Answer:
Explanation:
Given:
The two rods could be approximated as a fins of infinite length.
TA = 75 0C θA = (TA - T∞) = 75 - 25 = 50 0C
TB = 55 0C θB = (TB - T∞) = 55 - 25 = 30 0C
Tb = 100 0C θb = (Tb - T∞) = (100 - 25) = 75 0C
KA = 200 W/m · K
T∞ = 25 0C
Solution:
The temperature distribution for the infinite fins are given by
θ/θb=e⁻mx
θA/θb= e-√(hp/A.kA) x1 ....................(1)
θB/θb = e-√(hp/A.kB) x1.......................(2)
Taking natural log on both sides we get,
Ln(θA/θb) = -√(hp/A.kA) x1 ...................(3)
Ln(θB/θb) = -√(hp/A.kB) x1 .....................(4)
Dicving (3) and (4) we get
[ Ln(θA/θb) /Ln(θB/θb)] = √(KB/KA)
[ Ln(50/75) /Ln(30/75)] = √(KB/200)
Answer:
Here is the code in Matlab for the function.
I have also attached the m file for function as well as the test run of the code here and screenshot of the result.
Code:
function [ C ] = columnproduct( A, B )
% get the dimesnions of A
sizeA = size(A);
sizeB = size(B);
% check if columns of A are same as rows of B
if(sizeA(2) ~= sizeB(1))
error('matrix dimensions do not match')
end
% initialize resultant matrix
C = [];
for i = 1:sizeB(2)
% concatenating product of matrix A with each column of B
C = [C A*B(:,i)];
end
end
Answer:
(a) 148.148 lb/ft^2
(b) 62.245 ft/s
Explanation:
In this question, we are asked to calculate the divergence dynamic pressure at sea level and the divergence airspeed at sea level of a torsionally elastic wind tunnel of model of uniform wing.
Please check attachment for complete step by step solution