Question

Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be heated by solarheated hot air $\left(c_{p}=1010 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ in a double-pipe counterflow heat exchanger. Air enters the heat exchanger at \(90^{\circ} \mathrm{C}\) at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\), while water enters at $22^{\circ} \mathrm{C}\( at a rate of \)0.1 \mathrm{~kg} / \mathrm{s}$. The overall heat transfer coefficient based on the inner side of the tube is given to be $80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. The length of the tube is \)12 \mathrm{~m}\(, and the internal diameter of the tube is \)1.2 \mathrm{~cm}$. Determine the outlet temperatures of the water and the air.

Short answer

Based on the given information and calculations, the outlet temperatures of water and air in the double-pipe counterflow heat exchanger are: - Water outlet temperature: \(27.774^{\circ}\mathrm{C}\) - Air outlet temperature: \(82.039^{\circ}\mathrm{C}\)

Step-by-step solution

Calculate the heat capacity rates for water and air

Calculate the heat capacity rates (C) for both water and air using the formula: \(C = m \times c_p\), where \(m\) is the mass flow rate and \(c_p\) is the specific heat capacity. For water: \(C_w = m_w \times c_{p_{w}} = 0.1 \textrm{ kg/s} \times 4180 \textrm{ J/kgK} = 418 \textrm{ W/K}\) For air: \(C_a = m_a \times c_{p_{a}} = 0.3 \textrm{ kg/s} \times 1010 \textrm{ J/kgK} = 303 \textrm{ W/K}\)

Determine the minimum heat capacity rate and the heat capacity rate ratio

Identify the fluid with the minimum heat capacity rate (C_min) and calculate the heat capacity rate ratio (R) using the formula: \(R = \frac{C_{min}}{C_{max}}\) In this case, \(C_w < C_a\), so fluid with the minimum heat capacity rate is water. \(C_{min} = C_w = 418 \mathrm{~W/K}\) Now, calculate R: \(R = \frac{C_{w}}{C_{a}} = \frac{418}{303} = 1.3792\)

Calculate the heat transfer area

Calculate the heat transfer area (A) using the length and internal diameter of the tube. The formula for the heat transfer area of a tube is \(A = \pi d L\), where \(d\) is the internal diameter and \(L\) is the length of the tube. \(A = \pi(0.012 \mathrm{~m})(12 \mathrm{~m}) = 0.453 \mathrm{~m^2}\)

Calculate the NTU value

Calculate the Number of Transfer Units (NTU) using the formula: \(NTU = \frac{UA}{C_{min}}\), where U is the overall heat transfer coefficient, and A is the heat transfer area. \(NTU = \frac{80 \mathrm{~W/m^2K} \times 0.453 \mathrm{~m^2}}{418 \mathrm{~W/K}} = 0.0864\)

Determine the effectiveness

For a double-pipe counterflow heat exchanger, we can use the formula for calculating effectiveness: \(\epsilon = \frac{1 - e^{-NTU(1-R)}}{1 - R e^{-NTU(1-R)}}\) Substitute the values we obtained in the previous steps: \(\epsilon = \frac{1 - e^{-0.0864(1-1.3792)}}{1 - 1.3792 e^{-0.0864(1-1.3792)}} = 0.0844\)

Calculate the heat transfer

Calculate the heat transfer rate (Q) using the formula: \(Q = \epsilon \times C_{min} \times (\Delta T)_{max}\) In this case, the maximum possible temperature change is the difference in inlet temperatures of air and water: \((\Delta T)_{max} = T_{inlet_{air}} - T_{inlet_{water}} = 90^{\circ}C - 22^{\circ}C = 68 \mathrm{~K}\) \(Q = 0.0844 \times 418 \mathrm{~W/K} \times 68 \mathrm{~K} = 2412.312 \mathrm{~W}\)

Determine the outlet temperatures

Calculate the outlet temperature of water (T_outlet_water) using the formula: \(T_{outlet_{water}} = T_{inlet_{water}} + \frac{Q}{C_w}\) \(T_{outlet_{water}} = 22^{\circ}C + \frac{2412.312 \mathrm{~W}}{418 \mathrm{~W/K}} = 22^{\circ}C + 5.774 \mathrm{~K} = 27.774^{\circ}\mathrm{C}\) Calculate the outlet temperature of air (T_outlet_air) using the formula: \(T_{outlet_{air}} = T_{inlet_{air}} - \frac{Q}{C_a}\) \(T_{outlet_{air}} = 90^{\circ}C - \frac{2412.312 \mathrm{~W}}{303 \mathrm{~W/K}} = 90^{\circ}C - 7.961 \mathrm{~K} = 82.039^{\circ}\mathrm{C}\) The outlet temperatures of water and air are \(27.774^{\circ}\mathrm{C}\) and \(82.039^{\circ}\mathrm{C}\), respectively.