A long conducting rod of rectangular cross section (20 mm 30 mm) and thermal conductivity k 20 W/m K experiences uniform heat ge
neration at a rate q . 5 107 W/m3, while its surfaces are maintained at 300 K. Using a finite-difference method with a grid spacing of 5 mm, determine the temperature distribution in the rod.
1 answer:
Answer:
Explanation:
We are assuming that there is
a steady state two dimensional conduction
constant properties
uniform volumetric heat generation
From symmetry, we will be determining 6 unknown temperatures.
See attachment for calculation and and tabulation
With T(s) = 300 K, the set of equations were written directly into the IHT work space and solved for nodal temperatures.
The result is seen in the second attachment
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Answer:
For block A, a = 9.66 ft/s²
For block B, a = 15 ft/s²
Explanation:
A free body diagram for this force system is attached to this solution
Mass of block A = m₁ = 8 lb
Mass of block B = m₂ = 6 lb
Coefficient of kinetic friction = μ
Normal reaction on the blocks = N
Spring stiffness of the spring btw block A and B = k = 20 lb/ft
Compression of the spring = 0.2 ft
Analysing Block A first
The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 8 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 8/32.2 = 0.248 lbm
(20×0.2) - (0.2 × 8) = (0.248) aₓ
aₓ = 9.66 ft/s²
The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 6 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 6/32.2 = 0.186 lbm
(20×0.2) - (0.2 × 6) = (0.186) aₓ
aₓ = 15 ft/s²
