Answer:
Explanation:
Obtain the following properties at 6MPa and 600°C from the table "Superheated water".
![h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k](https://tex.z-dn.net/?f=h_1%3D3658.8KL%2FKg%5C%5Cs_1%3D7.1693kJ%2Fkg.k)
Obtain the following properties at 10kPa from the table "saturated water"
![h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K](https://tex.z-dn.net/?f=h_%7Bf2%7D%3D191.81KJ%2FKg.K%5C%5Ch_%7Bfg2%7D%3D2392.1KJ%2FKg%5C%5Cs_%7Bf2%7D%3D0.6492KJ%2FKg.K%5C%5Cs_%7Bfg2%7D%3D7.4996KJ%2FKg.K)
Calculate the enthalpy at exit of the turbine using the energy balance equation.
![\frac{dE}{dt}=Q-W+m(h_1-h_2)](https://tex.z-dn.net/?f=%5Cfrac%7BdE%7D%7Bdt%7D%3DQ-W%2Bm%28h_1-h_2%29)
Since, the process is isentropic process ![Q=0](https://tex.z-dn.net/?f=Q%3D0)
![0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg](https://tex.z-dn.net/?f=0%3D0-W%2Bm%28h_1-h_2%29%5C%5Ch_2%3Dh_1-%5Cfrac%7BW%7D%7Bm%7D%5C%5C%5C%5Ch_2%3D3658.8-%5Cfrac%7B2626%7D%7B2%7D%5C%5C%5C%5C%3D2345.8kJ%2Fkg)
Use the isentropic relations:
![s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87](https://tex.z-dn.net/?f=s_1%3Ds_%7B2s%7D%5C%5Cs_1%3Ds_%7Bf2%7D%2Bx_%7B2s%7Ds_%7Bfg2%7D%5C%5C7.1693%3D6492%2Bx_%7B2s%7D%287.4996%29%5C%5Cx_%7B2s%7D%3D87)
Calculate the enthalpy at isentropic state 2s.
![h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg](https://tex.z-dn.net/?f=h_%7B2s%7D%3Dh_%7Bf2%7D%2Bx_%7B2s%7D.h_%7Bfg2%7D%5C%5C%3D191.81%2B0.87%282392.1%29%5C%5C%3D2272.937kJ%2Fkg)
a.)
Calculate the isentropic turbine efficiency.
![\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%](https://tex.z-dn.net/?f=%5Ceta_%7Bturbine%7D%3D%5Cfrac%7Bh_1-h_2%7D%7Bh_1-h_%7B2s%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B3658.8-2345.8%7D%7B3658.8-2272.937%7D%3D0.947%3D94.7%25)
b.)
Find the quality of the water at state 2
since
at 10KPa <
<
at 10KPa
Therefore, state 2 is in two-phase region.
![h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9](https://tex.z-dn.net/?f=h_2%3Dh_%7Bf2%7D%2Bx_2%28h_%7Bfg2%7D%29%5C%5C2345.8%3D191.81%2Bx_2%282392.1%29%5C%5Cx_2%3D0.9)
Calculate the entropy at state 2.
![s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K](https://tex.z-dn.net/?f=s_2%3Ds_%7Bf2%7D%2Bx_2.s_%7Bfg2%7D%5C%5C%3D0.6492%2B0.9%287.4996%29%5C%5C%3D7.398kJ%2FKg.K)
Calculate the rate of entropy production.
![S=\frac{Q}{T}+m(s_2-s_1)](https://tex.z-dn.net/?f=S%3D%5Cfrac%7BQ%7D%7BT%7D%2Bm%28s_2-s_1%29)
since, Q = 0
![S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k](https://tex.z-dn.net/?f=S%3Dm%28s_2-s_1%29%5C%5C%3D2%5Cfrac%7Bkg%7D%7Bs%7D%287.398-7.1693%29kJ%2Fkg%5C%5C%3D0.4574kW%2Fk)
Answer:
Valvular stenosis , Valvular prolapse , Regurgitation,
Explanation:
Answer:
Divide the difference in tax by the amount of income from the investment, and you'll get the economic marginal tax rate from investing. Most people refer to marginal tax rates as being identical to tax brackets.
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Explanation:
Answer:
// Program is written in C++ Programming Language
// Comments are used for explanatory purpose
#include<iostream>
using namespace std;
int main ()
{
// Variable declaration
string name;
int numQuestions;
int numCorrect;
double percentage;
//Prompt to enter student's first and last name
cout<<"Enter student's first and last name";
cin>>name; // this line accepts input for variable name
cout<<"Number of question on test"; //Prompt to enter number of questions on test
cin>> numQuestions; //This line accepts Input for Variable numQuestions
cout<<"Number of answers student got correct: "; // Prompt to enter number of correct answers
cin>>numCorrect; //Enter number of correct answers
percentage = numCorrect * 100 / numQuestions; // calculate percentage
cout<<name<<" "<<percentage<<"%"; // print
return 0;
}
Explanation:
The code above calculates the percentage of a student's score in a certain test.
The code is extracted from the Question and completed after extraction.
It's written in C++ programming language
Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.
Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:
Without any extra adjustments or corrections, either 125% of the continuous load, OR
When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).
This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.
To know more about connectors click here:
brainly.com/question/16987039
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