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CS 2000 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 250 HP (186Kw) electric motor installed coupled to a pump that produces a flow of 200 GPM @ 2000 PSIG. The temperature differential of the oil entering the pump vs exiting the system is about 4.3°F. Even though the return line pressure oper- ates below 100 psi, calculate the system heat load potential (Q) based upon the prime movers (pump) capability. 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, proportional valves, etc…will increase the percentage required).

E xample a ) Q =200 x 210 x 4 .3°F = 180,600 btu / hr b ) Q =[(2000x200)/1714] x .30 x 2545 = 178,179 btu / hr

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

Constant for a given fluid ( CN )

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

c ) Q =250 x .30 x 2545 = 190,875 btu / hr d ) Q =186 x .30 x 3415 = 190,557 btu / hr

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 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.

F ormula

E xample  T = 190,875 BTU/hr ( from step 1,item c ) = 4.54°F =  T Rejected 210 CN x 200GPM

HOT FLUID  T =

Oil

CN x GPM

COLD FLUID  t =

BTU / h r

 t = 190,875 BTU/hr

= 3.81°F =  t Absorbed

Water

CN x GPM

500 CN x 100GPM ( for a 2:1 ratio )

T in = 104.54 °F T out = 100.0 °F t in = 90.0 °F t out = 93.81 °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

100.0°F -90.0°F = 10.0°F 104.54°F -93.81°F = 10.73°F

10.0°F 10.73°F

= .931

STEP 3: Calculate Log Mean Temperature Difference (LMTD) To calculate the LMTD please use the following method; LMTD i = L x M (L = Larger temperature difference from step 2) x (M = S/L number ( located in table A )) LMTD i = 10.73 x .964 ( from table A ) = 10.34 To correct the LMTD i for a multipass heat exchangers calculate R & K as follows:

F ormula

E xample

Locate the correction factor CF B ( from table B ) LMTD c =LMTD i x CF B LMTD c = 10.34 x .98 = 10.13

T in - T out t out - t in

104.54°F - 100°F 93.81°F - 90°F

4.54°F 3.81°F

R =

= {1.191=R}

R =

t out - t in T in - t in

93.81°F - 90°F 104.54°F - 90°F

3.81°F 14.54°F

K =

= {0.262=K}

K =

TABLE C

TABLE E- Flow Rate for Shell & Tube Shell Max. Liquid Flow - Shell Side

TUBE FLUID SHELL FLUID

Liquid Flow - Tube Side

400 350 100 300 90

Water Water Water

Water 50% E. Glycol Oil 50% E. Glycol Oil

dia .

Baffle Spacing

Code 2

12 Min. Max. Min. Max. Min. Max.

50% E. Glycol 50% E. Glycol

2000

160 240 320 500 90

650 45 320 25

160

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

355 American Industrial Drive LaCrosse, VA 23950

tel: 434-757-1800

fax: 434-757-1810

email: sales@aihti.com

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