Answer:
a. = 77.65%
b. bwr = 6.5%
c. 3538.986 kW
d. -163.169 kJ
Explanation:
a. The given property are;
P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa
p₄/p₁ = 10
P₂/P₁ = p₄/p₃ = √10
p₂ = 100·√10
= T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K
T₂ = T₁ + ( - T₁)/
= 300 + (416.85 - 300)/0.8 = 446.0625 K
p₄ = 10×p₁ = 10×100 = 1000 kPa
p₄/p₃ = √10 =
p₃ = 100·√10
T₃ = 300 K
T₃/ = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)
= T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K
T₄ = T₃ + ( - T₃)/
= 300 + (215.905- 300)/0.8 = 194.881 K
The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28
T₄ = 446.0625 K
T₆ = 1400 K
/T₆ = (1/√10)^(0.4/1.4)
= 1400×(1/√10)^(0.4/1.4) = 1007.6 K
T₇ = T₆ - (T₆ -
) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K
T₈ = 1400 K
T₉ = 1086.08 K
T₅ = T₄ + (T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K
=(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))
(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765
= 77.65%
b. Back work ratio, bwr =
((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))
40.9435/627.84 = 6.5%
c.
Power developed is given by the relation;
= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW
d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)
-163.169 kJ
The program was written in MATLAB. The program source file has an incomplete function, that needs to be completed.
Complete the function with the following code segment
function onTime = RunningLate (noTraffic, gasEmpty)
onTime = ((noTraffic) & not(gasEmpty))
end
The following, should be noted about the above code segment
I've added the image of the complete question (in a more presentable format), as an attachment.
As stated above;
After running the complete program; onTime will be true if noTraffic and gasEmpty are true and false, respectively.
Read more about MATLAB programs at;
Answer: This is EASY!
Explanation:
To make it easy, you would convert those binary numbers and to denary. And this gives:
84 104 105 115 32 105 115 32 69 65 83 89 33
Then, the denary numbers generated can be converted to ASCII code using this ASCII Table in the attachment below. And the result is: This is EASY!
Answer:
// This program is written in C++ programming language
// Comments are used for explanatory purpose
/* The aim of this program is to to remove all the white space,digits, punctuation, and other special characters, leaving only the letters. */
// Program starts here
#include <stdio.h>
#include<iostream>
using namespace std;
int main()
{
// Declare Variable of 100 characters
char word[100];
// Prompt user for input
cout<<"Your input goes here (max, 100 characters)";
cin>>word;
// Iterate through string to check for non alphabetic characters
for (int i = 0; word[i] != '\0'; ++i) {
// Check for uppercase and lowercase letters
while (!((word[i] >= 'a' && word[i] <= 'z') || (word[i] >= 'A' && word[i] <= 'Z') || word[i] == '\0')) {
for (int j = i; word[j] != '\0'; ++j) {
word[j] = word[j + 1];
}
word[j] = '\0';
}
}
cout<<"The resulting compressed string: "<<word;
return 0;
}
Answer:
w = str(input("input your values: "))
values = ' '.join(filter(str.isalpha, w))
while len(w) < 100:
print(values)
break
Explanation:
The code is written in python
w = str(input("input your values: "))
This code ask the user to input any string values with characters, numbers, line spaces , letters etc.
values = ' '.join(filter(str.isalpha, w))
This code filters the inputted value to bring only letters. All the letter are then joined together
while len(w) < 100:
The code check if the inputted value is less than 100 characters. While it is less than 100 characters. If it is less than 100 character the next code will function.
print(values)
This code prints the joined letters after checking with a while loop to confirm the length of character is less than 100
break
The break function breaks the code whether it print the values or not.
Generally, the letters will only be printed if the character inputted is less than 100 and later break the while loop or will not print any letter if the character is greater than 100 and later break.
Answer:
volume = 53.747 m3 = 14198.138 gal
weight = 526652 N = 118396.08 lbf
Explanation:
We know that volume of water
where A' = 61% of A
=1898.015 ft^3
=526652 N
Answer:
The bending stress of the face tooth is
Explanation:
From the question we are told that
The number of tooth of the pinion is
The velocity of rotation is given as
The number of tooth is of the gear is
The quality level is
The transmitted tangential load is =
The angle of the teeth is
The module is
The face width is
The diameter of the pinion is mathematically represented as
Substituting the values
The pitch line velocity is mathematically represented as
Substituting values
Generally the dynamic factor is mathematically represented as
Now B is a constant that is mathematically represented as
substituting values
A is also a constant that is mathematically represented as
Substituting values
Substituting these value into the equation for dynamic factor we have
The geometric bending factor for a 20° profile from table
"AGMA Bending Geometry Factor J for 20°, Full -Depth Teeth with HPSTC Loading , Table 2-9"
That corresponds to 55 tooth gear meshing with 26 pinion is
the diameter pitch can be mathematically represented as
Substituting values
The mathematically representation for gear tooth bending stress in the teeth face is as follows
Where is the tangential load
is the face width
is the application factor this is obtained from table "Application Factors, Table 12-17 " and the value is
= 1
is the load distributed factor
is the size factor
is the rim thickness factor which is obtained for M which has a value 1
is the idler
Substituting values into equation 1
Answer:
Q=127.66W
L=9.2mm
Explanation:
Heat transfer consists of the propagation of energy in the form of heat in different ways, these can be convection if it is through a fluid, radiation through electromagnetic waves and conduction through solid solids.
To solve any problem related to heat transfer, the general equation is used
Q = delta / R
Where
Q = heat
Delta = the temperature difference
R = is the thermal resistance by conduction, convection and radiation
to solve this problem we propose the previous equation
Q = delta / R
later we find R
Q=(25-(-5))/0.235=127.66W
part b
we use the same ecuation with Q=127.66
Q = delta / R
Δ