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Scilla [17]
3 years ago
12

A turboprop engine consists of a diffuser, compressor, combustor, turbine, and nozzle. The turbine drives a propeller as well as

the compressor. Air enters the diffuser with a volumetric flow rate of 83.7 m^3/s at 40 kPa, 240 K, and a velocity of 180 m/s, and decelerates essentially to zero velocity. The compressor pressure ratio is 10 and the compressor has an isentropic efficiency of 85%. The turbine inlet temperature is 11140 K, and its isentropic efficiency is 85%. The turbine exit pressure is 50 kPa. Flow through the diffuser and nozzle is isentropic. Using an air-standard analysis, determine the power delivered to the propeller, in MW. the velocity at the nozzle exit, in m/s.

Engineering
1 answer:
Yuri [45]3 years ago
7 0

Answer:

Power =10004.57kw

Velocity = 296.4m/s

Explanation:

Please see attachment

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‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Virty [35]

Answer:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

E=12\times 9.8\times 25\\\\E=2940\ J

So, the potential energy of the mass is 2940 J.

3 0
3 years ago
Write a program that prompts the user to enter time in 12-hour notation. The program then outputs the time in 24-hour notation.
Juliette [100K]

Answer:

THE CODE FOR THE PROGRAM IS GIVEN BELOW:

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int main()

{

convertTime convert;

int hr, mn, sc = 0;

 

cout << "Please input hours in 12 hr notation: ";

cin >> hr;

cout << "Please input minutes: ";

cin >> mn;

cout << "Please input seconds: ";

cin >> sc;

 

convert.invalidHr(hr);

convert.invalidMin(mn);

convert.invalidSec(sc);

convert.printMilTime();

 

system("Pause");

 

return 0;  

 

}

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int convertTime::invalidHr (int hour)

{

try{

 if (hour < 13 && hour > 0)

  {hour = hour + 12;

  return hour;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input hour again in correct 12 hour format: ";

  cin >> hour;

  invalidHr(hour);

  throw 10;

 }

   

}

catch (int c) { cout << "Invalid hour input!";}

}

int convertTime::invalidMin (int min)

{

try{

 if (min < 60 && min > 0)

  {return min;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input minutes again in correct 12 hour format: ";

  cin >> min;

  invalidMin(min);

  throw 20;

  return 0;

 }

   

}

catch (int e) { cout << "Invalid minute input!" << endl;}

}

int convertTime::invalidSec(int sec)

{

try{

 if (sec < 60 && sec > 0)

  {return sec;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input seconds again in correct 12 hour format: ";

  cin >> sec;

  invalidSec(sec);

  throw 30;

  return 0;

 }

   

}

catch (int t) { cout << "Invalid second input!" << endl;}

}

void convertTime::printMilTime()

{

cout << "Your time converted: " << hour << ":" << min << ":" << sec;

}

Explanation:

4 0
3 years ago
How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor
zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0
3 years ago
The theoretical maximum specific gravity of a mix at 5.0% binder content is 2.495. Using a binder specific gravity of 1.0, find
PSYCHO15rus [73]

Answer:

The theoretical maximum specific gravity at 6.5% binder content is 2.44.

Explanation:

Given the specific gravity at 5.0 %  binder content 2.495

Therefore

95 % mix + 5 % binder  gives S.G. = 2.495

Where the  binder is S.G. = 1, Therefore

Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495

(95 +5)/(Vx +5) = 2.495

2.495 × (Vx + 5) = 100

Vx =35.08 to 95

Or density of mix = Mx/Vx = 95/35.08 = 2.7081

Therefore when we have 6.5 % binder content, we get

Per 100 mass unit

93.5 Mass unit of Mx has a volume of

Mass/Density = 93.5/2.7081 = 34.526 volume units

Therefore we have

At 6.5 % binder content.

(100 mass unit)/(34.526 + 6.5) = 2.44

The theoretical maximum specific gravity at 6.5% binder content = 2.44.

3 0
2 years ago
1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b
Rainbow [258]

Answer:

1.  b. False

2. b. False

3.  b. False

4.  b. False

5. a. True

6. a. True

7.  b. False

8.  b. False

9. a. True

Explanation:

1. The fatigue properties of a material  are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6.  The bending fatigue could be handled with specific load requirements  for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the  maximum stress, as in torsional fatigue, which can be performed on  axial-type specially designed machines also, using the proper fixtures if  the maximum twist required is small, in which linear motion is changed to rotational motion.

7.  A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0
3 years ago
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