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The front door of a dishwasher of width \(580 \mathrm{~mm}\) has a vertical air vent that is \(500 \mathrm{~mm}\) in height with a \(20-\mathrm{mm}\) spacing between the inner tub operating at \(52^{\circ} \mathrm{C}\) and an outer plate that is thermally insulated. (a) Determine the heat loss from the tub surface when the ambient air is \(27^{\circ} \mathrm{C}\). (b) A change in the design of the door provides the opportunity to increase or decrease the \(20-\mathrm{mm}\) spacing by \(10 \mathrm{~mm}\). What recommendations would you offer with regard to how the change in spacing will alter heat losses?

Short Answer

Expert verified
(a) The heat loss from the tub surface when the ambient air is 27掳C is approximately 270.49 W. (b) Our recommendation is to increase the spacing by 10 mm to decrease heat losses from the tub surface.

Step by step solution

01

Find the area of the air vent

The given width and height are in millimeters, so first, we need to convert them into meters. We get a width of 580 mm / 1000 = 0.58 m and a height of 500 mm / 1000 = 0.5 m. We can now find the area by multiplying the width and height: area = 0.58 m * 0.5 m = 0.29 m^2.
02

Calculate the average temperature

The average temperature will be the average of the tub surface temperature and the ambient air temperature: T_avg = (52掳C+27掳C) / 2 = 39.5掳C.
03

Determine temperature difference and convert to Kelvin

The difference between the tub surface temperature and the ambient air temperature is: 螖T = 52掳C - 27掳C = 25掳C. Convert the average temperature and the temperature difference to Kelvin by adding 273.15: T_avg_K = 39.5掳C + 273.15 = 312.65 K and 螖T_K = 25掳C + 273.15 = 298.15 K.
04

Calculate the Grashof and Prandtl numbers

The Grashof number (Gr) and Prandtl number (Pr) are used to calculate the heat transfer coefficient. We will use air properties at the average temperature: - Air dynamic viscosity: 碌 = 1.9 x 10^{-5} Ns/m^2 - Air thermal conductivity: k = 0.0262 W/mK - Air coefficient of thermal expansion: 尾 = 1 / T_avg_K = 1 / 312.65 K = 3.199 x 10^{-3} K^{-1} - Air kinematic viscosity: v = 1.56 x 10^{-5} m^2/s - Air specific heat capacity: cp = 1006 J/kgK We can now find the Grashof number: Gr = (g * 尾 * 螖T_K * (0.02^3)) / (v^2) = (9.81 m/s^2 * 3.199 x 10^{-3} K^{-1} * 298.15 K * (0.02 m)^3) / (1.56 x 10^{-5} m^2/s)^2 = 9.3 x 10^7. And the Prandtl number: Pr = (cp * 碌) / k = (1006 J/kgK * 1.9 x 10^{-5} Ns/m^2) / 0.0262 W/mK = 0.715.
05

Determine the Nusselt number

For a vertical plate with natural convection, we will use the following correlations to determine the Nusselt number (Nu): If 10^4 <= Gr * Pr <= 10^9, then Nu = 0.59 * (Gr * Pr)^{1/4} In our case, Gr * Pr = 9.3 x 10^7 * 0.715 = 6.65 x 10^7. This value is within the range specified in the correlation, so we will use the correlation formula. Nu = 0.59 * (6.65 x 10^7)^{1/4} = 28.18
06

Calculate heat transfer coefficient

Using the Nusselt number, we can find the heat transfer coefficient (h) as follows: h = Nu * (k / spacing) = 28.18 * (0.0262 W/mK / 0.02 m) = 37.45 W/m^2K
07

Calculate heat loss

We can now calculate the heat loss from the tub surface using the heat transfer coefficient, the area of the air vent, and the temperature difference: q = h * area * 螖T = 37.45 W/m^2K * 0.29 m^2 * 25 K = 270.49 W (a) The heat loss from the tub surface when the ambient air is 27掳C is approximately 270.49 W. For part (b), we need to examine the effect of increasing and decreasing the spacing by 10 mm. We could follow the same process to calculate new heat losses in both cases. However, it is useful to observe that since the heat transfer coefficient is inversely proportional to the spacing (h 鈭 1/spacing), increasing the spacing would decrease the heat transfer coefficient and, as a result, decrease heat losses. Conversely, decreasing the spacing would increase the heat transfer coefficient and increase heat losses. (b) Our recommendation is to increase the spacing by 10 mm to decrease heat losses from the tub surface.

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Key Concepts

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

Natural Convection
Natural convection is a mode of heat transfer that occurs when fluid motion is instigated purely by buoyancy forces arising from temperature differences in the fluid. In other words, it does not need any external aid like a fan or a pump. Instead, the movement of the fluid鈥攁nd thus, the heat transfer鈥攔elies on differences in density that result from variations in temperature.
For example, when air heats up, it becomes less dense. As a result, warmer air rises while cooler, denser air sinks to take its place. This creates a flow of air known as a 'convection current'. In the context of the exercise with the dishwasher door, natural convection is crucial because it facilitates the transfer of heat from the inner tub to the ambient air. The way the air circulates within the vent affects how much heat is lost from the dishwasher.
Grashof Number
The Grashof Number ( ext{Gr}) is a dimensionless number that comes into play in free or natural convection. It serves as a measure of the ratio of buoyant force to viscous force within a fluid. A higher Grashof number indicates that the buoyant forces are significant relative to viscous forces, and thus convection is more dominant as a mode of heat transfer compared to conduction. This makes understanding Grashof essential in predicting the behavior of naturally convecting fluid flows.
To compute the Grashof number for the dishwasher problem, various physical properties of air such as viscosity and thermal expansion coefficient are used. These factors tell us how effectively air can move to transfer heat. When summarizing applications, a Grashof number within certain limits can help you predict whether increasing the spacing in the vent would result in more efficient heat dissipation due to natural convection.
Prandtl Number
The Prandtl Number ( ext{Pr}) helps relate the thickness of the thermal boundary layer to the velocity boundary layer in a fluid. It is also dimensionless and is defined as the ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity. In simple terms, it indicates whether heat conduction or momentum diffusion will dominate fluid movement.
In engineering contexts like our dishwasher example, knowing the Prandtl number complements the Grashof number to assess the fluid flow type. With a value typically under 1 for gasses like air, it suggests that the thermal diffusivity is more significant than momentum diffusivity, highlighting the conditions under which the bulk of the fluid would likely resist mixing compared to the heat diffusivity.
Nusselt Number
The Nusselt Number ( ext{Nu}) is a pivotal metric in analyzing convective heat transfer. It compares the total convective heat transfer to conduction across a boundary layer. A higher Nusselt number implies that convection significantly enhances heat transfer over pure conduction.
This number depends on the Grashof and Prandtl numbers; they are often used together to derive the Nusselt number, especially when addressing problems like air movement between surfaces. For a vertical surface within specific Grashof and Prandtl ranges, correlations allow us to estimate the Nusselt number for calculating the heat transfer coefficient. This coefficient is then integral in determining the dissipation of heat from one surface to the adjacent fluid鈥攍ike the dishwasher's ambient air, influencing systemic thermal management strategies.

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Most popular questions from this chapter

Consider the conveyor system described in Problem \(7.24\), but under conditions for which the conveyor is not moving and the air is quiescent. Radiation effects and interactions between boundary layers on adjoining surfaces may be neglected. (a) For the prescribed plate dimensions and initial temperature, as well as the prescribed air temperature, what is the initial rate of heat transfer from one of the plates? (b) How long does it take for a plate to cool from \(300^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) ? Comment on the assumption of negligible radiation.

Circuit boards are mounted to interior vertical surfaces of a rectangular duct of height \(H=400 \mathrm{~mm}\) and length \(L=800 \mathrm{~mm}\). Although the boards are cooled by forced convection heat transfer to air flowing through the duct, not all of the heat dissipated by the electronic components is transferred to the flow. Some of the heat is instead transferred by conduction to the vertical walls of the duct and then by natural convection and radiation to the ambient (atmospheric) air and surroundings, which are at equivalent temperatures of \(T_{\infty}=T_{\text {sur }}=20^{\circ} \mathrm{C}\). The walls are metallic and, to a first approximation, may be assumed to be isothermal at a temperature \(T_{s}\). (a) Consider conditions for which the electronic components dissipate \(200 \mathrm{~W}\) and air enters the duct at a flow rate of \(\dot{m}=0.015 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(T_{m, i}=20^{\circ} \mathrm{C}\). If the emissivity of the side walls is \(\varepsilon_{s}=0.15\) and the outlet temperature of the air is \(T_{m, o}=30^{\circ} \mathrm{C}\), what is the surface temperature \(T_{s}\) ? (b) To reduce the temperature of the electronic components, it is desirable to enhance heat transfer from the side walls. Assuming no change in the airflow conditions, what is the effect on \(T_{s}\) of applying a high emissivity coating \(\left(\varepsilon_{s}=0.90\right)\) to the side walls? (c) If there is a loss of airflow while power continues to be dissipated, what are the resulting values of \(T_{s}\) for \(\varepsilon_{s}=0.15\) and \(\varepsilon_{s}=0.90\) ?

A 50 -mm-thick air gap separates two horizontal metal plates that form the top surface of an industrial furnace. The bottom plate is at \(T_{h}=200^{\circ} \mathrm{C}\) and the top plate is at \(T_{c}=50^{\circ} \mathrm{C}\). The plant operator wishes to provide insulation between the plates to minimize heat loss. The relatively hot temperatures preclude use of foamed or felt insulation materials. Evacuated insulation materials cannot be used due to the harsh industrial environment and their expense. A young engineer suggests that equally spaced, thin horizontal sheets of aluminum foil may be inserted in the gap to eliminate natural convection and minimize heat loss through the air gap.

Consider a horizontal pin fin of 6- \(\mathrm{mm}\) diameter and \(60-\mathrm{mm}\) length fabricated from plain carbon steel \((k=57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \varepsilon=0.5)\). The base of the fin is maintained at \(150^{\circ} \mathrm{C}\), while the quiescent ambient air and the surroundings are at \(25^{\circ} \mathrm{C}\). Assume the fin tip is adiabatic. (a) Estimate the fin heat rate, \(q_{f}\). Use an average fin surface temperature of \(125^{\circ} \mathrm{C}\) in estimating the free convection coefficient and the linearized radiation coefficient. How sensitive is this estimate to your choice of the average fin surface temperature? (b) Use the finite-difference method of solution to obtain \(q_{f}\) when the convection and radiation coefficients are based on local, rather than average, temperatures for the fin. How does your result compare with the analytical solution of part (a)?

Consider a large vertical plate with a uniform surface temperature of \(130^{\circ} \mathrm{C}\) suspended in quiescent air at \(25^{\circ} \mathrm{C}\) and atmospheric pressure. (a) Estimate the boundary layer thickness at a location \(0.25 \mathrm{~m}\) measured from the lower edge. (b) What is the maximum velocity in the boundary layer at this location and at what position in the boundary layer does the maximum occur? (c) Using the similarity solution result, Equation \(9.19\), determine the heat transfer coefficient \(0.25 \mathrm{~m}\) from the lower edge. (d) At what location on the plate measured from the lower edge will the boundary layer become turbulent?

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