/*! 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 31 A heating panel is to be placed ... [FREE SOLUTION] | 91Ó°ÊÓ

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A heating panel is to be placed on the ceiling of a hospital room so that the patient will not be uncomfortable in \(290 \mathrm{~K}\) air. The panel is painted with an off-white matte paint of emittance \(0.88\). As a design criterion it is required that the patient's face-modeled as a dry, black adiabatic surface-attains a \(305 \mathrm{~K}\) equilibrium temperature. The panel temperature is \(355 \mathrm{~K}\) and is located \(2 \mathrm{~m}\) above the patient. The convective heat transfer coefficient for the face can be taken as \(4.0 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and it can be assumed that the room walls are black at the air temperature. If the panel is to be a circular disk, determine its diameter.

Short Answer

Expert verified
The diameter of the circular panel is determined using radiative and convective heat balance equations.

Step by step solution

01

Understand the Problem

The problem involves a heating panel in a hospital room where the patient is supposed to be comfortable at a specific equilibrium temperature under specific room conditions. We need to calculate the diameter of the circular panel required to achieve this.
02

Establish the Energy Balance

The energy balance on the patient's face involves the radiative heat transfer from the panel and surroundings to the face and the convective heat transfer from the face to the air. According to Stefan-Boltzmann Law and Newton's Law of Cooling, the total heat gained should equal the total heat lost when equilibrium is reached.
03

Calculate Radiative Heat Transfer

Use the Stefan-Boltzmann Law for radiative heat transfer: \[ Q_{rad} = ext{emittance} imes ext{area} imes ext{Stefan-Boltzmann constant} \times (T_p^4 - T_s^4) \]where the emittance is 0.88, the area is the panel area, and temperatures are given respectively.
04

Calculate Convective Heat Transfer

The convective heat transfer from the patient's face is given by Newton's Law of Cooling:\[ Q_{conv} = h imes A imes (T_s - T_{ ext{air}}) \]where \( h = 4.0 \, \mathrm{W/m^2 K} \), \( A \) is the patient's face area, and the temperature difference is \( T_s - T_{ ext{air}} \).
05

Equate Radiative and Convective Heat Transfer

At equilibrium, the heat gained by radiation and lost by convection are equal:\[Q_{rad} = Q_{conv}\]Insert the expressions for \( Q_{rad} \) and \( Q_{conv} \) and solve for the panel area \( A \).
06

Solve for Panel Diameter

The panel is a circle, so its area is \( A = \pi (d/2)^2 \). Solve for \( d \) (the diameter) given the area calculated in Step 5.

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Key Concepts

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

Stefan-Boltzmann Law
Radiative heat transfer is a key concept in understanding how heat is exchanged between objects without needing a medium, like air or water, to transfer it. One of the crucial components in radiative heat transfer is the Stefan-Boltzmann Law. This law tells us that the amount of energy radiated per unit surface area of a blackbody is proportional to the fourth power of the blackbody's absolute temperature. For practical purposes, when we consider surfaces that are not perfect blackbodies, we incorporate the concept of emittance. Emittance is a factor that represents how effective a surface is at emitting thermal radiation compared to a perfect blackbody.
  • The Stefan-Boltzmann Law is mathematically expressed as: \( Q_{rad} = ext{emittance} \times ext{area} \times ext{Stefan-Boltzmann constant} \times (T_p^4 - T_s^4) \).
  • In this case, emittance describes the efficiency of heat panel's painted surface, which is given as 0.88.
  • The Stefan-Boltzmann constant in this equation is a universal constant valued at \(5.67 \times 10^{-8} \, \mathrm{W/m^2K^4} \).
  • The temperature differences are significant as the power of four highlights how small changes in temperature lead to large changes in radiated energy.
Understanding these variables helps to solve the problem of determining how much energy is radiated by the panel to achieve the desired heating for the patient.
Convective Heat Transfer
Convective heat transfer describes the transfer of heat between a solid surface and a fluid (which might be a gas like air or even a liquid) in motion. In this exercise, we consider how the patient's face loses heat to the surrounding air via convection, which is quantified by Newton's Law of Cooling. This principle states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its environment, as well as the properties of the surface and fluid in contact.
  • The formula for convective heat transfer is: \( Q_{conv} = h \times A \times (T_s - T_{air}) \).
  • The convective heat transfer coefficient \( h \) accounts for the ease of heat transfer through convection, given here as \(4.0 \, \mathrm{W/m^2K} \).
  • Where \( A \) is the contact area and \((T_s - T_{air})\) is the temperature difference between the surface and the ambient air.
This convective process is essential in maintaining a balance where the heat loss through convection equals the heat gains through radiation, thereby stabilizing the temperature at the patient's face.
Energy Balance
The concept of energy balance plays a pivotal role in ensuring the patient's comfort in the hospital room. At equilibrium, the heat transfer through radiation and convection must be balanced. This means that the amount of energy the patient's face receives from the panel and surroundings is equal to the energy it loses to the air. Maintaining this equilibrium is crucial for achieving the desired temperature of 305 K, ensuring patient comfort.
  • For equilibrium, the radiative heat gained \( Q_{rad} \) must equal the convective heat lost \( Q_{conv} \).
  • By setting \( Q_{rad} = Q_{conv} \), you can solve for unknown variables, such as the area of the heating panel.
  • The solution involves calculating the required diameter of the panel so that the energy input and output are balanced at the given temperatures.
This energy balance ensures that the patient remains at the specific equilibrium temperature without becoming too hot or too cold. By carefully calculating and setting up these parameters, the design satisfies the required comfort conditions with precision.

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Most popular questions from this chapter

A \(10 \mathrm{~cm}-\mathrm{square}\) horizontal plate coated with a black paint \((\varepsilon=0.94)\) is exposed to a clear night sky when the ambient air is still at \(290 \mathrm{~K}, 1\) atm, and \(80 \% \mathrm{RH}\). If the backside of the plate is perfectly insulated, will dew form on the plate? Use the Brunt formula for sky emittance.

A long, square evacuated duct has \(1 \mathrm{~m}\) sides. Two adjacent sides are perfectly insulated and have an emittance of \(0.4\). The other two sides have an emittance of \(0.5\). One of these sides is cooled and has a thermostat temperature control; the other contains an embedded electrical heater. If the thermostat maintains the cooled surface at \(1000 \mathrm{~K}\), at what rate must power be supplied to the electrical heater to maintain the two insulated sides at \(2000 \mathrm{~K}\) ? Also, what is the temperature of the heated side? (i) Assume that the two insulated sides can be represented by a single isothermal surface. (ii) Can the problem be solved if the two insulated sides are represented by separate isothermal surfaces? (iii) Comment on the validity of the approach used in part (i).

Calculate the blackbody emissive power \(E_{b}\) for surfaces at \(300 \mathrm{~K}, 1000 \mathrm{~K}, 2000 \mathrm{~K}\), and \(5000 \mathrm{~K}\). Also estimate the peak wavelength \(\lambda_{\max }\) for each surface using: (i) \(\lambda_{\max } T=2898 \mu \mathrm{m} \mathrm{K}\) (from Eq. \(6.6\) and Fig. \(6.3 a\) ) (ii) \(\lambda_{\max } T=5100 \mu \mathrm{m} \mathrm{K}\) (from Fig. 6.3b)

Hot combustion gases flow in a duct whose walls are at \(600 \mathrm{~K}\). A thermocouple located in the center of the duct records a temperature of \(873 \mathrm{~K}\). If the emittance of the thermocouple is \(0.8\) and the heat transfer coefficient between the gas and the thermocouple is \(110 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), determine the true gas temperature. How much gain in accuracy is obtained if the thermocouple is surrounded by a radiation shield in the form of a cylinder \(1 \mathrm{~cm}\) in diameter and \(4 \mathrm{~cm}\) long, made from a thin-wall stainless steel tube with its axis in the flow direction? Take the emittance of the shield as \(0.6\) and the heat transfer coefficient between the gas and the shield as 40 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\)

A \(100 \mathrm{~W}, 10 \mathrm{~cm}\)-diameter disk heater is placed parallel to, and \(5 \mathrm{~cm}\) from, a 10 \(\mathrm{cm}\)-diameter disk receiver in an evacuated chamber. The backs of both disks are well insulated. If the disk surfaces have emittances of \(0.8\) and the walls of the chamber are black and maintained at \(350 \mathrm{~K}\), calculate the following quantities: (i) The heater temperature. (ii) The receiver temperature. (iii) The radiative heat transfer between the heater and receiver. (iv) The net radiative heat transfer to the chamber.

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