/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 It is desired to develop a simpl... [FREE SOLUTION] | 91Ó°ÊÓ

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It is desired to develop a simple model for predicting the temperature-time history of a plate during the drying cycle in a dishwasher. Following the wash cycle the plate is at \(T_{p}(t)=T_{p}(0)=65^{\circ} \mathrm{C}\) and the air in the dishwasher is completely saturated \(\left(\phi_{x}=1.0\right)\) at \(T_{x}=55^{\circ} \mathrm{C}\). The values of the plate surface area \(A_{s}\), mass \(M\), and specific heat \(c\) are such that \(M c / A_{s}=1600 \mathrm{~J} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Assuming the plate is completely covered by a thin film of water and neglecting the thermal resistances of the film and plate, derive a differential equation for predicting the plate temperature as a function of time. (b) For the initial conditions \((t=0)\) estimate the change in plate temperature with time, \(d T / d t\left({ }^{\circ} \mathrm{C} / \mathrm{s}\right)\), assuming that the average heat transfer coefficient on the plate is \(3.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Short Answer

Expert verified
The differential equation for the plate temperature as a function of time is: \[\frac{dT_p}{dt} = \frac{h A_s}{M c} (T_p - T_x)\] At the initial condition (t=0), the change in plate temperature with time is: \[\frac{dT_p}{dt}\Big|_{t=0} = 0.0021875 A_s\left(\frac{^\circ C}{s}\right)\]

Step by step solution

01

Derive the differential equation for plate temperature

\ First, we need to derive a differential equation to determine the plate's temperature as a function of time. We are given that the plate is initially at temperature \(T_p(0)\) and is completely covered by a thin film of water. The air in the dishwasher is saturated at temperature \(T_x\). Heat transfer from the plate can be described by the formula: \[Q = h A_s (T_p - T_x)\] where \(Q\) is the heat transferred, \(h\) is the heat transfer coefficient, and \(A_s\) is the plate surface area. Heat gained or lost by the plate may be expressed in terms of its mass and specific heat as: \[\frac{dQ}{dt} = M c \frac{dT_p}{dt}\] where \(M\) is the plate mass, \(c\) is the specific heat, and \(\frac{dT_p}{dt}\) represents the rate of change of the plate's temperature with time. From the given information, we have: \[\frac{M c}{A_s} = 1600 \frac{J}{m^2 \cdot K}\] Now, substituting the heat transfer equation to the temperature change equation: \[\frac{dQ}{dt} = h A_s (T_p - T_x)\] \[M c \frac{dT_p}{dt} = h A_s (T_p - T_x)\] Using the given relationship: \[\frac{dT_p}{dt} = \frac{h A_s}{M c} (T_p - T_x)\]
02

Estimate the change in plate temperature at initial conditions

\ At the initial conditions (t = 0), we are given the heat transfer coefficient \(h = 3.5 \frac{W}{m^2 \cdot K}\) and are asked to find the change in plate temperature with time, \(\frac{dT}{dt}\). Using the differential equation derived in step 1: \[\frac{dT_p}{dt}\Big|_{t=0} = \frac{h A_s}{M c} (T_p(0) - T_x)\] Substituting the given values: \[\frac{dT_p}{dt}\Big|_{t=0} = \frac{3.5 A_s}{1600} (65 - 55)\] Simplifying the equation: \[\frac{dT_p}{dt}\Big|_{t=0} = 0.0021875 A_s\] The change in plate temperature with time at the initial condition is directly proportional to the surface area of the plate, which is represented by the equation: \[\frac{dT_p}{dt}\Big|_{t=0} = 0.0021875 A_s\left(\frac{^\circ C}{s}\right)\]

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Key Concepts

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

Differential Equation
A differential equation is a mathematical equation that relates a function to its derivatives. In the context of heat transfer, it helps us understand how the temperature of an object changes over time. For this dishwasher scenario, the plate's temperature changes over time due to heat exchange with its surroundings. To derive the differential equation for our plate, we start by figuring out how much heat is being transferred. The heat transfer from the plate is captured by the equation:\[ Q = h A_s (T_p - T_x) \]Here, \( Q \) is the heat transferred, \( h \) is the heat transfer coefficient, \( A_s \) is the plate's surface area, \( T_p \) is the plate's temperature, and \( T_x \) is the temperature of the surrounding air.The rate of change of heat in terms of temperature for the plate is given by:\[ \frac{dQ}{dt} = M c \frac{dT_p}{dt} \]Where \( M \) is the plate's mass, \( c \) is its specific heat, and \( \frac{dT_p}{dt} \) is the rate of temperature change with time.By combining these equations, we obtain the differential equation describing how the plate's temperature changes:\[ \frac{dT_p}{dt} = \frac{h A_s}{M c} (T_p - T_x) \]This equation will help us predict the temperature timeline of the plate during its drying cycle.
Temperature-Time History
The temperature-time history of an object tells us how its temperature changes over a period. For our dishwasher plate, this means tracking its temperature as it dries. Initially, the plate has a temperature of \( 65^{\circ} \mathrm{C} \), and the environment (saturated air) is at \( 55^{\circ} \mathrm{C} \). As time progresses, heat is exchanged between the plate and the surrounding environment, resulting in a changing temperature on the plate. The rate of this change is captured by the differential equation we previously derived:\[ \frac{dT_p}{dt} = \frac{h A_s}{M c} (T_p - T_x) \]This equation is crucial because it tells us how the rate of temperature change (\( \frac{dT_p}{dt} \)) evolves over time as a response to the heat exchange with the surroundings. Understanding the temperature-time history can help in predicting how fast the plate will cool down and when it might reach a steady state, making it crucial for designing efficient drying processes in applications like dishwashers.
Heat Transfer Coefficient
The heat transfer coefficient \( h \) is a parameter that quantifies how easily heat is transferred between a solid surface and a fluid in contact with it. In this case, it tells us how effectively heat passes from the plate to the air and vice versa.In our scenario, we're given that the heat transfer coefficient is \( 3.5 \, \mathrm{W/m^2 \, K} \). This means for every square meter of surface area, and for every degree of temperature difference between the plate and the air, 3.5 watts of heat can be transferred.This coefficient is crucial in our differential equation for calculating how quickly the plate's temperature changes:\[ \frac{dT_p}{dt} = \frac{h A_s}{M c} (T_p - T_x) \]A higher heat transfer coefficient indicates a more efficient heat exchange, resulting in a faster change in temperature over time. It is influenced by factors like the properties of the plate material, the fluid's motion, and the characteristics of the surface involved.
Specific Heat
Specific heat \( c \) is a property that describes how much heat energy is needed to change the temperature of a unit mass of a substance by one degree Celsius. In simple terms, it tells us how resistant a material is to changing its temperature.For the plate in our exercise, specific heat helps in defining how quickly it will heat up or cool down in response to the heat being transferred.The rate of temperature change of the plate, as expressed in the differential equation, depends inversely on specific heat:\[ \frac{dT_p}{dt} = \frac{h A_s}{M c} (T_p - T_x) \]This means if the plate has a high specific heat, it would require more heat to change its temperature, hence slowing the rate of temperature change. The specific heat helps in determining the thermal inertia of the plate.When designing systems that involve heating or cooling, knowing the specific heat is essential for predicting how an object's temperature will respond to various amounts of heat being lost or gained.

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Most popular questions from this chapter

Forced air at \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(V=10 \mathrm{~m} / \mathrm{s}\) is used to cool electronic elements on a circuit board. One such element is a chip, \(4 \mathrm{~mm} \times 4 \mathrm{~mm}\), located \(120 \mathrm{~mm}\) from the leading edge of the board. Experiments have revealed that flow over the board is disturbed by the elements and that convection heat transfer is correlated by an expression of the form Estimate the surface temperature of the chip if it is dissipating \(30 \mathrm{~mW}\).

For laminar free convection from a heated vertical surface, the local convection coefficient may be expressed as \(h_{x}=C x^{-1 / 4}\), where \(h_{x}\) is the coefficient at a distance \(x\) from the leading edge of the surface and the quantity \(C\), which depends on the fluid properties, is independent of \(x\). Obtain an expression for the ratio \(\bar{h}_{x} / h_{x}\), where \(\bar{h}_{x}\) is the average coefficient between the leading edge \((x=0)\) and the \(x\)-location. Sketch the variation of \(h_{x}\) and \(\bar{h}_{x}\) with \(x\).

As a means of preventing ice formation on the wings of a small, private aircraft, it is proposed that electric resistance heating elements be installed within the wings. To determine representative power requirements, consider nominal flight conditions for which the plane moves at \(100 \mathrm{~m} / \mathrm{s}\) in air that is at a temperature of \(-23^{\circ} \mathrm{C}\). If the characteristic length of the airfoil is \(L=2 \mathrm{~m}\) and wind tunnel measurements indicate an average friction coefficient of \(\bar{C}_{f}=0.0025\) for the nominal conditions, what is the average heat flux needed to maintain a surface temperature of \(T_{s}=5^{\circ} \mathrm{C}\) ?

A 20 -mm-diameter sphere is suspended in a dry airstream with a temperature of \(22^{\circ} \mathrm{C}\). The power supplied to an embedded electrical heater within the sphere is \(2.51 \mathrm{~W}\) when the surface temperature is \(32^{\circ} \mathrm{C}\). How much power is required to maintain the sphere at \(32^{\circ} \mathrm{C}\) if its outer surface has a thin porous covering saturated with water? Evaluate the properties of air and the diffusion coefficient of the air-water vapor mixture at \(300 \mathrm{~K}\).

It is known that on clear nights the air temperature need not drop below \(0^{\circ} \mathrm{C}\) before a thin layer of water on the ground will freeze. Consider such a layer of water on a clear night for which the effective sky temperature is \(-30^{\circ} \mathrm{C}\) and the convection heat transfer coefficient due to wind motion is \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The water may be assumed to have an emissivity of \(1.0\) and to be insulated from the ground as far as conduction is concerned. (a) Neglecting evaporation, determine the lowest temperature the air can have without the water freezing. (b) For the conditions given, estimate the mass transfer coefficient for water evaporation \(h_{\mathrm{m}}(\mathrm{m} / \mathrm{s})\). (c) Accounting now for the effect of evaporation, what is the lowest temperature the air can have without the water freezing? Assume the air to be dry.

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