A counter flow heat exchanger is used to heat water from 20 degree C to 80 degre
ID: 2327099 • Letter: A
Question
A counter flow heat exchanger is used to heat water from 20 degree C to 80 degree C at a rate of 1.2 kg/s The heating is obtained by geothermal water available at 160 degree C at a mass flow rate of 2 kg/s. The inner tube is thin walled and has a diameter of 1.5 cm. The overall heat transfer coefficient is 640 W/m^2^+ K. Calculate the length of the heat exchanger required to achieve the desired heating by using effectiveness-NTU method. Take the specific heat of the geothermal water as 4.31 kj/kr. "K and that of ground water as 4.18 kj/kg. "K.Explanation / Answer
Heat capacity of cold fluid Cc = mcold*Cpcold
= 1.2 * 4.18
Cc = 5.016 kW / K
Heat capacity of hot fluid Ch = mhot * Cphot
= 2 * 4.31
Ch = 8.62 kW / K
Cmin = min (Ch, Cc) = 5.016 kW / K
Cmax = max (Ch, Cc) = 8.62 kW / K
Heat capacity ratio Cr = Cmin / Cmax
= 5.016 / 8.62
Cr = 0.5819
Heat transfer rate q = Cc * (T_coldout - T_coldin)
q = 5.016* (80 - 20)
q = 300.96 kW
Max possible heat transfer rate qmax = Cmin * (T_hotin - T_coldin)
= 5.016* (160 - 20)
= 702.24 kW
Effectiveness e = q / qmax
= 300.96 / 702.24
= 0.4286 or 42.86%
For a counterflow heat exchanger,
Effectiveness e = [1 - e(-NTU*(1 - Cr))] / [1 - Cr * e(-NTU*(1 - Cr))]
0.4286 = [1 - e(-NTU*(1 - 0.5819))] / [1 - 0.5819 * e(-NTU*(1 - 0.5819))]
Solving this we get, NTU = 0.652
Now, NTU = UA / Cmin
0.652 = 640 * A / 5.016*103
Heat transfer Area A = 5.11 m2
A = pi*D*L
5.11 = 3.14 * 1.5*10-2 * L
Length L = 108.49 m
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