1. Solved Problems

Example 1:Thermally Developing Flow Consider the flow of a gas with density 1 kg/m\(^3\), viscosity \(1.5 \times 10^{-5}\) kg/(m.s), specific heat \(C_p = 846\) J/(kg.K) and thermal conductivity \(k = 0.01665\) W/(m.K), in a pipe of diameter \(D = 0.01\) m and length \(L = 1\) m, and assume the viscosity does not change with temperature. The Nusselt number for a pipe with \((L/D)\) ratio greater than 10 and Reynolds number greater than 20000 is given by \[\text{Nu} = 0.026 \ \text{Re}^{0.8} \ \text{Pr}^{1/3}\] While the Nusselt number for a laminar flow for Reynolds number less than 2100 and \((\text{Re} \ \text{Pr} \ D/L) < 10\) is \[\text{Nu} = 1.86\ [\text{Re} \ \text{Pr} \ (D/L) ]^{1/3}\] If the gas flows through the pipe with an average velocity of 0.1 m/s, the heat transfer coefficient is (GATE-2005)

(a) 0.68 W/(m\(^2\).K) (b) 1.14 W/(m\(^2\).K)
(c) 2.47 W/(m\(^2\).K)
(d) 24.7=w= (m\(^2\).K)


Solution: From the given data, \[\begin{aligned} \text{Re} &= \frac{Dv\rho}{\mu} = \frac{0.01\times0.1\times1}{1.5\times10^{-5}} = 66.7 \\ \text{Pr} &= \frac{C_p\mu}{k} = \frac{846\times1.5\times10^{-5}}{0.01665} = 0.76 \\ \text{Re}\cdot \text{Pr}\cdot (D/L) &= 66.7\times0.76\times(0.01/1) = 0.507 \end{aligned}\] Hence, using the expression of \(\text{Nu}\) valid for this condition, we get \[\text{Nu} = 1.86\ [\text{Re} \ \text{Pr} \ (D/L) ]^{1/3} = 1.86\times(0.507)^{1/3} = 1.483\] By definition, \(\text{Nu} = hD/k\). Therefore, \[h = \frac{\text{Nu}\times k}{D} = \frac{1.483\times0.01665}{0.01} = 2.47 \text{ W/(m$^2$.K) } \tag*{{c}}\]