Question

Hot combustion gases of a furnace are flowing through a concrete chimney \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of rectangular cross section. The flow section of the chimney is $20 \mathrm{~cm} \times 40 \mathrm{~cm}\(, and the thickness of the wall is \)10 \mathrm{~cm}$. The average temperature of the hot gases in the chimney is \(T_{i}=280^{\circ} \mathrm{C}\), and the average convection heat transfer coefficient inside the chimney is \(h_{l}=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The chimney is losing heat from its outer surface to the ambient air at $T_{0}=15^{\circ} \mathrm{C}\( by convection with a heat transfer coefficient of \)h_{o}=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and to the sky by radiation. The emissivity of the outer surface of the wall is \(\varepsilon=0.9\), and the effective sky temperature is estimated to be \(250 \mathrm{~K}\). Using the finite difference method with \(\Delta x=\Delta y=10 \mathrm{~cm}\) and taking full advantage of symmetry, \((a)\) obtain the finite difference formulation of this problem for steady two-dimensional heat transfer, (b) determine the temperatures at the nodal points of a cross section, and \((c)\) evaluate the rate of heat loss for a \(1-m\)-long section of the chimney.

Short answer

Question: Determine the temperatures at the nodal points of a cross section and evaluate the rate of heat loss for a 1-meter-long section of the chimney. Answer: To determine the temperatures at the nodal points of a cross section (T_A and T_B), we need to solve the system of two nonlinear equations obtained in Step 2 using iterative methods like the Newton-Raphson method or employing a numerical solver. Once we have the nodal temperatures, we can calculate the convection and radiation heat losses separately and then sum them to find the total heat loss for a 1-meter-long section of the chimney, as demonstrated in Step 4. The total heat loss is given by q_total = q_conv1 + q_conv2 + q_rad.

Step-by-step solution

Set up the geometry and mesh

Using the symmetry of the problem, we will analyze only one quarter of the chimney's cross-section. Since the flow section of the chimney is \(20 \mathrm{~cm} \times 40 \mathrm{~cm}\) and the wall thickness is \(10 \mathrm{~cm}\), the dimensions of the one-quarter area are \(15 \mathrm{~cm} \times 20 \mathrm{~cm}\). Divide this area into a \(2 \times 2\) grid, resulting in the equal mesh spacings \(\Delta x = \Delta y = 10 \mathrm{~cm}\).

Obtain the finite difference formulation

Using the finite difference method for steady two-dimensional heat transfer, we can write the energy balance for the inner node, node A, as: $$ \frac{T_B - 2T_A + \left[\frac{k}{h_o \Delta x} + \frac{k}{h_l \Delta y}\right](T_{i}-T_0)}{(\Delta x)^2} = 0 $$ For the corner node, node B, the energy balance will include conduction, convection, and radiation. Radiation heat transfer is governed by Stefan-Boltzmann Law: $$ q_{rad} = \varepsilon \sigma (T_{s}^{4} - T_{sky}^{4}) $$ where \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_s\) and \(T_{sky}\) are the surface temperature of the wall and the sky temperature, respectively. Now, we can write the energy balance for node B as: $$\frac{T_C - 2T_B + T_A + \frac{k}{h_o \Delta x}(T_{i}-T_0) + \frac{2k\varepsilon\sigma}{h_o}(T_{s}^{3} - T_{sky}^{3})\Delta T}{(\Delta x)^2} = 0 $$

Solve the discretized equations for nodal temperatures

With the established energy balance equations for nodes A and B, we have a system of two nonlinear equations. We can solve them for the nodal temperatures \(T_A\) and \(T_B\) using iterative methods such as the Newton-Raphson method or by employing a numerical solver.

Calculate heat loss

We now know the nodal temperatures \(T_A\) and \(T_B\), which allows us to calculate the total heat loss for a 1-meter-long section of the chimney through convection and radiation. First, find the convection heat loss from the inner wall surface to the hot combustion gas, assuming the average temperature, \(T_{avg}\), for node A as the surface temperature: $$ q_{conv1} = h_l A_{conv1} (T_{i} - T_{avg}) $$ where \(A_{conv1}\) is the inner wall surface area of the 1-meter-long section. Next, find the convection heat loss from the outer wall surface to the ambient air for node B, again assuming \(T_{avg}\) as the surface temperature: $$ q_{conv2} = h_o A_{conv2} (T_{avg} - T_0) $$ where \(A_{conv2}\) is the outer wall surface area of the 1-meter-long section. Finally, to find the radiation heat loss from the outer wall surface, use the Stefan-Boltzmann Law: $$ q_{rad} = \varepsilon\sigma A_{rad}(T_{avg}^{4} - T_{sky}^{4}) $$ where \(A_{rad}\) is the radiation surface area of the 1-meter-long section. Adding these three components, we get the total heat loss for a 1-meter-long section of the chimney: $$ q_{total} = q_{conv1} + q_{conv2} + q_{rad} $$