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UCN, URCN, UCF & URCF Series selection Example [A] Calculate surface area required.

Step 4. Calculate the Capacity Factor F c for steam.

Q Btu/lb

2,500,000 Btu/hr 912 Btu/lb

Heat 50gpm fresh water from 70 O F to 170 O F, using saturated steam at 50psig. T s = Steam temperature O F t in = Cold Side entering fluid O F t out = Cold Side exiting fluid O F Step 1. Calculate the heat load Btu/hr [Q].

F c

=

= 5,050 lbs/hr steam required

= lbs/hr

Step 5. Select the proper diameter heat exchanger using the capac- ity graphs F or G and F c from step 4. Capacity = 5050 = 2000 series with 5/8” Tubes. Step 6. Select the proper size heat exchanger from the performance curves corresponding to the series selected using the capacity factor. Select the heat exchanger closest to the line landing on or above the calculated point. F s = 34,470 Btu/hr f = URCF 2084-6-TP Application [C] Calculate batch heating of a tank. Heat a 1000 gallon stainless steel tank of water from 50 O F to 150 O F in 1.5 hours using 40psig saturated steam, circulating at 30gpm. Tank size 6ft w x 6ft h x 6ft d. Ambient air temperature 60 O F worse case.

Q = GPM x CN x  T Q = 50gpm x 500 x (170 O F-70 O F) = 2,500,000 Btu/hr

Step 2 Acquire steam temperature and enthalpy from graphs T&L

50psig = 297 O F steam. From graph T (pg. 110). 50psig = 912 Btu/lb. From graph L (pg. 111).

Step 3. Calculate the mean temperature difference (MTD)

T s - t in T s - t out

297 O F - 70 O F 297 O F - 170 O F

227 O F (Larger) L 127 O F (Smaller) S

=

=

Step 1. Calculate the total heat load [Q] Btu/hr.

127 O F 227 O F

S L

=

= .559 Go to Table A .559 = .758

Q = Total Gallons x lbs/gallon x Specific heat Btu/lb x r T Q = 1000 x 8.34 x 1.0 x 100 o F = 834,000 Btu

Find the LMTD = [.758 x (L) 227] = 172.0

Corrected Q for time

834,000 x 60 min (1.5hours) x 60 min

=

= 556,00 Btu/hr

Step 4. Calculate the surface area required.

Step 2. Calculate the r T average (T a ) for the heated water.

Q (Btu/hr) LMTD x U

2,500,000 172 x 300

A s

=

=

= 48.4 sq. ft.

T f - T i 2

150 O F - 50 O F

r T a =

+ T

T a =

o F

= 116.7

o F

i

1.5 hours + 50

Step 5. Calculate the Capacity Factor [F c ] for steam.

Q Loss = Surface area tank sq.ft x .001 x r t a x 2545 Q Loss = 6x6x6 x .001 x (116.7 o F a - 60

o F) x 2545 = 31,169 Btu/hr

Q Btu/lb

2,500,000 Btu/hr 912 Btu/lb

F c

=

=

= 2,741.3 lbs/hr steam

Q t = Q + Q Loss = Q t 556,000 + 31,169 = 587,169 Btu/hr Step 3. Derive the steam temperature [T s ] from graph T. Derive the capacity factor from graph L. Given 40psig saturated steam = 287 O F steam acquired from graph T. Given 40psig saturated steam = 920 Btu/lb acquired from graph L.

Step 6. Select a the proper diameter heat exchanger using graphs F or G and F c from step 5. Capacity = 2741.3 @ 50psig = 1700 series 3/8 or 5/8 tubes from chart (G).

Use table D to determine the final heat exchanger size. 48.4 sq.ft. = URCF-1748-6-TP

Step 4. Calculate the mean temperature difference (MTD)

T s - T a T s - T e

287 O F - 100 O F 287 O F - 150 O F

187 O F 137 O F

Application [B] Calculate using the graphs.

=

=

Heat 70gpm fresh water from 50 O F to 180 O F using 65psig satu- rated steam.

S L

137 O F 187 O F

=

= .732 Goto Table A. .732 = .659

Step 1. Calculate the heat load Btu/hr [Q].

Q = GPM x CN x r T Q = 70 x 500 x (180 O F - 50 O F) = 4,550,000 Btu/hr Step 2. Derive the steam temperature [T s ] from the graph T . De- rive the capacity factor [F c ] from graph L .

Calculate the Log mean temperature difference LMTD

LMTD = [.659 x (L) 187 O F] = 123.2

Step 5. Calculate the required surface area.

Q LMTD x U

587,169 Btu/hr 123.2 x 300

65psig = 312 O F steam. From graph T 65psig = 901 Btu/lb. From graph L Step 3. Calculate the F s required.

A s

=

=

= 15.9 sq. ft.

Btu/hr [Q] T s - t out

4,550,000 312 O F - 180 O F

F s

=

=

= 34,470

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

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