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Chapter 10: Problem 132

The refrigerant in a household refrigerator is condensed as it flows through the coil that is typically placed behind the refrigerator. Heat transfer from the outer surface of the coil to the surroundings is by natural convection and radiation. Obtaining information about the operating conditions of the refrigerator, including the pressures and temperatures of the refrigerant at the inlet and the exit of the coil, show that the coil is selected properly, and determine the safety margin in the selection.

Short Answer

Expert verified
Question: Explain the process of determining whether the coil in a household refrigerator is properly selected and the safety margin in the selection. Answer: To determine if the coil in a household refrigerator is properly selected and the safety margin in the selection, we follow these steps: 1. Calculate the heat transfer rate by natural convection using the convective heat transfer coefficient, surface area of the coil, and the temperature difference between the coil and surroundings. 2. Calculate the heat transfer rate by radiation using the emissivity of the coil, Stefan-Boltzmann constant, surface area of the coil, and the temperature difference between the coil and surroundings in Kelvin. 3. Calculate the total heat transfer rate by adding the heat transfer rates by natural convection and radiation. 4. Evaluate if the coil is properly selected by comparing the calculated total heat transfer rate with the heat transfer rate required by the refrigerator found in specifications or from the energy balance. If the total heat transfer rate is greater than or equal to the required heat transfer rate, the coil is properly selected. 5. Determine the safety margin in the selection by calculating the difference between the total heat transfer rate and the required heat transfer rate, divided by the required heat transfer rate, and multiplied by 100. A positive safety margin indicates that the coil can transfer more heat than needed, while a negative value indicates that the coil is insufficient for the required heat transfer.

Step by step solution

01

Calculate the heat transfer rate by natural convection

First, we need to find the heat transfer rate by natural convection. Natural convection in this context can be described using the Nusselt number (Nu). To calculate the heat transfer rate by natural convection (Q_conv), you will need the following formula: Q_conv = h_conv * A * (T_coil - T_ambient) where: - h_conv is the convective heat transfer coefficient - A is the surface area of the coil - T_coil is the average temperature of the coil - T_ambient is the temperature of the surroundings To find h_conv, we will need the Nusselt number (Nu) relation, which depends on the geometry of the coil and the temperature difference between the coil and the surroundings. This relation might be provided in the problem statement or found in textbooks or heat transfer correlations for natural convection.
02

Calculate the heat transfer rate by radiation

Next, we need to determine the heat transfer rate by radiation (Q_rad). Radiation heat transfer can be calculated using the following formula: Q_rad = ε * σ * A * (T_coil^4 - T_ambient^4) where: - ε is the emissivity of the coil (a value between 0 and 1) - σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m²K^4) - A is the surface area of the coil - T_coil is the average temperature of the coil (in Kelvin) - T_ambient is the temperature of the surroundings (in Kelvin)
03

Calculate the total heat transfer rate

Now that we have the heat transfer rates by natural convection and radiation, we can calculate the total heat transfer rate (Q_total) using the following formula: Q_total = Q_conv + Q_rad
04

Evaluate if the coil is properly selected

To determine if the coil is properly selected, we need to compare the calculated total heat transfer rate (Q_total) with the heat transfer rate required by the refrigerator (Q_required), which can be found in the specifications or calculated from the energy balance. If Q_total >= Q_required, then the coil is properly selected.
05

Determine the safety margin in the selection

Finally, to calculate the safety margin (SM) in the coil selection, use the following formula: SM = (Q_total - Q_required) / Q_required * 100 The safety margin is expressed as a percentage, and a positive value indicates that the coil is capable of transferring more heat than needed, while a negative value indicates that the coil is insufficient for the required heat transfer. In conclusion, by following these steps, you can determine whether the coil in a household refrigerator is properly selected and the safety margin in the selection.

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Key Concepts

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

Natural Convection
Understanding how heat is transferred by natural convection is essential for fields such as HVAC (Heating, Ventilation, and Air-Conditioning) design and refrigeration. Natural convection occurs when a fluid such as air or water moves due to temperature differences within it—the warmer, less dense part of the fluid rises while the cooler, denser part sinks, creating a natural circulation.

Certain factors affect the rate of natural convection, including the properties of the fluid (like viscosity and thermal conductivity), the geometry of the surface, and the temperature difference between the surface and the fluid. For instance, in a refrigerator's context, heat is transferred from the warmer coil to the cooler air surrounding it through this process, which is crucial for the coil's efficiency.
Radiation Heat Transfer
In addition to natural convection, radiation is another mechanism through which heat can be transferred from the refrigerator coil. Radiation does not require a medium to travel through, which means it can occur even in a vacuum. It is the energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the emitting molecules or atoms.

The amount of heat transferred by radiation is proportional to the fourth power of an object's absolute temperature (as per the Stefan-Boltzmann law), making it a significant factor at higher temperatures. Materials with higher emissivity, a measure of a material's ability to emit thermal radiation, will be more effective at radiating heat. In household refrigerators, while the coils are not as hot as industrial equipment, the radiation can still contribute to the overall heat loss.
Nusselt Number
The Nusselt number (Nu) is a dimensionless parameter that is crucial in the study of convective heat transfer. It correlates the convective heat transfer to conduction heat transfer and is defined by the following equation:

\[\begin{equation}Nu = \frac{h \cdot L}{k}\end{equation}\]
where h is the convective heat transfer coefficient, L is the characteristic length, and k is the thermal conductivity of the fluid. A higher Nusselt number typically means a more efficient convection process. Engineers use correlations based on the Nusselt number to estimate the heat transfer coefficient, which then allows them to calculate how much heat is being transferred from surfaces like refrigerator coils.
Heat Transfer Coefficient
The heat transfer coefficient (h) is a measure of a material's or a system's ability to conduct heat. For the convective heat transfer from the coil of a refrigerator to the ambient air, this coefficient quantifies how good the coil is at transferring heat. The value of the heat transfer coefficient is influenced by the mode of heat transfer (natural or forced convection), the properties of the fluid (like thermal conductivity and viscosity), and the flow characteristics (such as velocity and surface roughness).

By knowing the heat transfer coefficient, one can calculate the convective heat transfer rate using the aforementioned formula. The larger the coefficient, the more effective the surface is at transferring heat, thus potentially allowing for a smaller or more efficient coil design.
Emissivity
Emissivity (\(\epsilon\))) is a key concept when dealing with radiation heat transfer. It is a dimensionless measure of a material's ability to emit energy as thermal radiation, and it varies between 0 and 1. A perfect black body, which is a theoretical object that absorbs all electromagnetic radiation that falls onto it, has an emissivity of 1. Real-world materials have emissivities less than 1.

For practical purposes, like in our refrigerator coil, emissivity affects how much of the heat from the coil's surface is emitted through radiation. Different materials and surface finishes affect the emissivity of a surface; for example, polished metal surfaces will have a lower emissivity than a matte, dark surface.

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Most popular questions from this chapter

An air conditioner condenser in an automobile consists of \(2 \mathrm{~m}^{2}\) of tubular heat exchange area whose surface temperature is \(30^{\circ} \mathrm{C}\). Saturated refrigerant-134a vapor at \(50^{\circ} \mathrm{C}\) \(\left(h_{f g}=152 \mathrm{~kJ} / \mathrm{kg}\right)\) condenses on these tubes. What heat transfer coefficent must exist between the tube surface and condensing vapor to produce \(1.5 \mathrm{~kg} / \mathrm{min}\) of condensate? (a) \(95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(640 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(727 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(799 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(960 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Discuss some methods of enhancing pool boiling heat transfer permanently.

Saturated water vapor at atmospheric pressure condenses on the outer surface of a \(0.1\)-m-diameter vertical pipe. The pipe is \(1 \mathrm{~m}\) long and has a uniform surface temperature of \(80^{\circ} \mathrm{C}\). Determine the rate of condensation and the heat transfer rate by condensation. Discuss whether the pipe can be treated as a vertical plate. Assume wavy-laminar flow and that the tube diameter is large relative to the thickness of the liquid film at the bottom of the tube. Are these good assumptions?

A 1-mm-diameter nickel wire with electrical resistance of \(0.129 \mathrm{\Omega} / \mathrm{m}\) is submerged horizontally in water at atmospheric pressure. Determine the electrical current at which the wire would be in danger of burnout in nucleate boiling.

Steam condenses at \(50^{\circ} \mathrm{C}\) on a \(0.8-\mathrm{m}\)-high and \(2.4-\mathrm{m}-\) wide vertical plate that is maintained at \(30^{\circ} \mathrm{C}\). The condensation heat transfer coefficient is (a) \(3975 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(5150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(8060 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(11,300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(14,810 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (For water, use \(\rho_{l}=992.1 \mathrm{~kg} / \mathrm{m}^{3}, \mu_{l}=0.653 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\), \(\left.k_{l}=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p l}=4179 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, h_{f g \oplus T_{\text {sat }}}=2383 \mathrm{~kJ} / \mathrm{kg}\right)\)
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