Question

In a manufacturing plant that produces cosmetic products, glycerin is being heated by flowing through a \(25-\mathrm{mm}-\) diameter and 10 -m-long tube. With a mass flow rate of \(0.5 \mathrm{~kg} / \mathrm{s}\), the flow of glycerin enters the tube at \(25^{\circ} \mathrm{C}\). The tube surface is maintained at a constant surface temperature of \(140^{\circ} \mathrm{C}\). Determine the outlet mean temperature and the total rate of heat transfer for the tube. Evaluate the properties for glycerin at \(30^{\circ} \mathrm{C}\).

Short answer

Answer: The outlet mean temperature is 56.1°C, and the total heat transfer rate for the tube is 13570 W.

Step-by-step solution

Evaluate the properties of glycerin

At 30°C, the properties of glycerin are: Density: \(\rho = 1260 \mathrm{~kg/m^3}\) Viscosity: \(\mu = 0.92 \times 10^{-3} \mathrm{~kg/(m \cdot s)}\) Specific heat capacity: \(c_{\mathrm{p}}= 2420\mathrm{~J/(kg \cdot K)}\) Thermal conductivity: \(k= 0.293\mathrm{~W/(m \cdot K)}\)

Determine the flow velocity of glycerin

We are given the mass flow rate (\(\dot{m}=0.5\mathrm{~kg/s}\)) and the diameter of the tube (\(D=25\mathrm{~mm}\)). We need to determine the flow velocity (\(v\)) for glycerin in the tube. The following formula can be used: \(v = \frac{\dot{m}}{\rho \cdot A}\) where \(A\) is the cross-sectional area of the tube, which is: \(A = \frac{\pi D^2}{4}\) \(v = \frac{0.5}{1260 \cdot \frac{\pi (0.025)^2}{4}}\) \(v \approx 1.01\mathrm{~m/s}\)

Calculate the Reynolds number

To determine the flow regime, we need to calculate the Reynolds number (\(Re\)): \(Re = \frac{\rho v D}{\mu}\) \(Re = \frac{1260 \cdot 1.01 \cdot 0.025}{0.92 \times 10^{-3}}\) \(Re \approx 34430\) Since the Reynolds number is greater than 2900, the flow is turbulent.

Calculate the Nusselt number

For turbulent flow, we can use the Gnielinski correlation for the Nusselt number: \(Nu = \frac{(f/8) (Re - 1000) Pr}{1 + 12.7\sqrt{f/8}(Pr^{2/3} - 1)}\) where \(f\) is the friction factor, given by the Petukhov equation: \(f = 0.0791 Re^{-0.25}\) \(Pr = \frac{c_{\mathrm{p}} \mu}{k} = \frac{2420 \times 0.92 \times 10^{-3}}{0.293} = 7.61\) \(f = 0.0791 (34430)^{-0.25} \approx 0.0089\) We can now plug in all the values to calculate the Nusselt number: \(Nu = \frac{(0.0089/8) (34430 - 1000) \cdot 7.61}{1 + 12.7\sqrt{0.0089/8}(7.61^{2/3} - 1)} \approx 108\)

Determine the heat transfer coefficient

Now we can determine the heat transfer coefficient (\(h\)) using the Nusselt number: \(h = \frac{k}{D} Nu\) \(h = \frac{0.293}{0.025} \cdot 108\) \(h \approx 1269\mathrm{~W/(m^2 \cdot K)}\)

Calculate the heat transfer rate

We can now calculate the total heat transfer rate (\(Q\)) using the heat transfer coefficient and the given temperature difference: \(Q = h A (T_{\mathrm{s}} - T_{\mathrm{in}}) \Delta x\) where \(\Delta x = 10\mathrm{~m}\) (tube length), \(T_{\mathrm{s}} = 140^\circ\mathrm{C}\) (surface temperature), and \(T_{\mathrm{in}} = 25^\circ\mathrm{C}\) (inlet temperature). \(Q = 1269 \cdot \frac{\pi(0.025)^2}{4} (140 - 25) \cdot 10\) \(Q \approx 13570\mathrm{~W}\)

Determine the outlet temperature

Finally, we can determine the outlet mean temperature (\(T_{\mathrm{out}}\)) by using the heat transfer rate, mass flow rate, and specific heat capacity: \(Q = \dot{m} c_\mathrm{p} (T_\mathrm{out} - T_\mathrm{in})\) \(T_\mathrm{out} = T_\mathrm{in} + \frac{Q}{\dot{m} c_\mathrm{p}}\) \(T_\mathrm{out} = 25 + \frac{13570}{0.5 \cdot 2420}\) \(T_\mathrm{out} \approx 56.1^\circ\mathrm{C}\) Thus, the outlet mean temperature is \(56.1^\circ\mathrm{C}\), and the total heat transfer rate for the tube is \(13570\mathrm{~W}\).