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During a plant visit, it was observed that a \(1.5-\mathrm{m}\)-high and \(1-m\)-wide section of the vertical front section of a natural gas furnace wall was too hot to touch. The temperature measurements on the surface revealed that the average temperature of the exposed hot surface was \(110^{\circ} \mathrm{C}\), while the temperature of the surrounding air was \(25^{\circ} \mathrm{C}\). The surface appeared to be oxidized, and its emissivity can be taken to be \(0.7\). Taking the temperature of the surrounding surfaces to be \(25^{\circ} \mathrm{C}\) also, determine the rate of heat loss from this furnace. The furnace has an efficiency of 79 percent, and the plant pays \(\$ 1.20\) per therm of natural gas. If the plant operates \(10 \mathrm{~h}\) a day, 310 days a year, and thus \(3100 \mathrm{~h}\) a year, determine the annual cost of the heat loss from this vertical hot surface on the front section of the furnace wall.

Step by step solution

01

Calculate heat loss due to radiation

First, let's find the heat loss from the hot surface due to radiation. Using the Stefan-Boltzmann law, the radiative heat transfer (Q_rad) from the surface can be calculated as follows: $$Q_\text{rad} = 蔚 \times 蟽 脳 A \times (T_\text{hot}^4 - T_\text{surround}^4)$$ Where: - 蔚: Surface emissivity (0.7) - 蟽: Stefan-Boltzmann constant (5.67 脳 10鈦烩伕 W/m虏K鈦�) - A: Surface area (1.5m 脳 1m = 1.5 m虏) - T_hot: Temperature of the hot surface in Kelvin (110掳C + 273.15 = 383.15 K) - T_surround: Temperature of the surrounding surfaces in Kelvin (25掳C + 273.15 = 298.15 K) Now, let's calculate the radiative heat transfer (Q_rad).

02

Calculate heat loss due to convection

Next, we will determine the heat transfer due to convection by using the following correlation for natural convection heat transfer coefficient (h). $$h_{\text{conv}} = 1.31 脳 (T_\text{hot} - T_\text{air})^{1/3}$$ Where: - T_air: Temperature of the surrounding air in Kelvin (25掳C + 273.15 = 298.15 K) Now, we calculate h_conv and use it to find the convective heat transfer (Q_conv) as follows: $$Q_\text{conv} = h_{\text{conv}} 脳 A 脳 (T_\text{hot} - T_\text{air})$$

03

Calculate total heat loss

Now that we have calculated the heat loss due to radiation (Q_rad) and convection (Q_conv), we can find the total heat loss (Q_total) by adding these two values: $$Q_\text{total} = Q_\text{rad} + Q_\text{conv}$$

04

Calculate heat input

To find the annual cost of the heat loss, we first need to calculate the heat input. Since we know the efficiency of the furnace (79%), we can use the following equation: $$Q_\text{input} = \frac{Q_\text{total}}{\text{Efficiency}}$$ Now, let's calculate the heat input (Q_input).

05

Calculate annual cost of heat loss

Now that we have calculated the heat input (Q_input), we can find the annual cost of the heat loss using the cost of natural gas given in the problem. The energy in one therm is equivalent to 100,000 BTU. To convert the heat input to BTU, we can use the following conversion factor: 1 W = 3.412 BTU/h We also know the plant operates 10 h a day, 310 days a year which equals 3100 h a year. Let's calculate the annual heat loss in BTU with the following equation: Annual Heat Loss (BTU) = Q_input 脳 Time 脳 Conversion Factor Now, let's calculate the annual cost of the heat loss using the cost of natural gas: Annual Cost of Heat Loss = (Annual Heat Loss 脳 Cost per therm) / 100,000 Finally, we can find the annual cost of the heat loss from the vertical hot surface on the front section of the furnace wall.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radiation Heat Transfer

Radiation heat transfer occurs when heat energy is emitted by a hot surface and travels through a medium or vacuum to be absorbed by another surface. This process follows the Stefan-Boltzmann law, which is a fundamental principle in thermal physics. According to this law, the rate at which energy is radiated by a surface is proportional to the fourth power of the surface's absolute temperature. This means that even small changes in temperature can lead to significant changes in radiative heat transfer.

**Key points about Radiation Heat Transfer:**

• Emissivity (\( \epsilon \)) of a surface is crucial, as it defines how effectively a surface can emit thermal radiation. An emissivity of 1 indicates a perfect black body, whereas the furnace in our example has an emissivity of 0.7, meaning it's quite effective at emitting heat.
• The Stefan-Boltzmann constant (\( \sigma \)) is used in calculations to determine the radiative heat transfer, equal to 5.67 脳 10鈦烩伕 W/m虏K鈦�, which is a universal constant.
• Surface area and temperature difference also play integral roles in determining the total heat transfer rate.

Convection Heat Transfer

Convection heat transfer involves the movement of heat between a solid surface and a fluid (liquid or gas) moving across it. This type of heat transfer can occur naturally due to buoyancy effects, like warm air rising, or it can be forced by fans or pumps.

In the exercise, the vertical hot surface's heat transfer involves natural convection. The heat transfer coefficient, denoted by (\(h_{\text{conv}}\)), is calculated using empirical correlations, which are based on temperature difference and fluid properties. **Fundamentals of Convection Heat Transfer:**

• Temperature Difference: A larger difference between the surface and surrounding air temperatures increases the convection rate.
• Convection Coefficient (\(h_{\text{conv}}\)): Calculated using specific formulas that consider fluid properties and temperature gradients (e.g., for air around a vertical surface).
• Fluid Movement: This can be natural, like in the exercise, or forced, depending on the situation.

Thermodynamics Efficiency

Thermodynamics efficiency is a measure of the effectiveness of a system in converting input energy into useful work or output. For a furnace, efficiency indicates how well it uses the fuel's energy to generate heat, with some of that energy invariably being lost through processes like radiation and convection.

The furnace efficiency is given as 79%, which means that 79% of the input energy is used for heating purposes, while the remainder is lost.

**Understanding Thermodynamics Efficiency:**

• Efficiency Calculation: Efficiency is calculated by comparing the useful output energy to the total input energy, often expressed as a percentage.
• Impact on Costs: Higher efficiency means less fuel is needed to achieve the same heating, leading to lower operational costs.
• Improvement Options: Insulation improvements, reducing surface emissions, and maintaining equipment properly can enhance efficiency.

Understanding these concepts helps students appreciate the comprehensive process of heat loss and gain a practical understanding of operational and cost efficiency in industrial systems.