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HEAT TRANSFER TEMPERATURE DIFFERENTIAL F (AVERAGE) S!?Fs;d 3 8 w

0t2 f:co UJ 0 0

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Sizing a steam trap for maximum load at maximumpressure is not a difficult task, but because steam pressurediminishes much more rapidly than steam temperature, it isdifficult to assure sufficient trap capacity at reduced loads.

Consider the heat transfer equation: Q=UA (AT).IQ represents the heat transferred in an hour: Btu/hr.U is the heat transfer co-efficient which varies with

the materials of construction, the media fromwhich the energy flows, and the media to which itflows: Btu/hr.-sq. ft.-Y.

"A" represents the area of the heat transfer surface:sq. ft.AT represents the temperature differentialacross the heat transfer surface: Fahrenheit

degrees.In a given heat transfer apparatus, the area of heat transfersurface does not vary, so "A"is a constant. Since the mediaand materials of construction do not change U is alsopretty much a constant. (It varies some with the velocity ofliquid or gas). The only variables then are Q and AT, andthat means that the heat transferred is directly proportionalto the temperature differential.

EXAMPLE: Air heater operating at full loadLets consider an air heater selected to function on full loadat 40 psig (287F. saturated). Lets say further, that the air isbeing heated from -10F. to 70F. at design conditions. Theair flow is 42,500 cfm.42,500 cubic x 60 min. x 0.018 Btu. x 8OF./919 Btu. = 4000#/Hr.

feet hr . OF-iC #min. foot

Draw in the 40 psi steam pressure line, 0. Draw in the 30F.average air temperature line, 0. From the intersection of 0and 0 draw a line horizontally to the left scale, 0, and read257F. average heat transfer differential. Its at this 257differential that the maximum load of 4000#/hr. occurs.

Air heater operating at one quarter loadNow lets explore what happens at quarter-load. At quarter- __

the right,@. Further, at one-fourth load, the air temperature risewill be 20F. instead of 80F. at maximum load. The rise willbe from 50 to 70 for an average air temperature of 60F. Solets intersect line 0 with the average air line of 60, 0.Dropping down with line 0, we read a steam temperatureof 124 which has an accompanying pressure just above-13 psi. Now draw a line through the two points we havelocated 0. This line is the curve for the air heating system we

just described. 8 If we rise from 0 psi on the bottom scaleto where it intersects the System Curve and then @ move tothe left from this intersection, we find that 168 is theaverage heat transfer differential. The maximum load of4000#/hr. occurred at a differential of 257, so 168represents a load of 2600#/hr. ([168/257] x 4000) or 65% offull load. This means 2600#/hr. is the maximum load thatmust be handled by the trap with no steam pressure, onlywhatever hydraulic head is available, to push thecondensate through the trap. We check our selection with al/2 psi differential. So if there is a 15 drop from the coil tothe level of the trap orifice, there is sufficient hydraulic headto keep the coil free of condensate under all conditions.(Dont forget, though, there must be a low differentialvacuum breaker at the coil out/et to assure this drainage.)Complete drainage is necessary to prevent coil damage byfreezing, corrosion and/or water hammer. Lets consideranother possibility, at 5 psi back pressure due to anoverhead return line or simply back pressure due to flash.Now a safety drain is in order. @ Draw in the 5 psi steamline to where it intersects the System Curve 0. Moving to

the left from this intersection, we read that the average heattransfer differential is 186. This means 72% of themaximum load (186/257) or 2900#/hr. The safety drain musthandle this at 5 psi differential as well as the previouslycalculated 2600#/hr. at l/2 psi differential.Please bear in mind that the foregoingdeals with a more orless ideal situation. Any safety factors applied to heatexchange equipment will increase the pressure drop and

load, the average heat transfer differential (AT) will be one-fourth that at maximum load. One-fourth of 257is 64? So A\& aggravate the condensate drainage problem. Note: Thecoils actual capacitymust be considered in drawing thelocate this on the left scale and draw a horizontal line to curve, not the anticipated load.

T hr ee Ri ve rs , Mi ch ig an 49 09 3 4 40 0 He rs ta l, B el gi um

Telephone (616) 273-1415 Telephone (041) 640867

art 1124 8M 8/87-0

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TEMPERATURE OFChart 1124

360

W 320a

40 80

@Copyright 1980 Armstrong Machine Works

1

Pi FOR STfing the modul;t less than ma

120

All HEATINGd steam presswm capacity.

160 200 2

-14 -13 -12 -10 - 5 0 IO 20

PRESSURE OF STE

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