A counter flow heat exchanger is used to heat water from 20

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.

Solution

Heat capacity of cold fluid Cc = m_cold*Cp_cold

= 1.2 * 4.18

Cc = 5.016 kW / K

Heat capacity of hot fluid Ch = m_hot * Cp_hot

= 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_cold_out - T_cold_in)

q = 5.016* (80 - 20)

q = 300.96 kW

Max possible heat transfer rate q_max = Cmin * (T_hot_in - T_cold_in)

= 5.016* (160 - 20)

= 702.24 kW

Effectiveness e = q / q_max

= 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