Absorbed

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

Water

CN x GPM

T in = 125.3 °F T out = 120.0 °F t in = 700 °F t out = 74.5 °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

120.0°F -70.0°F = 50.0°F = 50.0°F = .984

125.3°F -74.5°F = 50.8°F

50.8°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 ). LMTD i = L x M LMTD i = 50.8 x .992 ( from table A ) = 50.39 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

125.3°F - 120°F 74.5°F - 70°F

5.3°F 4.5°F

Locate the correction factor CF B ( from table B ) LMTD c =LMTD i x CF B LMTD c = 50.39 x 1 = 50.39

R =

{1.17=R}

R =

t out - t in T in - t in

74.5°F - 70°F 124.5°F - 70°F

4.5°F

K =

{0.081=K}

K =

55.4°F

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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