Top Banner
Chapter 18. Heat Chapter 18. Heat Transfer Transfer A PowerPoint Presentation by A PowerPoint Presentation by Paul E. Tippens, Professor Paul E. Tippens, Professor of Physics of Physics Southern Polytechnic State Southern Polytechnic State University University © 2007
26
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
  • Chapter 18. Heat TransferA PowerPoint Presentation byPaul E. Tippens, Professor of PhysicsSouthern Polytechnic State University 2007

  • TRANSFER OF HEAT is minimized by multiple layers of beta cloth. These and other insulating materials protect spacecraft from hostile environmental conditions. (NASA)

  • Objectives: After finishing this unit, you should be able to:Demonstrate your understanding of conduction, convection, and radiation, and give examples.Solve thermal conductivity problems based on quantity of heat, length of path, temperature, area, and time.Solve problems involving the rate of radiation and emissivity of surfaces.

  • Heat Transfer by ConductionConduction is the process by which heat energy is transferred by adjacent molecular collisions inside a material. The medium itself does not move.

  • Heat Transfer by ConvectionConvection is the process by which heat energy is transferred by the actual mass motion of a heated fluid.ConvectionHeated fluid rises and is then replaced by cooler fluid, producing convection currents.Convection is significantly affected by geometry of heated surfaces. (wall, ceiling, floor)

  • Heat Transfer by RadiationRadiation is the process by which heat energy is transferred by electromagnetic waves.No medium is required !

  • Kinds of Heat TransferConsider the operation of a typical coffee maker:Think about how heat is transferred by:Conduction?Convection?Radiation?

  • Heat CurrentThe heat current H is defined as the quantity of heat Q transferred per unit of time t in the direction from high temperature to low temperature.Typical units are: J/s, cal/s, and Btu/h

  • Thermal ConductivityThe thermal conductivity k of a material is a measure of its ability to conduct heat.

  • The SI Units for ConductivityTaken literally, this means that for a 1-m length of copper whose cross section is 1 m2 and whose end points differ in temperature by 1 C0, heat will be conducted at the rate of 1 J/s.In SI units, typically small measures for length L and area A must be converted to meters and square meters, respectively, before substitution into formulas.

  • Older Units for ConductivityTaken literally, this means that for a 1-in. thick plate of glass whose area is 1 ft2 and whose sides differ in temperature by 1 F0, heat will be conducted at the rate of 5.6 Btu/h.Older units, still active, use common measurements for area in ft2 time in hours, length in seconds, and quantity of heat in Btus.Glass k = 5.6 Btu in./ft2h F0

  • Thermal ConductivitiesExamples of the two systems of units used for thermal conductivities of materials are given below:

  • Examples of Thermal ConductivityComparison of Heat Currents for Similar Conditions: L = 1 cm (0.39 in.); A = 1 m2 (10.8 ft2); Dt = 100 C0

  • Example 1: A large glass window measures 2 m wide and 6 m high. The inside surface is at 200C and the outside surface is at 120C. How many joules of heat pass through this window in one hour? Assume L = 1.5 cm and that k = 0.8 J/s m C0. A = (2 m)(6 m) = 12 m2Q = 18.4 MJ

  • Example 2: The wall of a freezing plant is composed of 8 cm of corkboard and 12 cm of solid concrete. The inside surface is at -200C and the outside surface is +250C. What is the interface temperature ti?Note:

  • Example 2 (Cont.): Finding the interface temperature for a composite wall.Rearranging factors gives:

  • Example 2 (Cont.): Simplifying, we obtain:0.075ti + 1.50C = 250C - tiFrom which:ti = 21.90CKnowing the interface temperature ti allows us to determine the rate of heat flow per unit of area, H/A.The quantity H/A is same for cork or concrete:

  • Example 2 (Cont.): Constant steady state flow.Over time H/A is constant so different ks cause different DtsCork: Dt = 21.90C - (-200C) = 41.9 C0Concrete: Dt = 250C - 21.90C = 3.1 C0Since H/A is the same, lets just choose concrete alone:

  • Example 2 (Cont.): Constant steady state flow.Note that 20.7 Joules of heat per second pass through the composite wall. However, the temperature interval between the faces of the cork is 13.5 times as large as for the concrete faces.If A = 10 m2, the heat flow in 1 h would be ______?745 kW

  • RadiationThe rate of radiation R is the energy emitted per unit area per unit time (power per unit area).Rate of Radiation (W/m2):Emissivity, e : 0 > e > 1Stefan-Boltzman Constant : s = 5.67 x 10-8 W/mK4

  • Example 3: A spherical surface 12 cm in radius is heated to 6270C. The emissivity is 0.12. What power is radiated?A = 0.181 m2 T = 627 + 273; T = 900 KP = 808 WPower Radiated from Surface:

  • Summary: Heat TransferConvection is the process by which heat energy is transferred by the actual mass motion of a heated fluid.Conduction: Heat energy is transferred by adjacent molecular collisions inside a material. The medium itself does not move.Radiation is the process by which heat energy is transferred by electromagnetic waves.

  • Summary of Thermal Conductivity

  • Summary of Radiation

  • Summary of Formulas

  • CONCLUSION: Chapter 18Transfer of Heat