Answer:
True, True
Explanation:
1st Scenario: Purchase of new contamination equipment would indicate that the plant had been violating environmental regulations and reporting false data. So, Cathy and Henry must not purchase and install new contamination equipment unless the data indicates serious violatino of environmental regulations.
2nd Scenario: Company has the responsibiltiy to report contamination in plant water to regulatory agencies and Cathy Martin has been assigned this responsibility to report on behalf of the company. Reporting "adjusted" or inaccurate data is against the business ethics. So, professionally Cathy Martin is responsible for her reports to regulatory agencies.
Answer:
The rate of irreversible loss will be "55.22 MW".
Explanation:
The given values are:
Elevation,
h = 120 m
Flow of water,
Q = 100 m³/s
Efficiency,
= 80%
i.e,
= 0.8
Efficiency turbine,
= 50 MW
Now,
Without any loss,
The power generated by turbine will be:
⇒ 
On substituting the values, we get
⇒ 
⇒ 
Power generated in actual will be:
= 
= 
Hence,
Throughout the piping system,
The rate of irreversible loss is:
= 
= 
= 
Answer:
a. 81 kj/kg
b. 420.625K
c. 101.24kj/kg
Explanation:
![\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7Bt1%7D%20%3D%5B%5Cfrac%7Bp2%7D%7Bp1%7D%20%5D%5E%7B%5Cfrac%7By-1%7D%7By%7D%20%7D)
t1 = 360
p1 = 0.4mpa
p2 = 1.20
y = 1.13
substitute these values into the equation
![\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.20%7D%7B0.4%7D%20%5D%5E%7B%5Cfrac%7B1.13-1%7D%7B1.13%7D%20%7D)
![\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.2%7D%7B0.4%7D%20%5D%5E%7B0.1150%7D%5C%5C%5Cfrac%7Bt2%7D%7B360%7D%20%3D1.1347)
when we cross multiply
t2 = 360 * 1.1347
= 408.5
a. the work required in the firs compressor
w=c(t2-t1)
c=1.67x10³
t1 = 360
t2 = 408.5
w = 1670(408.5-360)
= 1670*48.5
= 80995 J
= 81KJ/kg
b. 
n = 80%
t2 = 408.5
t1 = 360
0.80 = 408.5-360 ÷ t'2-360

cross multiply to get the value of t'2
0.80(t'2-360) = 48.5
0.80t'2 - 288 = 48.5
0.8t'2 = 48.5+288
0.8t'2 = 336.5
t'2 = 336.5/0.8
= 420.625
this is the temperature at the exit of the first compressor
c. cooling requirement
w = c(t2-t1)
= 1.67x10³(420.625-360)
= 1670*60.625
= 101243.75
= 101.24kj/kg
You need to attach the image
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