Heat flux The heat flow vector field for conducting objects is F = – k ▿ T , where T ( x , y , z ) is the temperature in the object and k > 0 is a constant that depends on the material. Compute the outward flux of F across the following surfaces S for the given temperature distributions. Assume k = 1. 61. T ( x , y , z ) = 100 e − x − y ; S consists of the faces of the cube | x | ≤ 1 , | y | ≤ 1 , | z | ≤ 1 .
Heat flux The heat flow vector field for conducting objects is F = – k ▿ T , where T ( x , y , z ) is the temperature in the object and k > 0 is a constant that depends on the material. Compute the outward flux of F across the following surfaces S for the given temperature distributions. Assume k = 1. 61. T ( x , y , z ) = 100 e − x − y ; S consists of the faces of the cube | x | ≤ 1 , | y | ≤ 1 , | z | ≤ 1 .
Solution Summary: The author explains how to compute the outward flux of F across the surface S.
Heat fluxThe heat flow vector field for conducting objects isF = –k▿T, where T(x, y, z) is the temperature in the object and k > 0 is a constant that depends on the material. Compute the outward flux ofFacross the following surfaces S for the given temperature distributions. Assume k = 1.
61.
T
(
x
,
y
,
z
)
=
100
e
−
x
−
y
; S consists of the faces of the cube
|
x
|
≤
1
,
|
y
|
≤
1
,
|
z
|
≤
1
.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
Fourier's Law of heat transfer (or heat conduction) states that the heat flow vector F at a point is proportional to the negative gradient of the temperature: that is, F = − kVT, which means that heat
energy flows from hot regions to cold regions. The constant k is called the conductivity, which has metric units of J/m-s-K or W/m-K. A temperature function T for a region D is given below. Find the
SSF
FondSk
-KSS
VT n dS across the boundary S of D. It may be easier to use the Divergence Theorem and evaluate a triple integral. Assume that k = 1.
S
S
T(x,y,z) = 65e¯x² - y² − z²;
net outward heat flux
D is the sphere of radius a centered at the origin.
Fourier's Law of heat transfer (or heat conduction) states that the heat flow vector F at a point is proportional to the negative gradient of the temperature;
that is, F = -KVT, which means that heat energy flows from hot regions to cold regions. The constant k is called the conductivity, which has metric units
SS
S
of J/m-s-K or W/m-K. A temperature function T for a region D is given below. Find the net outward heat flux
boundary S of D. It may be easier to use the Divergence Theorem and evaluate a triple integral. Assume that k = 1.
T(x,y,z) = 100 - 5x+ 5y +z; D = {(x,y,z): 0≤x≤5, 0≤y≤4, 0≤z≤ 1}
The net outward heat flux across the boundary is
(Type an exact answer, using as needed.)
-KSS
S
F.ndS = -k
VT n dS across the
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