Question

A double-pipe parallel-flow heat exchanger is to heat water $\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( from \)25^{\circ} \mathrm{C}\( to \)60^{\circ} \mathrm{C}\( at a rate of \)0.2 \mathrm{~kg} / \mathrm{s}$. The heating is to be accomplished by geothermal water \(\left(c_{p}=4310 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) available at \(140^{\circ} \mathrm{C}\) at a mass flow rate of $0.3 \mathrm{~kg} / \mathrm{s}\(. The inner tube is thin-walled and has a diameter of \)0.8 \mathrm{~cm}$. If the overall heat transfer coefficient of the heat exchanger is \(550 \mathrm{~W} / \mathrm{m}^{2}\), \(\mathrm{K}\), determine the length of the tube required to achieve the desired heating.

Short answer

Answer: The required length of the tube is approximately 43.25 meters.

Step-by-step solution

Identify given values

We are given the following information: - the mass flow rate of cold water (\(\dot{m}_c=0.2 ~\text{kg/s}\)) - the mass flow rate of hot water (\(\dot{m}_h=0.3 ~\text{kg/s}\)) - specific heat at constant pressure of cold water (\(c_{p_c}=4180 ~\text{J/kg}\cdot\text{K}\)) - specific heat at constant pressure of hot water (\(c_{p_h}=4310 ~\text{J/kg}\cdot\text{K}\)) - the overall heat transfer coefficient of the heat exchanger (\(U=550 ~\text{W/m}^2\cdot\text{K}\)) - the diameter of the inner tube (\(D_i=0.008 ~\text{m}\), converted from 0.8 cm)

Calculate the heat transfer rate

We can find the heat transfer rate (Q) by using the mass flow rate and specific heat of the cold water stream before and after it reaches the desired final temperature: $$Q = \dot{m}_c c_{p_c} (T_{c, out} - T_{c, in})$$ where \(T_{c, in} = 25^\circ\text{C}\) and \(T_{c, out} = 60^\circ\text{C}\). $$Q = 0.2 ~\text{kg/s} \times 4180 ~\text{J/kg}\cdot\text{K} \times (60-25) ~\text{K} = 29380 ~\text{W}$$

Calculate change in hot water temperature

Since energy is conserved, the energy transferred to heat up the cold water must come from the hot water stream. Thus, we have $$Q = \dot{m}_h c_{p_h} (T_{h, in} - T_{h, out})$$ Solving for \(T_{h, out}\), we get: $$T_{h, out} = T_{h, in} - \frac{Q}{\dot{m}_h c_{p_h}}$$ $$T_{h, out} = 140^\circ\text{C} - \frac{29380 ~\text{W}}{0.3 ~\text{kg/s} \times 4310 ~\text{J/kg}\cdot\text{K}} = 123.94^\circ\text{C}$$

Logarithmic mean temperature difference

We compute the logarithmic mean temperature difference (LMTD) between the hot and cold water streams as follows: $$\Delta T_1 = T_{h, in} - T_{c, out}$$ $$\Delta T_2 = T_{h, out} - T_{c, in}$$ $$LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})}$$ $$LMTD = \frac{(140-60) - (123.94-25)}{\ln(\frac{140-60}{123.94-25})} = 64.43 ~\text{K}$$

Determine the length of the tube

Finally, we can use the heat transfer rate equation with the given overall heat transfer coefficient and the calculated LMTD to determine the required length of the inner tube: $$Q = U \cdot A \cdot LMTD$$ The area (A) can be calculated using the tube diameter: $$A = \pi D_i \cdot L$$ Substituting this area into the equation, we get: $$Q = U \cdot \pi D_i \cdot L \cdot LMTD$$ Solving for the length (L), we have: $$L = \frac{Q}{U \cdot \pi D_i \cdot LMTD}$$ $$L = \frac{29380 ~\text{W}}{550 ~\text{W/m}^2\cdot\text{K} \times \pi \times 0.008 ~\text{m} \times 64.43 ~\text{K}} = 43.25 ~\text{m}$$ The length of the tube required to achieve the desired heating is approximately 43.25 meters.