In the previous problem, it may be necessary to scale up or scale down by as much as \(50 \%\) from the original operating conditions. Prepare a performance curve similar to the one in Figure 21. 1 for operation of the condenser over this range.

Figure 21.1

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Tout = 300C Steam, mkg/h Flow Ratio: M TB=253C 6.0 PB = 42 bar 1.3 =253C 5.5 = m(kg/h) 55,300M 1.2 1.1 Steam Temperature = 253C (sat.) 5.0 1.0 Q (GJ/h) Area: A 22.6 m 0.9 4.5. 4.0 Light Oil Duty, 3.78 m = 55,300 kg/h Tin=325C 3.5 3.43 3.0 Boiler Feed Water m = 1280 kg/h 2.5 Tw= 27C Equipment Data: 2.0 310 320 0.9 Total Heat Transfer Area: A = 25.1 m Fraction of Tubes Covered: L/L Area Covered: A= 25.1 (0.9) = 22.6 m Thermodynamic Data: Density: p 750 kg/m Light Oil Heat Capacity: 2.48 kJ/kg C H-Hw=H (steam out) - H (water in) = 2680 kJ/kg 325 332 330 340 350 360 Temperature. T (C) QmCAT 55,300 (2.48) (325-300) = 3.43 GJ/h 9.53 x 10W AT im = U=Q/AAT = Im (300C-253C)-(325C-253C) In [(300C-253C)/(325C -253C}] = 58.6C = 3.43 x 10/(22.6)/(58.6) = 2590 kJ/h m2 c m=Q(H-H)= 3.43 x 10/2680 1280 kg/h = 719 W/m C