This manometer is used to measure the difference in water level between the two tanks.

Calculate the availability of a system where the mean time between failures is 900 hours and the mean time to repair is 100 hour

Debora [2.8K]

Answer:

The availability of system will be 0.9

Explanation:

We have given mean time of failure = 900 hours

Mean time [to repair = 100 hour

We have to find availability of system

Availability of system is given by $\frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}$

So availability of system $=\frac{900}{900+100}=\frac{900}{1000}=0.9$

So the availability of system will be 0.9

7 0

3 years ago

Which of the following is not one of the systems required to ensure the safe and correct operation of an engine?

velikii [3]

Brake system

Explanation: the engine doesn’t need to be running to make the brake system work the brake system it’s independent

6 0

3 years ago

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago

Create a program named PaintingDemo that instantiates an array of eight Room objects and demonstrates the Room methods. The Room

Serggg [28]

Answer:

Explanation:

Code used will be like

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

namespace PaintingWall

{

class Room

{

public int length, width, height,Area,Gallons;

public Room(int l,int w,int h)

{

length = l;

width = w;

height = h;

}

private int getLength()

{

return length;

}

private int getWidth()

{

return width;

}

private int getHeight()

{

return height;

}

public void WallAreaAndNumberGallons()

{

Area = getLength() * getHeight() * getWidth();

if (Area < 350)

{

Gallons = 1;

}

else if (Area > 350)

{

Gallons = 2;

}

Console.WriteLine ("The area of the Room is " + Area);

Console.WriteLine("The number of gallons paint needed to paint the Room is " + Gallons);

}

class PaintingDemo

{

static void Main(string[] args)

{

int l, w, h;

Room[] r = new Room[8];

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room "+(i+1));

Console.Write("Enter Length : ");

l = Convert.ToInt32(Console.ReadLine() );

Console.Write("Enter Width : ");

w = Convert.ToInt32(Console.ReadLine());

Console.Write("Enter Height : ");

h= Convert.ToInt32(Console.ReadLine());

r[i] = new Room(l,w,h);

Console.WriteLine();

}

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room " + (i + 1));

r[i].WallAreaAndNumberGallons();

}

Console.ReadKey();

}

3 0

3 years ago

Car insurance incentives and discounts are available depending on _____. A. school attendance and driver skill B. vehicle type a

WARRIOR [948]

Answer:D. Location, vehicle type, and driving habits

5 0

3 years ago

Read 2 more answers