Acounter-flow double-piped heat exchange is to heat water from 20oc to 80oc at a rate of 1.2 kg/s. the heating is to be accomplished by geothermal available at 160oc 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 640 w/m2.oc, determine the length of the heat exchanger required to achieve the desired heating using the effectiveness-ntu method.

110 m or 11,000 cm

Explanation:

Mc = 1.2 kg/s and Mh = 2 kg/s

Cpc = 4.18 kj/kg °c and Cph = 4.31 kj/kg °c

Using effectiveness-NUT method

Cc = Mc × Cpc = 1.2 kg/s  × 4.18 kj/kg °c = 5.016 kW/°c

Ch = Mh × Cph = 2 kg/s  × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= Cmin/Cmax

C = 5.016/8.62 = 0.582

          .2 Second, we determine the maximum heat transfer rate, Qmax

Qmax = Cmin (Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Qmax = (5.016 kW/°c)(160 - 20) °c

Qmax = (5.016 kW/°c)(140) °c = 702.24 kW

          .3 Third, we determine the actual heat transfer rate, Q

Q = Cmin (outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Qmax = (5.016 kW/°c)(60) °c = 303.66 kW

            .4 Fourth, we determine Effectiveness of the heat exchanger, ε

ε = Q/Qmax

ε = 303.66 kW/702.24 kW

ε = 0.432

           .5 Fifth, using appropriate  effective relation for double pipe counter flow to determine NTU for the heat exchanger

NTU =

NTU =

NTU = 0.661

          .6 sixth, we determine Heat Exchanger surface area, As

From the question, the overall heat transfer coefficient U = 640 W/m²

As =

As =

As = 5.18 m²

            .7 Finally, we determine the length of the heat exchanger, L

L =

L =

L= 109.91 m

L ≅ 110 m = 11,000 cm

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