In an oil cooler, oil $(m= 2500kg/hr and C_p= 1.9 kJ/kg-K)$ at $160^0C$ is cooled by water$(m= 1500kg/hr \ and \ C_p= 4.187 kJ/kg-K)$ entering at $35^0C$. Determine Capacity ratio, NTU and effectiveness if the overall heat transfer coefficient is $300 W/m^2K$. Assume parallel flow.

Given,

ṁh=2500 kg/hr =0.694 kg/s (Mass flow rate of Hot fluid)

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

th1=160℃=433K (Entry Temperature of Hot fluid )

ṁc=1500 kg/hr = 0.416 kg/s (Mass flow rate of Cold Fluid)

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

tc1=35℃=308K (Entry temperature of cold fluid)

U=300W/m2K ( Overall heat transfer coefficient)

A=1m2 (Common area between two fluids)

Find:

(1) C (Capacity Ratio)

(2) NTU (Number of transfer unit)

(3) ε (Effectiveness)

Solution:

Heat Capacity of Hot fluid

Ch=ṁh.Cph=0.694×1900

Ch=1318.6 J/kgK

Heat Capacity of Cold fluid

Cc=ṁc.Cpc=0.416×4187

Cc=1741.792 J/kgK

Heat Capacity Ratio

C===1318.61741.792

C=0.757

Number of Transfer Units

NTU=U.ACmin=300×11318.6

NTU=0.227

Effectiveness (For parallel flow arrangement)

ε=1-e-NTU (1+C)1+C=1-e-0.227(1+0.757)1+0.757

ε=0.187