A double pipe heat exchanger consisting of a 1 inch schedule

A double pipe heat exchanger consisting of a 1 inch schedule 40 pipe inside of a 3 inch pipe is used to heat 2.5 kg/s of olive oil from 20 degree C to 60 degree C. Geothermal water is available at 85 oC and 2 kg/s as the heating medium. The heat transfer coefficient on inside of pipe is 1000 W/m^2 C and 700 W/m^2 C on outside of pipe. Determine the area of heat exchanger required for this process.

Solution

Tc1= 20°©, Tc2= 60°©, Th1= ?, Th2=85°©

Cpc= 1.97 kj/Kg C, Cph= 4.2 kj/Kg C,

Mc= 2.5 kg/s, Mh= 2 kg/s, Hi = 1000 W/m2°C, Ho = 700 W/m2°C

The heat transfer rate= Q = Mc * Cpc * (Tc2 - Tc1)

   = 2.5*1.97*(60-20) = 197 kj/s

Heat lost by hot geotehermal water = Heat gained by olive oil

Mh * Cph * (Th1 - Th2) = Mc * Cpc * (Tc2 - Tc1)

= 2*4.2*(Th1-85) = 2.5* 1.97 * (60-20)

= Th1 = 108.5 °C

Logarithmic mean temperature difference given by =

LMTD = (Th1 - Tc1) - ( Th2 - Tc2)/ ln [(Th1 - Tc1)/(Th2 - Tc2)]

= (108.5 - 20) - (85 - 60) / ln [(108.5 - 20)/(85-60)]

= (88.5-25)/ln(88.5/25)

= 63.5/ln(3.54)

= 50.23°C

Overall heat transfer coefficient is given by U

= 1/U = ( 1/Hi) + (1/Ho)

= 1/U = (1/1000) + (1/700)

= U = 413 W/m2°C

We know that Q = U* A* LMTD

= 197* 1000 = 413 * A * 50.23

=A = 9.49 m2 ( Answer)