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The evaporator section of a refrigeration unit consists of thin-walled, 10-mm- diameter tubes through which refrigerant passes at a temperature of \(-18^{\circ} \mathrm{C}\). Air is cooled as it flows over the tubes, maintaining a surface convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of \(-3^{\circ} \mathrm{C}\), what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator's defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range \(0 \leq \delta \leq 4 \mathrm{~mm}\). Frost may be assumed to have a thermal conductivity of \(0.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which \(T_{\infty}=20^{\circ} \mathrm{C}\) and natural convection maintains a convection coefficient of \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long will it take for the frost to melt? The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

Step by step solution

01

Use the heat transfer formula

The formula of heat transfer through convection is given by: \[q = h A (T_s - T_{\infty})\] where \(q\) = heat extraction rate (W), \(h\) = surface convection coefficient (W/m²K), \(A\) = surface area (m²), \(T_s\) = surface temperature (°C), \(T_{\infty}\) = fluid temperature (°C). Since we are asked to calculate the rate of heat extraction per unit tube length, we can rewrite the area as: \[A = \pi D L\] where \(D\) = diameter of the tube (m), \(L\) = length of the tube (m).

02

Plug in the values and calculate the heat extraction rate per unit length

Using the given values: \(h = 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(D = 0.01 \mathrm{~m}\), \(T_s = -18^{\circ}\mathrm{C}\), \(T_{\infty} = -3^{\circ}\mathrm{C}\). The heat extraction rate per unit length is: \[\frac{q}{L} = h \pi D (T_s - T_{\infty})\] Plugging in the values, we get: \[\frac{q}{L} = 100 \times \pi \times 0.01 \times (-18 - (-3))\] \[\frac{q}{L} = 47.12 \mathrm{~W} / \mathrm{m}\] (b) Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range \(0 \leq \delta \leq 4 \mathrm{~mm}\)

03

Determine the additional thermal resistance due to frost layer

The additional thermal resistance due to the frost layer is given by: \[R_f = \frac{\delta}{k_f A_f}\] where \(R_f\) = frost thermal resistance (K/W), \(\delta\) = frost layer thickness (m), \(k_f\) = frost thermal conductivity (W/mK), \(A_f\) = outer surface area (m²). The outer surface area is: \[A_f = \pi (D+2\delta) L\] Now we can express the total thermal resistance (\(R_{total}\)) as the sum of the frost layer resistance and the convection resistance: \[R_{total} = R_f + \frac{1}{hA}\]

04

Calculate the new rate of heat extraction for different frost layer thicknesses

Using the given parameters, we get: \[\frac{q_{new}}{L} = \frac{\pi (D + 2\delta) (T_s - T_{\infty})}{R_{total}}\] For the given range of frost layer thicknesses (\(0 \leq \delta \leq 4 \mathrm{~mm}\)), we must calculate \(R_{total}\) and plug it in above formula to find the new rate of heat extraction. (c) Find the time required for the frost to melt

05

Use the energy balance equation to find the time

We know that, to melt the frost, the energy required is given by \(Q = m_L \cdot L_f\), where \(m_L\) is the mass of frost per unit length and \(L_f\) is the latent heat of fusion. We can also relate the energy input with the heat transfer from the surroundings, where \(q_{melt}\) is the rate of heat transfer from the surrounding air per unit length, then: \[q_{melt} \Delta t = m_L \cdot L_f\] We know \(q_{melt} = h_m A_{melt} (T_\infty - T_s)\), where \(\Delta t\) is the time required to melt the frost, \(h_m\) is convection coefficient value, \(A_{melt}\) is the area where the melting process takes place. The mass of frost per unit length can be written as \(m_L = \rho L\delta\), where \(\rho\) is the mass density of the frost. Plugging in the given values and the calculated values, we can find \(\Delta t\), the time required for the frost to melt.

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Key Concepts

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

Conduction

Conduction is a mode of heat transfer through which energy flows due to the temperature gradient within a solid object or between solid objects in direct contact. It primarily occurs without the movement of the material itself. This mechanism is driven by the interaction of particles vibrating at different energy levels.
Inside a refrigeration unit, conduction plays a significant role when considering the flow of heat within the tubes carrying the refrigerant. The tube material's conductivity influences how efficiently heat can be transferred from the tube surface to the refrigerant inside it.
The basic formula governing conduction is Fourier's Law: \[ q = -kA \frac{dT}{dx} \] where:

• \( q \) is the heat transfer rate through conduction (Watt),
• \( k \) is the thermal conductivity of the material (W/mK),
• \( A \) is the cross-sectional area perpendicular to the direction of heat flow (\( m^2 \)),
• \( \frac{dT}{dx} \) is the temperature gradient over the distance \( x \)

To assess the system's efficiency, understanding conduction is fundamental, as it influences not only the cooling capacity but also determines the necessity to control material properties like thickness and thermal conductivity.

Convection

Convection is all about the transportation of heat through fluids, which can be liquids or gases. It combines the aspects of environmental interaction and flow within a given system. In the context of heat transfer in the refrigeration unit, the air moving over the tube surface is responsible for the convection heat transfer process.
This heat transfer can be characterized by Newton's Law of Cooling: \( q = hA(T_s - T_{\infty}) \)where:

• \( q \) is the rate of heat transfer through convection (Watt),
• \( h \) is the surface convection heat transfer coefficient (W/m²K),
• \( A \) is the surface area over which convection takes place (\( m^2 \)),
• \( T_s \) is the temperature of the surface (°C),
• \( T_{\infty} \) is the temperature of the bulk fluid (°C)

When frost forms on the tubes, it may affect the convection coefficient \( h \), as the surface texture changes. This can lead to reduced cooling efficiency. Also, different conditions, such as forced or natural convection, impact the heat transfer rate, particularly during the defrosting process.

Thermal Resistance

Thermal resistance is a concept that quantifies an object’s ability to resist the flow of heat. It is central when layers of materials with different thermal conductivities are involved, like when frost builds up on the refrigeration tubes.
This resistance can be thought of similarly to electrical resistance where high resistance implies less heat flows. The formula for thermal resistance in the context of the frost on tubes is given by: \[ R = \frac{\delta}{kA} \]where:

• \( R \) is the thermal resistance (K/W),
• \( \delta \) is the thickness of the frost layer (m),
• \( k \) is the thermal conductivity of the frost (W/mK),
• \( A \) is the area through which heat is being transferred (\( m^2 \))

As frost accumulates, the total thermal resistance of the tube and the frost layer increases, reducing the rate of heat transfer, which impacts the cooling capacity of the system. Effectively managing layers and accounting for their resistances are crucial for maintaining efficient thermal systems.

Latent Heat of Fusion

Latent heat of fusion is the amount of heat needed to convert a solid into a liquid at its melting point, without changing its temperature. This concept is key in understanding the energy requirements for melting frost in a refrigeration unit.
When frost accumulates and needs to be melted, the energy needed is calculated using the frost's mass and its latent heat of fusion. The basic relation is:\[ Q = mL_f \]where:

• \( Q \) is the heat energy required (Joules),
• \( m \) is the mass of the frost (kg),
• \( L_f \) is the latent heat of fusion (J/kg)

In the exercise, given the frost's thickness and density, one can compute how much energy is needed to melt it. Understanding this helps in estimating the time it takes for natural or assisted thawing processes to complete, impacting practical operations of cooling systems after defrosting issues occur.