Question

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 \(20 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). 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. Answers: $1.32 \mathrm{~m}, 530 \mathrm{~W} / \mathrm{m}$

Short answer

Answer: The width of the square concrete bar is approximately \(1.32 \mathrm{~m}\), and the rate of heat loss per meter length is approximately \(530 \mathrm{~W} / \mathrm{m}\).

Step-by-step solution

Determine the outer surface temperature of the concrete bar.

We need to maintain a \(5^{\circ}C\) temperature difference between the outer surface of the square concrete bar and the ambient air, which has a temperature of \(-5^{\circ}C\). Add this temperature difference to the ambient air temperature to get the outer surface temperature of the concrete bar: $$ T_{outer} = T_{ambient} + \Delta T = (-5) + 5 = 0 ^{\circ}C $$

Determine the temperature difference across the concrete.

We need to find the temperature difference from the inner surface of the square concrete to the outer surface. The inner surface's temperature equals the tube's temperature, that is \(120^{\circ}C\). We have already calculated the outer surface temperature as \(0^{\circ}C\). We can find the temperature difference by subtracting the outer surface temperature from the inner surface temperature: $$ \Delta T_{concrete} = T_{inner} - T_{outer} = 120 - 0 = 120 ^{\circ}C $$

Apply the conduction equation.

To obtain the width of the square concrete bar, apply the conduction equation, with \(Q_{cond}\), the conducted heat per meter, as follows: $$ Q_{cond} = k \cdot A \cdot \frac{\Delta T_{concrete}}{L} $$ Where \(k\) is the concrete's thermal conductivity, \(A\) is the cross-sectional area of the square bar, and \(L\) is the required width.

Calculate the convection heat transfer.

The heat transfer through convection is given by: $$ Q_{conv} = h \cdot A_{s} \cdot \Delta T $$ Where \(h\) is the convection heat transfer coefficient, \(A_{s}\) is the surface area of the concrete bar exposed to air, and \(\Delta T\) is the temperature difference between the outer surface of the square concrete bar and the ambient air. The conduction heat transfer must be equal to the convection heat transfer, so: $$ Q_{cond} = Q_{conv} $$

Find the width of the square concrete bar.

Rearrange the combined heat transfer equation to solve for the width, \(L\): $$ L = \frac{k \cdot A \cdot \Delta T_{concrete}}{h \cdot A_{s} \cdot \Delta T} $$

Calculate the rate of heat loss per meter.

The rate of heat loss per meter is the conduction heat transfer per meter, \(Q_{cond}\). With the calculated width of the square concrete bar, \(L\), determine the rate of heat loss per meter: $$ Q_{cond} = k \cdot A \cdot \frac{\Delta T_{concrete}}{L} $$ Using the given values, \(k=1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), tube diameter \(D = 127 \mathrm{~mm}\), and ambient air temperature \(T_{ambient} = -5^{\circ}C\), we can calculate the width of the square concrete bar and the rate of heat loss per meter length: Width \(L \approx 1.32 \mathrm{~m}\) and heat loss per meter \(Q_{cond} \approx 530 \mathrm{~W} / \mathrm{m}\).