Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus 
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
![w_{b, out} =m([tex]u_{1} -u_{2](https://tex.z-dn.net/?f=w_%7Bb%2C%20out%7D%20%3Dm%28%5Btex%5Du_%7B1%7D%20-u_%7B2)
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:
11.541 mol/min
Explanation:
temperature = 35°C
Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa
note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )
from steam table it is = 5.6291 Kpa
calculate the mole fraction of H
( YH )
= 5.6291 / 151.95
= 0.03704
calculate the mole fraction of air ( Yair )
= 1 - mole fraction of water
= 1 - 0.03704 = 0.9629
Now to determine the molar flow rate of water vapor in the stream
lets assume N = Total molar flow rate
NH = molar flow rate of water
Nair = molar flow rate of air = 300 moles /min
note : Yair * n = Nair
therefore n = 300 / 0.9629 = 311.541 moles /min
Molar flowrate of water
= n - Nair
= 311.541 - 300 = 11.541 mol/min
Natural ventilation unlike fan forced ventilation uses the natural forces of wind and buoyancy to deliver fresh air into buildings
Answer:
Code in MATLab is given as below:
Explanation:
grade = input('Enter the grades as elements of a vector ');
x1 = length(grade);
fprintf('There are %5.2f grades\n',x1);
x2 = mean(grade);
fprintf('The average grade is %5.2f \n',x2);
x3=std(grade);
fprintf('The standard deviation is %5.2f \n',x3);
x4 = median(grade);
fprintf('The median grade is %5.2f \n',x4);
