{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Thermodynamics HW Solutions 93

Thermodynamics HW Solutions 93 - Chapter 2 Heat Conduction...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 2 Heat Conduction Equation 2-46E A 1.5-kW resistance heater wire is used for space heating. Assuming constant thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem is to be obtained for steady operation. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3 Heat is generated uniformly in the wire. Analysis The heat flux at the surface of the wire is & & & (. q Q A G rL s s == = = s 2 W in)(15 in) .2 W / in 2 1200 20 0 6 212 0 ππ Noting that there is thermal symmetry about the center line and there is uniform heat flux at the outer surface, the differential equation and the boundary conditions for this heat conduction problem can be expressed as 0 1 = + k g dr dT r dr d r & 2 kW L = 15 in D = 0.12 in dT dr k dT r dr q s () & 0 0 212 0 = −= = .2 W / in 2 2-47 Heat conduction through the bottom section of an aluminum pan that is used to cook stew on top of
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online