Question

Consider a 10-m-long smooth rectangular tube, with \(a=50 \mathrm{~mm}\) and \(b=25 \mathrm{~mm}\), that is maintained at a constant surface temperature. Liquid water enters the tube at \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.01 \mathrm{~kg} / \mathrm{s}\). Determine the tube surface temperature necessary to heat the water to the desired outlet temperature of $80^{\circ} \mathrm{C}$.

Short answer

Answer: The tube surface temperature necessary to heat the water to the desired outlet temperature is 63.7°C.

Step-by-step solution

Find the specific heat capacity of water.

The specific heat capacity of water is given as \(c_p = 4.18 kJ/(kg \cdot K)\). We will need this value to compute the energy transfer necessary to achieve the desired temperature difference.

Calculate the energy required to raise the water temperature.

We can determine the energy required using the formula: \(Q = m \cdot c_p \cdot \Delta T\), where \(Q\) is the energy transfer, \(m\) is the mass flow rate of water, \(c_p\) is the specific heat capacity of water, and \(\Delta T\) is the temperature difference. Given: \(m = 0.01 kg/s\), \(\Delta T = 80^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 60K\) Then, \(Q = 0.01 kg/s \cdot 4.18 kJ/(kg \cdot K) \cdot 60K = 2.508 kJ/s\)

Calculate the heat transfer rate per unit area.

We can calculate the heat transfer rate per unit area (\(q'\)) using the formula: \(q'= Q / A\), where \(A\) is the area of the tube's surface. The cross-sectional area is \(A_c = ab = 50mm \times 25mm = 0.05m \times 0.025m\). The area of the tube's surface can be calculated as \(A = 2(a + b) L\), where \(L\) is the length of the tube. Given \(L=10m\), the surface area \(A = 2(0.05m + 0.025m)\cdot 10m = 1.5m^2\). Then, \(q' = \dfrac{2.508 kJ/s}{1.5m^2} = 1.672 \dfrac{kJ}{s \cdot m^{2}}\)

Determine the tube surface temperature.

To find the surface temperature of the tube, we will use the convection heat transfer equation: \(q' = h \cdot (T_s - T_f)\), where \(h\) is the heat transfer coefficient, \(T_s\) is the surface temperature, and \(T_f\) is the average fluid temperature. As we don't have the value of \(h\), we will use the expression: \(h = Nu \cdot \dfrac{k}{D_h}\), where \(Nu\) is the Nusselt number (assumed to be 5 for this problem in laminar flow), \(k\) is the thermal conductivity of liquid water (\(0.609 W/m\cdot K\)), and \(D_h\) is the hydraulic diameter, given by \(D_h = \dfrac{4A_c}{P}\), where \(P\) is the perimeter. Calculating \(D_h\) and \(h\), we get \(D_h = \dfrac{4\cdot0.05m\cdot0.025m}{0.1m + 0.05m} = 0.025m\) and \(h = 5\cdot \dfrac{0.609 W/m\cdot K}{0.025m} = 121.8 \dfrac{W}{m^2 \cdot K}\). Finally, using the convection equation, we get: \(T_s = T_f + \dfrac{q'}{h}\) Considering the average fluid temperature, \(T_f = \dfrac{20^{\circ} \mathrm{C} + 80^{\circ} \mathrm{C}}{2} = 50^{\circ} \mathrm{C}\), we get: \(T_s = 50^{\circ} \mathrm{C} + \dfrac{1.672 \dfrac{kJ}{s \cdot m^2} \cdot 1000 \dfrac{J}{kJ}}{121.8 \dfrac{W}{m^2 \cdot K}} = 50^{\circ} \mathrm{C} + 13.7^{\circ} \mathrm{C} = 63.7^{\circ} \mathrm{C}\). Therefore, the tube surface temperature necessary to heat the water to the desired outlet temperature of \(80^{\circ} \mathrm{C}\) is \(63.7^{\circ} \mathrm{C}\).