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lara31 [8.8K]
3 years ago
15

A closed vessel of volume 80 litres contain gas at a gauge pressure of 150 kPa. If the gas is compressed isothermally to half it

s volume, determine the resulting pressure.
Engineering
1 answer:
horsena [70]3 years ago
6 0

Answer:

The resulting pressure is 300 kilopascals.

Explanation:

Let consider that gas within the closed vessel behaves ideally. By the equation of state for ideal gases, we construct the following relationship for the isothermal relationship:

P_{1}\cdot V_{1} = P_{2} \cdot V_{2} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kilopascals.

V_{1}, V_{2} - Initial and final volume, measured in litres.

If we know that \frac{V_{1}}{V_{2}} = 2 and P_{1} = 150\,kPa, then the resulting pressure is:

P_{2} = P_{1}\times \frac{V_{1}}{V_{2}}

P_{2} = 300\,kPa

The resulting pressure is 300 kilopascals.

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Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 280
stiks02 [169]

Answer:

Minimum Allowable diameter for each of the 5 bolts = 0.0394 m = 3.94 cm

Explanation:

Maximum Working Stress = Ultimate Shear Stress/factor of safety

Maximum Working Stress = 280 MPa/3.8 = 73.68 MPa

Working Stress = Applied load/Minimum allowable Area = L/A

Minimum Allowable Area = Applied Load/Maximum Working Stress

A = 450000/73680000 = 0.00611 m²

This area is supplied by 5 bolts, so each bolt supplies A/5 = 0.0061/5 = 0.00122 m²

Cross sectional Area of bolts = πD²/4

0.00122 = πD²/4

D² = 4 × 0.00122/π = 0.00155

D = √0.00155 = 0.0394 m = 3.94 cm

Each of the five bolt can have a minimum diameter of 3.94 cm

Hope this Helps!!!

8 0
3 years ago
If gas costs $3.50 per gallon, how much would it cost to drive 500 miles in a city in a car that is 58.3 km/L
Akimi4 [234]
1 liter = .264 gallon
1 km = .621 mile

this means that 58.3km/L is equal to 137.13mpg

so

500/137.13 = 3.65 gallons of gas

3.65 x 3.5 = $12.78
5 0
3 years ago
A point in the x-y plane is represented by its x-coordinate and y-coordinate. Design the class Point that can store and process
Black_prince [1.1K]

Answer:

#include <iostream>

#include <iomanip>

using namespace std;

class pointType

{

public:

pointType()

{

x=0;

y=0;

}

pointType::pointType(double x,double y)

{

this->x = x;

this->y = y;

}          

void pointType::setPoint(double x,double y)

{

this->x=x;

this->y=y;

}

void pointType::print()

{

cout<<"("<<x<<","<<y<<")\n";

}

double pointType::getX()

{return x;

}

double pointType::getY()

{return y;

}

private:

   double x,y;

};

int main()

{

pointType p2;

double x,y;

cout<<"Enter an x Coordinate for point ";

cin>>x;

cout<<"Enter an y Coordinate for point ";

cin>>y;

p2.setPoint(x,y);

p2.print();

system("pause");    

return 0;

}

5 0
3 years ago
A two-phase mixture of water and steam with a quality of 0.63 and T = 300F expands isothermally until only saturated vapor rema
VMariaS [17]

Answer:

Explanation:

Hello!

To solve this problem you must follow the following steps, which are fully registered in the attached image.

1. Draw the complete outline of the problem.

2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties.

3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.

3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.

4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values ​​and finds the work done.

5. draw the pV diagram using the 300F isothermal line

5 0
3 years ago
Under the right conditions, it is possible, due to surface tension, to have metal objects float on water. Consider placing a sho
stiv31 [10]

Answer:

D = 0.060732 in

Explanation:

given data

sp. wt. = 500 lb/ft³

diameter = 0.036 in

solution

we get here maximum diameter of rod that is express as

D = \sqrt{\frac{8 \sigma }{\pi y}}   ......................1

here \sigma surface tension of water at 60⁰f  = 5.03 × 10^{-3}  lb/ft and y = 500 lb/ft³

so put here value and we will get

D = \sqrt{\frac{8 \times 5.03 \times 10^{-3} }{\pi \times 500}}

D = 0.005061 ft

D = 0.060732 in

4 0
3 years ago
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