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A counter flow heat exchanger is employed to cool 0.55 kg/s (Cp = 2.45 KJ/KgC) of oil from 115 C to 40 C by the use of water The inlet and outlet temperature of cooling water 15C and 75C respectively.

The overall heat transfer coefficient is expected to be 1450 W/m2C. Using NTU method, calculate the following:

i) The mass flow rate of water

ii) The effectiveness of the heat exchanger

iii) The surface area required.

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Given:

mh=0.55Kg⁄s,Cp=2.45KJ⁄KgC,T(h,1)=115℃,T(h,2)=40℃,Tc1=15℃,T(c,2)=75℃,

U=1450W(m2C)

The mass flow rate of water:

Q=ϵCminT(h.1)T(c,1)

To find mc,use energy balance equation

0.55× 2450 (115 - 40)= mc× 4186 (75 - 15)

mc = 0.40kg/s

Effectiveness of heatexchanger

Heat capacity rate of hot fluid: Ch=mh Cph=0.55×2.45=1.35KW⁄K

Heat capacity rate of cold fluid:Cc=mcCpc=0.40×4.186=1.67KW/K

$C_h …

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