/*! 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 96 Unique characteristics of biolog... [FREE SOLUTION] | 91Ó°ÊÓ

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Unique characteristics of biologically active materials such as fruits, vegetables, and cther products require special care in handling. Following harvest and separation from producing plants, glucose is catabolized to produce carbon diexide, water vapor, and heat, with attendant internal energy generation. Consider a carton of apples, each of 80-mm diameter, which is ventilated with air at \(5^{\circ} \mathrm{C}\) and a velocity of \(0.5 \mathrm{~m} / \mathrm{s}\). The corresponding value od the heat transfer coefficient is \(7.5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Within each apple thermal energy is uniformly generated at a total rate of \(4000 \mathrm{~J} / \mathrm{kg}\) - diry. The density and thermal conductivity of the apple are \(840 \mathrm{~kg} / \mathrm{m}^{3}\) and \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively.(a) Determine the apple center and surface temperatures. (b) For the stacked arrangement of apples within the crate, the convection coefficient depends on the velocity as \(h=C_{1} v^{0.02 s}\), where \(C_{1}=10.1 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K} \cdot(\mathrm{m} / \mathrm{s})^{0.05}\). Compute and plot the center and surface temperatures as a function of the air velocity for \(0.1 \leq V \leq 1 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
(a) Use heat balance to find temperatures. (b) Vary velocity to see effects on temperatures.

Step by step solution

01

Understanding the Problem

We have a carton of apples that generates internal heat and is ventilated to remove this heat. We need to find the center and surface temperatures of an apple when the air velocity is known, and also determine how these temperatures change as air velocity varies.
02

Defining Constants and Known Parameters

Given: Diameter of apple, \( D = 0.08 \) m; Ambient temperature, \( T_\infty = 5^\circ C \); Velocity, \( v = 0.5 \ m/s \); Heat transfer coefficient, \( h = 7.5 \ W/m^2 \cdot K \); Heat generation rate, \( q = 4000 \ J/kg \cdot s \); Density, \( \rho = 840 \ kg/m^3 \); Thermal conductivity, \( k = 0.5 \ W/m\cdot K \).
03

Calculating Heat Generation in the Apple

Calculate the volumetric heat generation, \( \dot{q} = \frac{q \cdot \rho}{t} = 4000 \times 840 \). This provides the rate at which heat is generated throughout the body of the apple.
04

Applying the Lumped Capacitance Method

Given the geometry and uniform generation, apply the lumped capacitance method to estimate temperatures. Assume the Biot number \( Bi = \frac{hD}{k} \) to be much less than 1, allowing simplification to regard the apple as a lump mass.
05

Using the Heat Balance Equation

For a steady-state heat balance, the generated heat must equal the heat lost via convection: \[ \dot{q} \times \frac{4}{3} \pi \left(\frac{D}{2}\right)^3 = hA(T_{apple} - T_\infty) \]. Solving this will yield \( T_{apple} \), since \( A \) is the surface area \( A = \pi D^2 \).
06

Calculating Steady-State Temperatures

Solve for \( T_{center} \) and \( T_{surface} \) using the heat balance equation. Use above expressions to input the known constants and perform the numerical calculations.
07

Modeling Temperature Variation with Velocity

Use \( h = C_1 v^{0.05} \) to determine the heat transfer coefficient for various velocities. Apply the results in the heat balance equation to find how temperatures vary with \( v \), produce a plot based on these computations.

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Key Concepts

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

Thermal Conductivity in Fruits
Thermal conductivity in fruits like apples is a measure of their ability to conduct heat. An apple's composition—mainly water, carbohydrates, and fibers—affects how heat moves through it. For apples, the thermal conductivity is given as \( 0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \). This is crucial because it influences how quickly heat generated internally can be transferred to its surface and then to the surrounding environment. Higher conductivity would mean that heat moves through the apple more efficiently, reducing temperature gradients inside it.In essence, thermal conductivity is the bridge for heat transfer within the apple, impacting cooling or heating rates when subjected to temperature changes. For biological materials, this property is vital in understanding and controlling processes like ripening and spoilage.
Convection Coefficient Dependency
The convection coefficient \( h \) illustrates how effectively heat is transferred from the surface of an object—in this case, an apple—to the surrounding fluid, typically air. Dependency on air velocity is an interesting aspect here.In the problem, the convection coefficient is defined by the relationship \( h = 10.1 v^{0.05} \), where \( v \) is air velocity in \( \mathrm{m/s} \). This indicates that as air velocity increases, the convection coefficient also increases, facilitating better heat transfer.This dependency is key in designing storage and transport systems for fruits like apples. By optimizing air flow, one can enhance the cooling or heating process, maintaining desired temperatures efficiently.
Biological Heat Generation
Fruits generate heat through biological processes like respiration. This heat generation can affect storage conditions since it leads to self-heating.In our problem, apples generate thermal energy at a rate of \( 4000 \mathrm{~J}/\mathrm{kg} \), derived from metabolic processes converting glucose to energy. This internal energy must be accounted for in storage, as it can lead to temperature increases, affecting quality and shelf life.Understanding such heat generation helps in designing systems to manage and dissipate this heat, especially in storage where a constant temperature is crucial for preserving the fruit's integrity and freshness.
Lumped Capacitance Method
The lumped capacitance method is a simplification technique used for analyzing transient heat transfer problems. For it to be applicable, the Biot number \( Bi = \frac{hD}{k} \) must be much less than 1, indicating uniform temperature throughout the object.In the case of an apple, this method helps approximate the center and surface temperatures under the assumption that thermal gradients inside the apple are negligible. This allows for easier calculations using a heat balance.This tool is vital in simplifying complex conduction problems in biological materials, providing a quick and less computationally intensive way to estimate temperatures when detailed spatial temperature distribution is not required.

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Most popular questions from this chapter

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between shect metal pancls. Consider a wall made from fiberglass insulation of thermal conductivity \(k_{1}=0.046 \mathrm{~W} / \mathrm{m}-\mathrm{K}\) and thickness \(I_{-}=50 \mathrm{~mm}\) and steel parels, each of thermal conductivity \(k_{p}=60\) W/m \(: \mathrm{K}\) and thickness \(L_{7}=3 \mathrm{~mm}\). If the wall separates refrigerated air at \(T_{x, j}=4^{\circ} \mathrm{C}\) from ambient air at \(T_{2,0}=25^{\circ} \mathrm{C}\), what is the heat gain per unit surface area? Cocfficicnts associated with natural convection at the inner and outer surfaces may be approximated as \(h_{y}=h_{g}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A 2-mm-diameter electrical wire is insulated by a 2 -mm-thick rubberized sheath \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and the wirc/sheath interface is characterized by a thermal contuct resistance of \(R_{t,}^{\prime \prime}=3 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The convection heat transfer coefficient at the outer surface of the sheath is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the tempenature of the ambient air is \(20^{\circ} \mathrm{C}\). If the temperature of the insulation may not exceed \(50^{\circ} \mathrm{C}\), what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

A hollow aluminem sphere, with an clectrical heater in the center, is used in tests to determine the thermal conductivity of insulating materials. The inner and outer radii of the sphere are \(0.15\) and \(0.18 \mathrm{~m}\), respectively. and testing is done under steady-state conditions with the inner surface of the aluminum maintained at \(250^{\circ} \mathrm{C}\). In a particular test, a spherical shell of insulation is cast on the outer surface of the sphere to a thickness of \(0.12 \mathrm{~m}\). The system is in a roorn for which the air temperature is \(20^{\circ} \mathrm{C}\) and the convection cocfficient at the outer surface of the insulation is \(30 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). If \(80 \mathrm{~W}\) are dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?

Aluminum fins of triangular profile are attached to a plane wall whose surface temperature is \(250^{\circ} \mathrm{C}\). The fin base thickness is \(2 \mathrm{~mm}\), and its length is \(6 \mathrm{~mm}\). The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) What is the heat dissipated per unit width by a single fin?

The evaporator section of a refrigeration unit consists of thin-walled, 10-mm- diameter tubes through which refrigcrant passes at a temperature of \(-18^{\circ} \mathrm{C}\). Air is cooled as it flows over the tubes, maintaining a surface convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2}\) - \(\mathrm{K}\), 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 laycr 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 \(2 \mathrm{~mm}\)-thick layer of frost has formed. If the tubes are in ambient air for which \(T_{\mathrm{w}}=20^{\circ} \mathrm{C}\) 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 assamed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a later heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

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