Question
Asked Sep 22, 2019
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Cooling Oil by Water in an Exchanger. Oil flowing at the rate of 5.04 kg/s (cpm
2.09 kJ/kg K) is cooled in a 1-2 heat exchanger from 366.5 K to 344.3 K by 2.02 kg/s
of water entering at 283.2 K.The overall heat-transfer coefficient U, is 340 W/m2. K.
Calculate the area required. (Hint: A heat balance must first be made to determine
the outlet water temperature.)
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Cooling Oil by Water in an Exchanger. Oil flowing at the rate of 5.04 kg/s (cpm 2.09 kJ/kg K) is cooled in a 1-2 heat exchanger from 366.5 K to 344.3 K by 2.02 kg/s of water entering at 283.2 K.The overall heat-transfer coefficient U, is 340 W/m2. K. Calculate the area required. (Hint: A heat balance must first be made to determine the outlet water temperature.)

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Expert Answer

Step 1

Heat exchanger is a mechanical device which is used to transfer heat between two fluids. One fluid is heated or cooled with the help of another fluid. Given heat exchanger is 1-2(one shell and two tubes) heat exchanger. These numbers show number of shells and number of tubes respectively.

Step 2

Applying energy balance on heat exchanger to find out outlet temperature of water

Amount of heat lose by hot fluid = amount of heat gained by cold fluid

Where TH1 = Inlet temperature of hot (oil) fluid

             TH2 =Outlet temperature of hot (oil) fluid

             TC1 = Inlet temperature of cold (water) fluid

             TC2= Outlet temperature of cold (water) fluid

              m∙oil = mass flow rate of oil

               m∙water = mass flow rate of water

               Cp(oil) =specific heat of oil

              Cp(water) =specific heat of water

                               = 4.186 kJ/kg ∙ K

 

Step 3

Plugging the ...

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,(TH-TH) .(Tc,-Te) Т. -Т. Pol С. Pwater m, oil water kg 5.04 x2.09 kg kJ -x(366.5-344.3)K-2.02x4.186 kg.K kJ -x(T -283.2)K S ) 233.845-8.455x(T -283.2 233.845 (Tc -283.2 8.455 =27.655 Te310.855K

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