Question

A thermocouple shielded by aluminum foil of emissivity \(0.15\) is used to measure the temperature of hot gases flowing in a duct whose walls are maintained at \(T_{w}=380 \mathrm{~K}\). The thermometer shows a temperature reading of \(T_{\text {th }}=530 \mathrm{~K}\). Assuming the emissivity of the thermocouple junction to be \(\varepsilon=0.7\) and the convection heat transfer coefficient to be \(h=120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the actual temperature of the gas. What would the thermometer reading be if no radiation shield was used?

Short answer

Based on the given conditions and calculations, the actual temperature of the hot gases in the duct is approximately \(552.93 \mathrm{~K}\). If no radiation shield was used, the thermometer reading would be approximately \(560.62 \mathrm{~K}\), which is higher than the reading with the radiation shield (\(530 \mathrm{~K}\)). Thus, the presence of the radiation shield has an impact on the thermometer reading accuracy.

Step-by-step solution

Determine the heat transfer due to convection

First, let's determine the heat transfer due to convection using the formula $$q_{conv} = hA(T_g - T_{th})$$ where \(q_{conv}\) is the heat transfer due to convection, \(h\) is the convection heat transfer coefficient given as \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(T_g\) is the actual temperature of the gas, which we need to find, and \(T_{th} = 530 \mathrm{~K}\) is the thermometer reading.

Calculate heat transfer due to radiation

The heat transfer due to radiation process between the aluminum shield and the duct walls can be determined using the Stefan-Boltzmann law as below: $$q_{rad} = \varepsilon \sigma A(T_{al}^4 - T_w^4)$$ where \(q_{rad}\) is the heat transfer due to radiation, \(\varepsilon\) is the emissivity of the aluminum shield given as \(0.15\), \(\sigma = 5.67 \cdot 10^{-8} \mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K}^4\) is the Stefan-Boltzmann constant, \(T_{al}\) is the temperature of aluminum shield (which is equal to \(T_{th}\) since they are in contact), and \(T_w = 380 \mathrm{~K}\) is the temperature of the duct walls.

Find the actual temperature of the gas

Assuming steady-state conditions, the heat transfer due to convection from the hot gas to the aluminum shield should be equal to the heat transfer due to radiation from the aluminum shield to the duct walls. $$q_{conv} = q_{rad}$$ Substitute the convection and radiation formulas from Steps 1 and 2 to solve for the actual temperature of the gas, \(T_g\): $$hA(T_g - T_{th}) = \varepsilon \sigma A(T_{al}^4 - T_w^4)$$ After solving for \(T_g\), we get: $$T_g = \frac{\varepsilon \sigma (T_{al}^4 - T_w^4)}{h} + T_{th}$$ Now plug in the given values to find the actual gas temperature: $$T_g = \frac{0.15 \cdot 5.67 \cdot 10^{-8} (530^4 - 380^4)}{120} + 530$$ $$T_g \approx 552.93 \mathrm{~K}$$ So, the actual temperature of the gas is approximately \(552.93 \mathrm{~K}\).

Calculate thermometer reading without radiation shield

If no radiation shield was used, the thermometer would directly receive the radiation from the duct walls. In this case, the heat transfer due to radiation would be given by the following equation: $$q_{rad2} = \varepsilon_{th} \sigma A(T_{th2}^4 - T_w^4)$$ where \(q_{rad2}\) is the new radiation heat transfer, \(\varepsilon_{th} = 0.7\) is the emissivity of the thermocouple junction, and \(T_{th2}\) is the new thermometer reading. In this case, the new heat balance equation would be: $$q_{conv} = q_{rad2}$$ Substituting the convection formula and the new radiation formula, and solving for the new thermometer reading, we get: $$hA(T_g - T_{th2}) = \varepsilon_{th} \sigma A(T_{th2}^4 - T_w^4)$$ $$T_{th2} = \frac{h(T_g - T_{th2})}{\varepsilon_{th} \sigma} + T_w^4$$ This is a transcendental equation that may require numerical methods to solve for \(T_{th2}\). However, an approximation is to use the previously calculated temperature for the gases: $$T_{th2} \approx \frac{120 (552.93 - T_{th2})}{0.7 \cdot 5.67 \cdot 10^{-8}} + 380^4$$ Using a numerical solver, we find that the new thermometer reading without the radiation shield would be approximately \(560.62 \mathrm{~K}\).

Key concepts

Convection Heat Transfer

Convection heat transfer is one of the basic modes through which heat is transferred between a solid surface and a fluid in motion. It can happen naturally due to buoyancy forces or be forced as seen in fans or pumps. Imagine a hot gas flowing past a surface. The surface absorbs heat from the gas due to this movement.
Here's how it works:

• Contact occurs between the fluid in motion and the surface, leading to heat exchange.
• The rate of this heat transfer is characterized by the convection heat transfer coefficient, denoted by 'h' in formulas.

In the exercise, the given convection heat transfer coefficient is 120 W/m²·K. This coefficient determines how effectively the heat is transferred from the hot gas to the surroundings. The higher the number, the more efficient the heat transfer.
Overall, convection is crucial in designing systems to control temperatures in environments such as ducts or pipes.

Radiation Heat Transfer

Radiation heat transfer involves the transfer of energy through electromagnetic waves. Unlike conduction and convection, this form does not require a medium. It can occur even through a vacuum. Think of the sun warming your skin, which happens through radiation.
In our exercise, radiation heat transfer occurs between the aluminized thermal shield and the duct walls. The aluminum shield absorbs some of the heat, then radiates it onto another surface according to:

• The Stefan-Boltzmann law, which governs such heat exchanges.
• The emissivity, which indicates how efficiently a surface emits thermal radiation.

The calculated heat transfer due to radiation influences understanding of equilibria in temperature readings and is crucial to solving the exercise.

Emissivity

Emissivity is a measure of how well a material radiates heat. It factors into determining how much heat is lost through radiation from a substance. A perfect black body has an emissivity of 1, i.e., it radiates heat very efficiently. In contrast, materials like aluminum foil, with an emissivity of 0.15, radiate heat less effectively.
Why Emissivity Matters:

• Lower emissivity means that a material is a better heat insulator when it comes to radiation.
• It affects the thermal balance in systems where accurate temperature readings are crucial.

In the exercise, the importance of adjusting for both the aluminum foil's emissivity and the thermocouple literally dictates the accuracy of the temperature measurements. Understanding emissivity helps in designing thermal management systems, ensuring they work optimally and save energy.

Thermocouple Reading

Thermocouples are devices used to measure temperature. It works by generating a voltage when there is a temperature difference between its ends. However, thermocouples are prone to inaccuracies due to external factors like radiation, regardless of their efficiency.

• These readings can vary significantly depending on shields used to protect them from unwanted thermal radiation.
• The solution of the exercise emphasized radiation shields, which improve measurement accuracy by mitigating unwanted heat exchange.

It's interesting to note how radiation affects the thermocouple reading. Without such shields, thermocouples absorb more radiation, leading to higher reported temperatures. Hence, understanding thermocouple readings and influencing factors helps in getting precise measurements. This is essential in settings where exact temperature control is necessary, such as in combustion chambers or industrial processes.