Question

A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg/s. The heating is to be accomplished by geothermal water available at 160 degrees Celsius at a mass flow rate of 2 kg/s. The inner tube is thin-walled and has a diameter of 1.5 cm. if the overall heat transfer coefficient of the heat exchanger is 649 W/m2K, determine the length of the heat exchanger required to achieve the desired heating.

Answer 1

`L=107.6m`

Step-by-step explanation:

Cold water in:
`m_(c)=1.2kg/s, C_(c)=4.18kJ/kg\°C, T_(c,in)=20\°C, T_(c,out)=80\°C`

Hot water in:
`m_(h)=2kg/s, C_(h)=4.18kJ/kg\°C, T_(h,in)=160\°C, T_(h,out)=?\°C`

`D=1.5cm=0.015m, U=649W/m^(2)K, LMTD=?\°C, A_(s)=?m^(2),L=?m`

Step 1: Determine the rate of heat transfer in the heat exchanger

`Q=m_(c)C_(c)(T_(c,out)-T_(c,in))`

`Q=1.2*4.18*(80-20)`

`Q=1.2*4.18*(80-20)`

`Q=300.96kW`

Step 2: Determine outlet temperature of hot water

`Q=m_(h)C_(h)(T_(h,in)-T_(h,out))`

`300.96=2*4.18*(160-T_(h,out))`

`T_(h,out)=124\°C`

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

`dT_(1)=T_(h,in)-T_(c,out)`

`dT_(1)=160-80`

`dT_(1)=80\°C`

`dT_(2)=T_(h,out)-T_(c,in)`

`dT_(2)=124-20`

`dT_(2)=104\°C`

`LMTD = (dT_(2)-dT_(1))/(ln((dT_(2))/(dT_(1))))`

`LMTD = (104-80)/(ln((104)/(80)))`

`LMTD = (24)/(ln(1.3))`

`LMTD = 91.48\°C`

Step 4: Determine required surface area of heat exchanger

`Q=UA_(s)LMTD`

`300.96*10^(3)=649*A_(s)*91.48`

`A_(s)=5.07m^(2)`

Step 5: Determine length of heat exchanger

`A_(s)=piDL`

`5.07=pi*0.015*L`

`L=107.57m`