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vredina [299]
3 years ago
10

Implement a program that requests four numbers (integer or floating-point) from the user. Your program should compute the averag

e of the first three numbers and compare the average to the fourth number. If they are equal, your program should print 'Equal' on the screen. >>>
Engineering
1 answer:
maks197457 [2]3 years ago
3 0

Answer:

#include<iostream>

int main() {

  float num_1, num_2, num_3, num_4, average;

  //Taking input for four numbers

  std::cout << "Enter first number(integer or floating-point)";

  std::cin >> num_1;

  std::cout << "Enter second number(integer or floating-point)";

  std::cin >> num_2;

  std::cout << "Enter third number(integer or floating-point)";

  std::cin >> num_3;

 

  std::cout << "Enter fourth number(integer or floating-point)";

  std::cin >> num_4;

  average= (num_1+num_2+num_3)/3;

 // Comparing average with fourth number

 if (average==num_4)

  {

  std::cout << "Equal";

 }

  return 0;

 }

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4 years ago
A rectangular steel alloy A-36 (structural steel) plate is hanging vertically and supporting a hanging weight of 90 kN. The plat
KengaRu [80]

Answer:

a) Final length of bar = 0.5 + 0.4838 *10^-3 = 0.5004838 M

b)Final Thickness = 6- -1.739 * 10^-3 mm = 5.998260mm\

c)  % Reduction in area = (450-449.7391/450  )  = 0.58 %.

Explanation:

a) Change in length = Pl /AE = 90*1000*0.5*1000/75*207*6*10^3

= 0.4838mm Expansion.

Final length of bar = 0.5 + 0.4838 *10^-3 = 0.5004838 M

b )Change in width = - μpt/AE = -(0.3*90*1000*6/ 75*207*6*10^3)

= -1.739 * 10^-3 mm

Final Thickness = 6- -1.739 * 10^-3 mm = 5.998260mm

c )

New C/s area = 74.97827 *5.998260 = 449.7391 mm^2

% Reduction in area = (450-449.7391/450  )  = 0.58 %.

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4 years ago
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3 years ago
Air at 2.5 bar, 400 K is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchange
Shkiper50 [21]

Answer:

a) Power developed by the turbine = 132.89 kW

b) magnitude of the rate of heat transfer from the air to the ambient, in kw = 251.25 kW

Explanation:

b) The process is a constant pressure process (Isobaric process)

The constant pressure specific heat of air, c_{p} = 1.005 kJ/kg -K

Specific heat ratio for air, \gamma = 1.4

The mass flow rate of air, \dot{m} = 2.5 kg/s

P₁ = 2.5 bar, T₁ = 400 K

P₂ = 2.5 bar, T₂ = 300 K

Using the steady flow energy equation:

Q_{1-2}  = \dot{m} c_{p} (T_{2} - T_{1} \\Q_{1-2}  = 2.5 * 1.005 * (300 - 400)\\Q_{1-2}  = -251.25 kW

Therefore, the magnitude of the rate of heat transfer from the air to the ambient, in kw, Q_{1-2} = 251.25 kW

a) For the isentropic process:

Power developed by the turbine is given by the relation \dot{W} = \dot{M}  c_{p} (T_{2} - T_{3})

Isentropic efficiency, \eta_{t} = 80%

P₂ = 2.5 bar, T₂ = 300 K

P₃ = 1 bar, T_{3s} = ? where T_{3s} is the isentropic temperature at 100% efficiency

The isentropic relation is given by:

\frac{T_{3s} }{T_{2} } = (\frac{P_{3} }{P_{2} }) ^{\frac{\gamma - 1}{\gamma} } \\\frac{T_{3s} }{300 } = (\frac{1 }{2.5 }) ^{\frac{1.4 - 1}{1.4 }

T_{3s} = 230.9 K

To get the temperature at 80% efficiency, we will use the relation:

\eta_{t} = \frac{T_{2} - T_{3}  }{T_{2} - T_{3s} } \\0.8= \frac{300 - T_{3}  }{300 - 230.9 }

T₃ = 244.72 K

Power developed by the turbine is given by the relation:

\dot{W} = \dot{M}  c_{p} (T_{2} - T_{3})\\ \dot{W} = 2.5 * 1.005* (300-244.72)\\ \dot{W} = 138.89 kW

4 0
3 years ago
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Effectus [21]

Answer:

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