Introduction to Heat Transfer
6th Edition
ISBN: 9780470501962
Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine
Publisher: Wiley, John & Sons, Incorporated
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Chapter 2, Problem 2.29P
(a)
To determine
The expression for
(b)
To determine
The expression for
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Chapter 2 Solutions
Introduction to Heat Transfer
Ch. 2 - Assume steady-state, one-dimensional heat...Ch. 2 - Assume steady-state, one-dimensional conduction in...Ch. 2 - A hot water pipe with outside radius r1 has a...Ch. 2 - A spherical shell with inner radius r1 and outer...Ch. 2 - Assume steady-state, one-dimensional heat...Ch. 2 - A composite rod consists of two different...Ch. 2 - A solid, truncated cone serves as a support for a...Ch. 2 - To determine the effect of the temperature...Ch. 2 - Prob. 2.9PCh. 2 - A one-dimensional plane wall of thickness 2L=100mm...
Ch. 2 - Consider steady-state conditions for...Ch. 2 - Consider a plane wall 100 mm thick and of thermal...Ch. 2 - Prob. 2.13PCh. 2 - In the two-dimensional body illustrated, the...Ch. 2 - Consider the geometry of Problem 2.14 for the case...Ch. 2 - Steady-state, one-dimensional conduction occurs in...Ch. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Consider a 300mm300mm window in an aircraft. For a...Ch. 2 - Prob. 2.20PCh. 2 - Use IHT to perform the following tasks. Graph the...Ch. 2 - Calculate the thermal conductivity of air,...Ch. 2 - A method for determining the thermal conductivity...Ch. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - At a given instant of time, the temperature...Ch. 2 - Prob. 2.27PCh. 2 - Uniform internal heat generation at q.=5107W/m3 is...Ch. 2 - Prob. 2.29PCh. 2 - The steady-state temperature distribution in a...Ch. 2 - The temperature distribution across a wall 0.3 m...Ch. 2 - Prob. 2.32PCh. 2 - Prob. 2.33PCh. 2 - Prob. 2.34PCh. 2 - Prob. 2.35PCh. 2 - Prob. 2.36PCh. 2 - Prob. 2.37PCh. 2 - One-dimensional, steady-state conduction with no...Ch. 2 - One-dimensional, steady-state conduction with no...Ch. 2 - The steady-state temperature distribution in a...Ch. 2 - One-dimensional, steady-state conduction with no...Ch. 2 - Prob. 2.42PCh. 2 - Prob. 2.43PCh. 2 - Prob. 2.44PCh. 2 - Beginning with a differential control volume in...Ch. 2 - A steam pipe is wrapped with insulation of inner...Ch. 2 - Prob. 2.47PCh. 2 - Prob. 2.48PCh. 2 - Two-dimensional, steady-state conduction occurs in...Ch. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - A chemically reacting mixture is stored in a...Ch. 2 - A thin electrical heater dissipating 4000W/m2 is...Ch. 2 - The one-dimensional system of mass M with constant...Ch. 2 - Consider a one-dimensional plane wall of thickness...Ch. 2 - A large plate of thickness 2L is at a uniform...Ch. 2 - Prob. 2.57PCh. 2 - Prob. 2.58PCh. 2 - A plane wall has constant properties, no internal...Ch. 2 - A plane wall with constant properties is initially...Ch. 2 - Consider the conditions associated with Problem...Ch. 2 - Prob. 2.62PCh. 2 - A spherical particle of radius r1 experiences...Ch. 2 - Prob. 2.64PCh. 2 - A plane wall of thickness L=0.1m experiences...Ch. 2 - Prob. 2.66PCh. 2 - A composite one-dimensional plane wall is of...Ch. 2 - Prob. 2.68PCh. 2 - The steady-state temperature distribution in a...
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- 5.10 Experiments have been performed on the temperature distribution in a homogeneous long cylinder (0.1 m diameter, thermal conductivity of 0.2 W/m K) with uniform internal heat generation. By dimensional analysis, determine the relation between the steady-state temperature at the center of the cylinder , the diameter, the thermal conductivity, and the rate of heat generation. Take the temperature at the surface as your datum. What is the equation for the center temperature if the difference between center and surface temperature is when the heat generation is ?arrow_forwardHeat is generated uniformly in the fuel rod of a nuclear reactor. The rod has a long, hollow cylindrical shape with its inner and outer surfaces at temperatures of TiandTo, respectively. Derive an expression for the temperature distribution.arrow_forward1.3 A furnace wall is to be constructed of brick having standard dimensions of Two kinds of material are available. One has a maximum usable temperature of 1040°C and a thermal conductivity of 1.7 W/(m K), and the other has a maximum temperature limit of 870°C and a thermal conductivity of 0.85 W/(m K). The bricks have the same cost and are laid in any manner, but we wish to design the most economical wall for a furnace with a temperature of 1040°C on the hot side and 200°C on the cold side. If the maximum amount of heat transfer permissible is 950 , determine the most economical arrangement using the available bricks.arrow_forward
- Consider a solid sphere of radius R with a fixed surface temperature, TR. Heat is generated within the solid at a rate per unit volume given by q = ₁ + ₂r; where ₁ and ₂ are constants. (a) Assuming constant thermal conductivity, use the conduction equation to derive an expression for the steady-state temperature profile, T(r), in the sphere. (b) Calculate the temperature at the center of the sphere for the following parameter values: R=3 m 1₁-20 W/m³ TR-20 °C k-0.5 W/(m K) ₂-10 W/m³arrow_forwardlook into the attached filearrow_forwardA solid cylinder of radius R and length L is made from material with thermal conductivity 2. Heat is generated inside the cylinder at a rate S (energy per unit volume per unit time). (a) Neglecting conduction along the axis of the cylinder, find the steady-state temperature distribution in the cylinder, given that the surface temperature is Ts. (b) Consider a crude approximation of a mouse modeled as a cylinder of radius 1 cm and length 5 cm. If the ambient air temperature is 10°C and the internal rate of heat generation in the animal is 10-³ W/cm³, find the skin temperature (Ts) for the mouse. The external heat-transfer coefficient is h = 0.2 W/m².K. (You can neglect conduction along the axis of the mouse, as in part a.)arrow_forward
- Question 5: Assume steady-state, one-dimensional heat conductionthrough the symmetric shape shown in Figure 1.Assuming that there is no internal heat generation, derivean expression for the thermal conductivity k(x) for theseconditions: A(x) = (1 -x), T(x) = 300(1 - 2x -x3),and q = 6000 W, where A is in square meters, T inkelvins, and x in meters. Consider x= 0 and 1arrow_forward2. The slab shown is embedded in insulating materials on five sides, while the front face experiences convection off its face. Heat is generated inside the material by an exothermic reaction equal to 1.0 kW/m'. The thermal conductivity of the slab is 0.2 W/mk. a. Simplify the heat conduction equation and integrate the resulting ID steady form of to find the temperature distribution of the slab, T(x). b. Present the temperature of the front and back faces of the slab. n-20- 10 cm IT- 25°C) 100 cm 100 cmarrow_forwardOne more time. PLEASE explain how the integral is formed, dT/dr doesn't make sense. Why we are replacing L with dr? dr is in radial direction and L is in the vertical direction.arrow_forward
- energy generation is öccuring inner radius r, and outer radius r2. Under what condi- a Spic tion is the linear temperature distribution shown in Problem 2.38 possible? 2.40 The steady-state temperature distribution in a one- dimensional wall of thermal conductivity k and thick- ness L is of the form T= ax +bx+ cx + d. Derive expressions for the heat generation rate per unit volume in the wall and the heat fluxes at the two wall faces (x = 0, L). 2.41 One-dimensional, steady-state conduction with no energy generation is occurring in a plane wall of con- stant thermal conductivity. 120 100arrow_forwardProblem 4 A cork board (k = 0.039W/m K) 3 cm thick is exposed to air with an average temperature T.. = 30°C with a convection heat transfer coefficient, h = 25 W/m².K. The other surface of the board is held at a constant temperature of 15°C. A volumetric heat generation of 5 W/m³ is occurring inside the wall. Assuming one dimensional conduction, write the governing equation and express the boundary conditions for the wall. (solve the equation is not required) Problomarrow_forwardAfter a thorough derivation by Doraemon to establish an equation for cylindrical fuel rod of a nuclear reactor. Here he was able to come up an equation of heat generated internally as shown below. 9G = 9. where qG is the local rate of heat generation per unit volume at radius r, ro is the outside radius, and qo is the rate of heat generation per unit volume at the centre line. Calculate the temperature drop from the centre line to the surface for a 2.5 cm outer diameter rod having k = 25 W/m K, if the rate of heat removal from the surface is 1650 kW/m² А) 619°C В 719 °C C) 819 °C D) 919 °C E 1019 °C F None of thesearrow_forward
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