Fuel oil is to be heated in a heat exchanger using steam under the following con

Question

Fuel oil is to be heated in a heat exchanger using steam under the following conditions: Inlet temperature of oil = 40°C. Outlet temperature of oil = 120°C. Flow rate of oil = 420 kg h–1. Mean specific heat capacity of oil = 2.05 kJ kg–1 K–1. Temperature of steam = 165°C (dry saturated). Temperature of steam condensate = 130°C 8% of the heat is lost to the atmosphere (i.e. process is 92% efficient). (a) Using steam tables, find: (i) the specific latent heat of vaporization of steam (hfg) at 165°C in kJ kg–1 (ii) the difference in enthalpy of liquid water (hf) between 165°C and 130°C in kJ kg–1. (b) Calculate the quantity of this steam required per hour to perform this duty.

Explanation / Answer

heat duty = mass flow rate of oil* specific heat of oil* temperature difference= 420 kg/hr*2.05 Kj/Kg.K*(120-40)=68880 KJ/hr

There is 8% heat losses, So heat to be supplied = 68880/0.92= 74870 Kj/hr

Enthalpy og steam at 165 deg.c and saturated conditions = 2570.4 Kj/Kg

Assuming the steam is leaving as saturated liquid at 130 deg.c =546 Kj/Kg

at 165 deg.c, it it is liquid , the enthalpy is =695 Kj/Kg

hence the difference= (695-546)= 149 Kj/Kg

Change in enthalpy of steam during condensation, change in enthalpy = 2570.4-546=2024.4 Kj/Kg

steam required = heat duty/ Change in enthalpy of steam = 74870/2024.4 Kg/hr-36.98 kg/hr

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