Question

A process fluid having a specific heat of 3500 J/kg∙K and flowing at 2 kg/s is to be cooled from 80 °C to 50 °C with chilled water, which is supplied at a temperature of 15 °C and a flow rate of 2.5 kg/s.

Assuming an overall heat transfer coefficient of 2000 W/m2∙K, calculate the required heat transfer areas for the following exchanger configurations:

(a) Parallel flow;

(b) Counter flow;

(c) a 1-2 shell and tube exchanger with the water on the shell side.

Answer 1

a)
`A_(s)=5.02m^(2)`

b)
`A_(s)=4.75m^(2)`

c)
`A_(s)=4.87m^(2)`

Step-by-step explanation:

Water in:
`m_(w)=2.5kg/s, C_(w)=4.2kJ/kg\°C, T_(w,in)=15\°C, T_(w,out)=?\°C`

Process fluid in:
`m_(pf)=2kg/s, C_(pf)=3.5kJ/kg\°C, T_(pf,in)=80\°C, T_(pf,out)=50\°C`

`U=2000W/m^(2)K, LMTD=?\°C, A_(s)=?m^(2)`

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

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

`Q=2*3.5*(80-50)`

`Q=420kW`

Step 2: Determine outlet temperature of water

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

`420=2.5*4.2*(15-T_(w,out))`

`T_(w,out)=25\°C`

Part a) Parallel Flow

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

`dT_(1)=T_(pf,in)-T_(w,in)`

`dT_(1)=80-15`

`dT_(1)=65\°C`

`dT_(2)=T_(pf,out)-T_(w,out)`

`dT_(2)=50-25`

`dT_(2)=25\°C`

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

`LMTD = (65-25)/(ln((65)/(25)))`

`LMTD = (40)/(ln(2.6))`

`LMTD = 41.86\°C`

Step 4: Determine required surface area of heat exchanger

`Q=UA_(s)LMTD`

`420*10^(3)=2000*A_(s)*41.86`

`A_(s)=5.02m^(2)`

Part b) Counter Flow

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

`dT_(1)=T_(pf,in)-T_(w,out)`

`dT_(1)=80-25`

`dT_(1)=55\°C`

`dT_(2)=T_(pf,out)-T_(w,in)`

`dT_(2)=50-15`

`dT_(2)=35\°C`

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

`LMTD = (55-35)/(ln((55)/(35)))`

`LMTD = (20)/(ln(1.57))`

`LMTD = 44.25\°C`

Step 4: Determine required surface area of heat exchanger

`Q=UA_(s)LMTD`

`420*10^(3)=2000*A_(s)*44.25`

`A_(s)=4.75m^(2)`

Part c) 1-2 shell and tube exchanger with the water on the shell side

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

`dT_(1)=T_(pf,in)-T_(w,out)`

`dT_(1)=80-25`

`dT_(1)=55\°C`

`dT_(2)=T_(pf,out)-T_(w,in)`

`dT_(2)=50-15`

`dT_(2)=35\°C`

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

`LMTD = (55-35)/(ln((55)/(35)))`

`LMTD = (20)/(ln(1.57))`

`LMTD = 44.25\°C`

Step 4: Determine LMTD Correction Factor and Corrected LMTD

`P=(T_(pf,out)-T_(pf,in))/(T_(w,in)-T_(pf,in))`

`P=(50-80)/(15-80)`

`P=(-30)/(-65)`

`P=0.462`

`R=(T_(w,in)-T_(w,out))/(T_(pf,out)-T_(pf,in))`

`R=(15-25)/(50-80)`

`R=(-10)/(-30)`

`R=0.333`

Correction factor F can be determined from attached graph using P and F values calculated above

`F=0.975`

`LMTD_(corrected)=F*LMTD`

`LMTD_(corrected)=0.975*44.25`

`LMTD_(corrected)=43.14\°C`

Step 5: Determine required surface area of heat exchanger

`Q=UA_(s)LMTD_(corrected)`

`420*10^(3)=2000*A_(s)*43.14`

`A_(s)=4.87m^(2)`