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A mercury-in-glass thermometer used to measure the air temperature in an enclosure reads \(15^{\circ} \mathrm{C}\). The enclosure walls are all at \(0^{\circ} \mathrm{C}\). Estimate the true air temperature if the convective heat transfer coefficient for the thermometer bulb is estimated to be \(12 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

Short Answer

Expert verified
The true air temperature is approximately 15.06°C.

Step by step solution

01

Identify the Known Values

We are provided the following values:- Observed thermometer reading: \( T_t = 15^{\circ} \mathrm{C} \)- Enclosure wall temperature: \( T_s = 0^{\circ} \mathrm{C} \)- Convective heat transfer coefficient: \( h = 12 \, \mathrm{W/m}^2\, \mathrm{K} \).
02

Understand the Energy Balance Equation

The thermometer reading is influenced by the radiation exchange with the walls. We need to find the air temperature, \( T_a \). Applying the energy balance on the thermometer bulb, we have the equation:\[ h(T_a - T_t) = \sigma \varepsilon (T_t^4 - T_s^4) \]where \( \sigma \) is the Stefan-Boltzmann constant (≈ 5.67 \( \times 10^{-8} \mathrm{W/m^2 \cdot K^4} \)) and \( \varepsilon \) is the emissivity. Assume emissivity is close to 1 for the initial estimation.
03

Solve for Actual Air Temperature

Substitute the known values into the energy balance equation:\[ 12 (T_a - 15) = 5.67 \times 10^{-8} \times 1 \times ((15+273)^4 - (0+273)^4) \]Calculate the Stefan-Boltzmann term:\[ 5.67 \times 10^{-8} ((288)^4 - (273)^4) \approx 5.67 \times 10^{-8} \times (68874752 - 55678864) \approx 5.67 \times 10^{-8} \times 13195888 \approx 0.747 \mathrm{W/m^2} \]Thus:\[ 12 (T_a - 15) = 0.747 \]Now, solve for \( T_a \):\[ T_a - 15 = \frac{0.747}{12} \approx 0.06225 \]\[ T_a \approx 15 + 0.06225 \approx 15.06225^{\circ} \mathrm{C} \]
04

Conclusion of Calculated Air Temperature

After solving the equation, we estimate the true air temperature to be approximately 15.06225°C when considering convective and radiative effects on the thermometer bulb.

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Key Concepts

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

Thermodynamics
Thermodynamics is the study of energy, heat, and work, and their interactions in physical systems. It's a fundamental branch of physics that describes how energy is transformed and transferred. In this context, thermodynamics helps us understand how the thermometer interacts with its environment—how heat is transferred from inside the thermometer to the walls of the enclosure.

This process involves the principles of energy conversion, where the internal energy of the thermometer changes due to heat exchange. The first law of thermodynamics, which is essentially an energy balance equation, states that energy cannot be created or destroyed, only converted from one form to another. This principle is crucial to determine the actual air temperature in the given problem, as it takes into consideration both convective and radiative heat transfers in the energy balance equation.

In practical terms, the aim is to find out where heat flows from and to, so we can calculate the accurate temperature readings taking all influencing factors into consideration.
Convective Heat Transfer
Convective heat transfer is one of the primary modes of heat transfer. It occurs through the movement of fluid, such as air or liquid, flowing over a surface. In this exercise, it deals with the heat exchange between the thermometer bulb and the surrounding air.

When the thermometer reads a certain temperature, it may not be identical to the actual air temperature due to convective heat transfer. Instead, the heat from the air is transferred to the thermometer bulb by convection, depending on the convective heat transfer coefficient, represented as "h" in the formula. This coefficient indicates how effectively heat is transferred through this process and can vary based on the flow conditions and the surface properties.

In the given problem, the estimated value for the convective heat transfer coefficient is 12 W/m²K. This value helps us understand how much heat is being imparted to or taken from the thermometer by the air, adjusting our interpretation of the recorded temperature compared to the ambient temperature.
Radiative Heat Transfer
Radiative heat transfer is another essential mechanism of heat transfer, where energy is emitted or absorbed in the form of thermal radiation. This takes place without the need for a medium, unlike conduction and convection.

In our scenario, radiative heat transfer occurs between the thermometer bulb and the enclosure walls. The walls emit or absorb thermal radiation, impacting the bulb temperature. This process is captured mathematically using the Stefan-Boltzmann law, which states that the energy radiated by a black body in thermal equilibrium is proportional to the fourth power of its temperature (\( T^4 \)).

The key variables influencing radiative transfer in this exercise include the Stefan-Boltzmann constant and the emissivity coefficient, which indicates how effectively a surface emits thermal radiation. When computing the true air temperature, the equation must account for the energy balance between convective and radiative heat transfer effects on the thermometer bulb.
Energy Balance Equation
The energy balance equation in thermodynamics is a tool that relates the exchange of energy between a system and its surroundings. In this case, the system is the thermometer bulb, and the surroundings include the air and the enclosure's walls.

The general form of the energy balance equation in this scenario is the sum of the convective and radiative heat transfers, set equal to zero, as the bulb isn't changing temperature over time:\[h(T_a - T_t) = \sigma \varepsilon (T_t^4 - T_s^4)\]This equation represents a balance where the net heat flow is zero, aligning with the first law of thermodynamics. The convective component accounts for heat transferred via air movement, while the radiative component involves thermal emissions from the surrounding walls to the bulb.

Solving this equation involves substituting known values, including the convective coefficient and radiative constants, to find the true air temperature. Essentially, it corrects the observed temperature for the combined effect of convective and radiative processes, providing a more precise measurement.

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Most popular questions from this chapter

A low-pressure heat exchanger transfers heat between two helium streams, each with a flow rate of \(m=5 \times 10^{-3} \mathrm{~kg} / \mathrm{s}\). In a performance test the cold stream enters at a pressure of \(1000 \mathrm{~Pa}\) and a temperature of \(50 \mathrm{~K}\), and exits at \(730 \mathrm{~Pa}\) and \(350 \mathrm{~K}\). (i) If the flow cross-sectional area for the cold stream is \(0.019 \mathrm{~m}^{2}\), calculate the inlet and outlet velocities. (ii) If the exchanger can be assumed to be perfectly insulated, determine the rate of heat transfer in the exchanger. For helium, \(c_{p}=5200 \mathrm{~J} / \mathrm{kg} \mathrm{K}\).

A \(1 \mathrm{~cm}\)-diameter sphere is maintained at \(60^{\circ} \mathrm{C}\) in an enclosure with walls at \(35^{\circ} \mathrm{C}\) through which air at \(40^{\circ} \mathrm{C}\) circulates. If the convective heat transfer coefficient is \(11 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), estimate the rate of heat loss from the sphere when its emittance is (i) \(0.05\). (ii) \(0.85 .\)

Constant delivery of low-vapor-pressure reactive gases is required for semiconductor fabrication. In one process, tungsten fluoride \(\mathrm{WF}_{6}\) (normal boiling point \(17^{\circ} \mathrm{C}\) ) is supplied from a \(80 \mathrm{~cm}\)-diameter spherical tank containing liquid \(\mathrm{WF}_{6}\) under pressure. The tank is located in surroundings at \(21^{\circ} \mathrm{C}\). After connecting a full tank to the gas delivery system, supply at a rate of \(2500 \mathrm{sccm}\) (standard cubic centimeters per minute) commences. In order to supply the required heat of vaporization, the liquid \(\mathrm{WF}_{6}\) temperature drops until a steady state is reached, for which the heat transferred into the tank from the surroundings balances the heat of vaporization required. (i) Estimate the steady-state temperature of the liquid \(\mathrm{WF}_{6}\). (ii) Estimate how long it will take for the liquid to approach within \(1^{\circ} \mathrm{C}\) of its steady value. Property values for liquid \(\mathrm{WF}_{6}\) include \(\rho=3440 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1000 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), \(h_{f g}=25.7 \times 10^{3} \mathrm{~kJ} / \mathrm{kmol}\), and for steel, \(c=434 \mathrm{~J} / \mathrm{kg} \mathrm{K}\). The weight of the empty tank is \(30 \mathrm{~kg}\), and the heat transfer coefficient for convection and radiation to the \(\operatorname{tank}\) is \(h=8 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

A \(2.5 \mathrm{~m}\)-diameter, \(3.5 \mathrm{~m}\)-high milk storage tank is located in a dairy factory in Onehunga, New Zealand, where the ambient temperature is \(30^{\circ} \mathrm{C}\). The tank has walls of stainless steel \(2 \mathrm{~mm}\) thick and is insulated with a \(7.5 \mathrm{~cm}\)-thick layer of polyurethane foam. The tank is filled with milk at \(4^{\circ} \mathrm{C}\) and is continuously stirred by an impeller driven by an electric motor that consumes \(400 \mathrm{~W}\) of power. What will the milk temperature be after 24 hours? For the milk, take \(\rho=1034 \mathrm{~kg} / \mathrm{m}^{3}\). \(c=3894 \mathrm{~J} / \mathrm{kg} \mathrm{K} ;\) for the insulation, \(k=0.026 \mathrm{~W} / \mathrm{m} \mathrm{K} ;\) and for the outside heat transfer coefficient, \(h=5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). The impeller motor efficiency can be taken as \(0.75\).

A shell-and-tube condenser for an ocean thermal energy conversion and fresh water plant is tested with a water feed rate to the tubes of \(4000 \mathrm{~kg} / \mathrm{s}\). The water inlet and outlet conditions are measured to be \(P_{1}=129 \mathrm{kPa}, T_{1}=280 \mathrm{~K}\); and \(P_{2}=108 \mathrm{kPa}, T_{2}=285 \mathrm{~K}\). (i) Calculate the rate of heat transfer to the water. (ii) If saturated steam condenses in the shell at \(1482 \mathrm{~Pa}\), calculate the steam condensation rate. For the feed water, take \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{v}=4192 \mathrm{~J} / \mathrm{kg} \mathrm{K}\). (Steam tables are given as Table A. \(12 a\) in Appendix A.)

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