Page 406 - Handbook of Energy Engineering Calculations

FIGURE 1 Temperature relations in typical parallel-flow and
counterflow heat exchangers.

For this parallel-flow exchanger, G = T – t = 382 – 60 = 322°F (179°C),
1
1
2
where 382°F (194°C) = the temperature of 200-lb/in (abs) (1379-kPa)
saturated steam, from a table of steam properties. Also, L = T – t = 382 –
2
2
120 = 262°F (145.6°C), where the condensate temperature = the saturated
steam temperature because there is no subcooling of the condensate. Then
LMTD = G – L/ln (G/L) = (322 – 262)/ln (322/262) = 290°F (16°C).

3. Compute the required heat-transfer area
2
Use the relation A = Q/U × LMTD, where A = required heat-transfer area, ft ;
2
U = overall coefficient of heat transfer, Btu/(ft · h · °F). Thus, A =