In a combined heat and power (CHP) generation process, by-product heat is used for domestic or industrial heating purposes. Hot steam is carried from a CHP generation plant by a tube with diameter of \(127 \mathrm{~mm}\) centered at a square crosssection solid bar made of concrete with thermal conductivity of \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface temperature of the tube is constant at \(120^{\circ} \mathrm{C}\), while the square concrete bar is exposed to air with temperature of \(-5^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the temperature difference between the outer surface of the square concrete bar and the ambient air is to be maintained at \(5^{\circ} \mathrm{C}\), determine the width of the square concrete bar and the rate of heat loss per meter length.

Step by step solution

01

Write down the given information

The given information is as follows: - Tube diameter: \(D = 127\;\text{mm}\) - Surface temperature of tube: \(T_s = 120^{\circ}\, \text{C}\) - Ambient air temperature: \(T_{\infty} = -5^{\circ}\, \text{C}\) - Convection heat transfer coefficient: \(h = 20\displaystyle\frac{W}{m^2\cdot K}\) - Concrete bar thermal conductivity: \(k = 1.7\displaystyle\frac{W}{m\cdot K}\) - Temperature difference between outer surface and air: \(\Delta T = 5^{\circ}\, \text{C}\)

02

Find the middle temperature of the bar

Since we know the difference in temperature across the concrete bar is \(5^{\circ}\, \text{C}\), we need to find the midway temperature between the outer surface and the ambient air: \(T_m = T_{\infty} + \Delta T = -5 + 5 = 0^{\circ}\, \text{C}\)

03

Calculate the heat transfer rate by convection

We can find the heat transfer rate by convection between the outer surface of the concrete bar and the air using Newton's Law of cooling: \(q''_{conv} = h (T_m - T_{\infty}) = 20 (0 - (-5)) = 100\displaystyle\frac{W}{m^2}\)

04

Apply Fourier's Law of conduction for the concrete

We can find the heat transfer rate by conduction within the concrete bar using Fourier's Law of conduction: \(q''_{cond} = k \displaystyle\frac{T_s - T_m}{L}\) Where \(L\) represents the thickness of the concrete bar from the tube's surface to the outer surface.

05

Equate the heat transfer rates and solve for L

Since the heat transfer rates by convection and conduction must be equal, we can set them equal and solve for the thickness \(L\): \(q''_{conv} = q''_{cond} \Rightarrow 100 = 1.7 \displaystyle\frac{120 - 0}{L}\) Solving for \(L\), we find that \(L = \displaystyle\frac{1.7 \times 120}{100} = 2.04\; \text{mm}\)

06

Determine the width of the square concrete bar

To find the width of the square concrete bar, we should add the thickness \(L\) to the diameter of the tube: \(w = D + 2L = 127 + 2 \times 2.04 = 131.08\; \text{mm}\)

07

Calculate the rate of heat loss per meter length

Now, we should multiply the heat transfer rate \(q''_{conv}\) by the perimeter of the square cross-section, to calculate the heat loss per meter length: \(q' = q''_{conv} \times P\), where the perimeter \(P = 4w = 4 \times 131.08 = 524.32\; \text{mm} = 0.52432\; \text{m}\) Hence, \(q' = 100\displaystyle\frac{\text{W}}{m^2} \times 0.52432\; \text{m} = 52.432\displaystyle\frac{\text{W}}{m}\) So, the width of the square concrete bar is approximately \(131.08\mathrm{~mm}\), and the rate of heat loss per meter length is about \(52.432\displaystyle\frac{\text{W}}{\text{m}}\).

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