Water $(mass = 1.4kg/s, C_p= 4.187kJ/kg-K)$ is heated from $40^0C \ to \ 70^0C$ by an oil $(mass = 2kg/s, C_p= 1.9kJ/kg-K)$ entering at $110^0c$ in a counter flow heat exchanger. If overall heat transfer coefficient is $350W/m^2K$, calculate the surface area required.

ṁc=1.4 kg/s (Mass flow rate of cold fluid)

Cpc=4.187 kJ/kgK = 4187 J/kgK (Specific Heat of Cold Fluid)

tc1=40℃=313K (Entry temperature of cold fluid)

tc2=70℃=343K (Exit temperature of cold fluid)

ṁh=2 kg/s (Mass flow rate of Hot fluid)

Cph=1.9 kJ/kgK = 1900 kJ/kgK (Specific Heat of Hot fluid)

th1=110℃=383K (Entry Temperature of Hot fluid )

U=350W/m2K (Overall heat transfer coefficient)

Counter flow arrangement.

Find:

(1) A (Surface area)

Solution:

Energy balance equation

Heat rejected by Hot fluid = Heat gained by Cold fluid

ṁh.Cph.(th1-th2)=ṁc.Cpc.(tc2-tc1)

2×1900(383-th2)=1.4×4187.(343-313)

∴ th2=336.722K

Heat Transfer Rate

Q=ṁc.Cpc.(tc2-tc1)

Q=1.4×4187×(343-313)

Q=175854 W

Also,

Q=U.A.θm …(1)

where,

θm=θ1-θ2ln (θ1θ2) →Logarithmic Mean Temperature Difference(LMTD)

θm=(383-343)-(336.72-313)ln (383-343336.72-313)

θm=31.154

∴From (1)

175854=350×A×31.1