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 \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). 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.

Short answer

Answer: The width of the square concrete bar is approximately 131.08 mm, and the rate of heat loss per meter length is about 52.432 W/m.

Step-by-step solution

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: \(\Delta T = 5^{\circ}\, \text{C}\)

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: \(T_m = T_{\infty} + \Delta T = -5 + 5 = 0^{\circ}\, \text{C}\)

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: \(q''_{conv} = h (T_m - T_{\infty}) = 20 (0 - (-5)) = 100\displaystyle\frac{W}{m^2}\)

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.

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 \(L = \displaystyle\frac{1.7 \times 120}{100} = 2.04\; \text{mm}\)

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: \(w = D + 2L = 127 + 2 \times 2.04 = 131.08\; \text{mm}\)

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 \(52.432\displaystyle\frac{\text{W}}{\text{m}}\).

Key concepts

CHP Generation

Combined Heat and Power (CHP) generation, also known as cogeneration, is a highly efficient process that simultaneously generates electricity and useful heat from a single fuel source. In a CHP system, the by-product heat from electricity generation is captured and utilized for either domestic or industrial heating purposes. This dual-purpose use of energy results in a significant increase in overall energy efficiency, compared to traditional methods, where heat is often wasted. CHP systems are particularly beneficial in applications where there is a substantial and consistent demand for both electricity and thermal energy, such as in urban heating systems or industrial processing facilities.
- Efficient energy use: By using the waste heat, CHP systems can achieve efficiency levels of up to 80% or more, effectively reducing fuel consumption and greenhouse gas emissions. - Versatility: CHP systems can be powered by various fuels, including natural gas, biomass, and coal, allowing flexibility in energy production.
In CHP systems, such as the one described in the exercise, careful design is pivotal to ensure heat is effectively transferred and used without excessive losses.

Convection Heat Transfer

Convection heat transfer occurs when heat is carried away by fluid movement, which can be in the form of a liquid or a gas. In the context of the exercise, air flowing around the square concrete bar removes heat from the system, demonstrating convection. Convection is driven by the difference in temperature between the surface and the fluid surrounding it. There are two types of convection processes:

• Natural Convection: Occurs due to natural causes such as temperature differences causing fluid movement, like hot air rising or cold air sinking.
• Forced Convection: Results from external forces, such as fans or pumps, moving the fluid across the surface.

The heat transfer rate by convection can be calculated using Newton's Law of Cooling:\( q''_{conv} = h (T_s - T_{ ext{ambient}}) \)where:

• \( h \) is the convection heat transfer coefficient.
• \( T_s \) is the surface temperature.
• \( T_{ ext{ambient}} \) is the ambient air temperature.

This equation helps determine how much heat is lost, which is essential for calculating and maintaining the necessary thermal properties in systems like the CHP bar.

Fourier's Law of Conduction

Fourier's Law of Conduction is pivotal in understanding how heat moves through materials. It states that the heat transfer rate through a solid is proportional to the negative gradient of temperature and the area through which the heat flows. Mathematically, it is expressed as:\[ q''_{cond} = -k \frac{dT}{dx} \]where:

• \( q''_{cond} \) is the heat flux or the heat transfer per unit area.
• \( k \) is the thermal conductivity of the material.
• \( dT/dx \) is the temperature gradient in the direction of heat flow.

In practical terms, this law helps calculate how efficiently heat can travel through a material, like the concrete bar surrounding the steam-carrying tube in the exercise. The higher the thermal conductivity \( k \), the better the material can conduct heat. By applying Fourier's Law, engineers can ensure that materials chosen for heat transfer applications will meet the necessary efficiency and safety standards, as well as control the heat flow accurately. In the exercise, setting conduction heat equal to convection helps in solving for important dimensions like the thickness of the concrete bar to minimize heat loss.