16.5 kg/s of the product at 6500C (Cp= 3.55 kJ/kg K), in a chemical plant are to be used to heat 20.5 kg/s of the incoming fluid from 1000C (Cp= 4.2 kJ/kg K). If the overall heat transfer coefficient is 0.95 kW/m2K and the installed heat transfer surface is 44 m2. Calculate the fluid outlet temperature for the counter flow and parallel flow arrangements.

Given,

ṁh=16.5 kg/s (Mass flow rate of Hot fluid)

th1=650℃=923K (Entry Temperature of Hot fluid )

Cph=3.55 kJ/kgK = 3550 kJ/kgK (Specific Heat of Hot fluid)

ṁc=20.5 kg/s (Mass flow rate of Cold Fluid)

tc1=100℃=373K (Entry temperature of cold fluid)

Cpc=4.2 kJ/kgK = 4200 J/kgK (Specific Heat of Cold Fluid)

U=0.95KW/m2K = 950W/m2K (Overall heat transfer coefficient)

A=44m2 (Common area between two fluids)

Find:

For counter flow and parallel flow arrangements

(1) th2 (Exit temperature of Hot fluid)

(2) tc2 (Exit temperature of Cold fluid)

Solution:

Case (1) Counter flow arrangement

Heat Capacity of Hot fluid

Ch=ṁh.Cph=16.5×3550

Ch=58575 J/kgK

Heat Capacity of Cold fluid

Cc=ṁc.Cpc=20.5×4200

Cc=86100 J/kgK

Heat Capacity Ratio

C===5857586100

C=0.68

Number of Transfer Units

NTU=U.ACmin=950×4458575

NTU=0.713

Effectiveness (For counter flow arrangement)

ε=1-e-NTU (1-C)1-C.e-NTU (1-C)=1-e-0.713(1-0.68)1-0.68×e-0.713(1-0.68)

ε=0.444

as Ch=Cmin

ε=th1-th2th1-tc1

∴ 0.444=923-th2923-373

∴ th2=678.8K=405.8℃

Energy balance equation

Heat rejected by Hot fluid = Heat gained by Cold fluid

ṁh.Cph.(th1-th2)=ṁc.Cpc.(tc2-tc1)

16.5×3550×(923-678.8)=20.5×4200×(tc2-373)

∴ tc2=539.132K=266.132℃

Case (2) Parallel flow arrangement

Effectiveness (For parallel flow arrangement)

ε=1-e-NTU (1+C)1+C=1-e-0.713(1+0.68)1+0.68

ε=0.415

also,

ε=th1-th2th1-tc1

∴ 0.415=923-th2923-373

∴ th2=694.75K=421.75℃

Energy balance equation

Heat rejected by Hot fluid = Heat gained by Cold fluid

ṁh.Cph.(th1-th2)=ṁc.Cpc.(tc2-tc1)

16.5×3550×(923-694.75)=20.5×4200×(tc2-373)

∴ tc2=528.281K=255.281℃