Page 405 - Handbook of Energy Engineering Calculations

1. Compute the heat-transfer rate of the heater
With a flow rate of 100 gal/min (6.3 L/s) or (100 gal/min)(60 min/h) = 6000
gal/h (22,710 L/h), the weight flow rate of the oil, using the weight of water
of specific gravity 1.0 as 8.33 lb/gal, is (6000 gal/h) (0.9 specific gravity)
(8.33 lb/gal) = 45,000 lb/h (20,250 kg/h), closely.

Since the temperature of the oil rises 120 – 60 = 60°F (33.3°C) during
passage through the heat exchanger and the oil has a specific heat of 0.50,
find the heat-transfer rate of the heater from the general relation Q = wc Δt,

where Q = heat-transfer rate, Btu/h; w = oil flow rate, lb/h; c = specific heat
of the oil, Btu/(lb · °F);Δt = temperature rise of the oil during passage
through the heater. Thus, Q = (45,000)(0.5)(60) = 1,350,000 Btu/h (0.4 MW).

2. Compute the heater logarithmic mean temperature difference
The LMTD is found from LMTD = (G – L)/ln (G/L), where G = greater

terminal temperature difference of the heater, °F; L = lower terminal
temperature difference of the heater, °F; ln = logarithm to the base e. This
relation is valid for heat exchangers in which the number of shell passes
equals the number of tube passes.

In general, for parallel flow of the fluid streams, G = T – t and L = T –
1
1
2
t , where T = heating fluid inlet temperature, °F; T = heating fluid outlet
2
1
2
temperature, °F; t = heated fluid inlet temperature, °F; t = heated fluid
2
1
outlet temperature, °F. Figure 1 shows the maximum and minimum terminal
temperature differences for various fluid flow paths.