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CS & STC Series selection
WMP 1: Calculate the heat load
RPP-24544 REV Id
The heat load in BTU/HR or (Q) can be de rived 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.
Terms
GPM . GallonsPer Minute
CN . Constant Number fora given fluid
AT = Temperature di0'ercndal across d= potential
PSI . Pounds per Square Inch (p=um) of the operating side of the system
MIT . Horsepower of the electric motor ddving the hydraulic pump Q - arm / l m
r w a wowaa twos x two)
Tt° a Not fluid entering tempe rature in OF
T. . Hot fluid exiting temperature in OF
ti. • Cold fluid iemperawn: ent ering in OF
t aie a Cold fluid temperature exiting in OF
For example purposes, a hydrau lic system has a 125 HP (93Kw) elec tric motor installed coupled to a pump that p roduces a flow of 80 GPM @
2500 PS1G. The temperature differential of the oil entering the pump vs exiting the system Is about 53°F. Even though our return line pressure
ope ra tes below 100 psi, we must calculate the system heat load potential (Q) based upon the p rime movers (pump) capability. We can use one
of the fo ll
owing equations to accomplish this:
To de rive the required heat load (Q) to be removed by the heat exchanger, apply ONE of the following. Note: The cal culated but loads may di[
fer slightly from one formula to the next. This is due to assump ti ons trade when es timating heat removal requirements. The factor (v) represents
the percentage of the overall input energy to be rejected by the beat exchanger. The (v) fac tor is generally about 30% for most hydrau lic systems,
. however it can range from 20'%-70% depending upon the installed system components and heat being gene ra ted (e. servo valves, proportional
valves, ete—will increase the percentage required).
FORMULA EX4.%ME Constant for a given fluid ( CN )
A) Q - GPM x CN x actu al AT A) Q =80 x 210 x 53°F = 89,040 "R
e) Q - [ (PSI x GPM) / 1714) x (v) x 2545 a) Q -[(7500x80)117141 x 30 x 2545 = 89,090 eruf 1) Oil
e) Q - MIIP x (v ) x 2545 _ tin 2) Water..M.._....._._._...CN =
D) Q = Kw to be removed x 3415 e) Q =125 x.30 x 2545 = 95,347 sTu4 t 3) 50%!? Glycol..._._.. = 4 0
_ x) Q • HP to be removed x 2545 D) Q =28 x 3415 = 95,620 smltm
•
STEP 2: Calculate the Mean Tempera ture Difference
When calcula ting the MTD you will be requi red to choose a li quid flow rate to derive the cold side AT. 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 cepplica
tiots of water to water or 50 % Ethylene Glycol to water, a 1:1 ratio is common.
FORMMA
HOT FLUID AT = Q
Oil CN x GPM
I xAmnE
AT = 89,090^f "' uep 1em
Jt a) = 5.3°F = AT Rejected
COLA FLUID A t = BTU/hr
89 090 BTUfhr
At
= 4.45°P = ^TAbsorbed
Water CN xGPM
x
'D lora2:l ra tio)
T,, . Ilot Fluid entering temperature in degrees F TI =1253 T
Tee, • Hot Fluid exiting temperature in degrees F T. = 120.0 OF
tin a Cold Fluid entering temperature in degrees F L, -70.09F
ter„ = Cold Fluid exiting temperatu re in degrees F ter, a 74.5 OF 0 p
T, a • La a S[smaller temperature difference) = rS 120.0017-70.0°17- 50.0617 a 500°F a 3M
l
TI, . tei1 L (larger tempe ra ture differencel \ L / 1253°F-743°F = 50.8°F 30.8°F ^{ £
STEP 3: Calculate Log Mean Temperature Diffe rence Q MTD) - D
Attachment
9
To calculate the LMTD please use the following method; Calm No.: 145579-V-CA-004
L- Luger temperature difference from step 2.
M a S/L number (LoeATm err TAat a A).
Rev. No.:
Sheet 4
1
of 10 0
m
LMTDI =L%M LMTDI=50.8x.992 (nom TAmmA)=5039 t
To comet the LMT'D, for a multipass heat exchangers calculate R & K as fo ll
ows: Q U
FotcatuIA EXAMPLE
B•n
R=
Th. -Tow
t -^
1253°F-120°F 5.3°F
R=
4 - F = 7V = (L17=R)
the wrrec6on factor CFa
r(ROM rO'
=CHID, z CFa
K=
er
Tin-t„
_745°F-70°F
K-
124.5°F-70°F
= 43°F
55.4°F
=
(0.081=K)
e=5039x1.5039
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