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cricket20 [7]
3 years ago
12

A heat exchanger takes compressed liquid water and converts it into steam with a temperature and pressure of 20 Mpa and 480 C; r

espectively. Steam is used to generate work �! in a turbine with an isentropic efficiency of 78%. Steam that exits the turbine at 14 Mpa is used in a second identical turbine (same efficiency) after reheating it to 500 C (with the same heating exchanger). Once the steam exits the second turbine it passes through a condenser (working at a pressure of 6 Kpa) before being pumped (with an internal reversible and adiabatic pump) back to the heat exchanger.
A. Work done by the first turbine (KJ/Kg).
B. Work done by the second turbine (KJ/Kg).
C. Quality of steam leaving the second turbine (5pts).
D. Cycle thermal efficiency (5pts).

Engineering
1 answer:
belka [17]3 years ago
6 0

Answer:

See explaination

Explanation:

Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.

Please kindly check attachment for the step by step solution of the given problem.

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Explanation:

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A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r
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3 years ago
C#: Arrays - Ask the user how many students names they want to store. You will create two parallel arrays (e.g. 2 arrays with th
zhenek [66]

Answer:

  1. using System;      
  2. public class Program
  3. {
  4. public static void Main()
  5. {
  6.  Console.WriteLine("Enter number of students: ");
  7.  int num = Convert.ToInt32(Console.ReadLine());
  8.  string [] firstName = new string[num];
  9.  string [] lastName = new string[num];
  10.  
  11.  for(int i=0 ; i < num; i++){
  12.   Console.WriteLine("Enter first name: ");
  13.   firstName[i] = Console.ReadLine();
  14.    
  15.   Console.WriteLine("Enter last name: ");
  16.   lastName[i] = Console.ReadLine();
  17.  }
  18.  
  19.  for(int j=0; j < num; j++){
  20.   Console.WriteLine(lastName[j] + "," + firstName[j]);
  21.  }
  22. }
  23. }

Explanation:

Firstly, prompt user to enter number of student to be stored (Line 6- 7). Next, create two array, firstName and lastName with num size (Line 8-9).

Create a for-loop to repeat for num times and prompt user to enter first name and last name and then store them in the firstName and lastName array, respectively (Line 11 - 17).

Create another for loop to traverse through the lastName and firstName array and display the last name and first name by following the format given in the question (Line 19 - 21).

4 0
3 years ago
An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve
lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is  that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa  against the claim Ha: u< 3.1 MPa

The alternate hypothesis is  that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68  falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0
3 years ago
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