Ask question

Ask question

All categories

3 years ago

14

The gage pressure measured as 2.2 atm, the absolute pressure of gas is 3.2 bar. Please determine the local atmospheric pressure

in kPa.

1 answer:

LiRa [457]3 years ago

8 0

Answer:

$97.085\ \text{kPa}$

Explanation:

$P_{g}$ = Gauge pressure = 2.2 atm = $2.2\times 101325=222915\ \text{Pa}$

$P_{abs}$ = Absolute pressure = $3.2\ \text{bar}=3.2\times 10^5\ \text{Pa}$

$P_{atm}$ = Local atmospheric pressure

Absolute pressure is given by

$P_{abs}=P_{atm}+P_g\\\Rightarrow P_{atm}=P_{abs}-P_g\\\Rightarrow P_{atm}=3.2\times 10^5-222915\\\Rightarrow P_{atm}=97085\ \text{Pa}=97.085\ \text{kPa}$

The local atmospheric pressure is $97.085\ \text{kPa}$ .

You might be interested in

Fill in the correct answer.

mestny [16]

Answer:

Explanation:

A conservator with a(n) guardianship background and knowledge of foreign languages will have more job opportunities because they broaden the scope of work a conservator can undertake.

6 0

3 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Ilya [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

8 0

3 years ago

Find the largest number. The process of finding the maximum value (i.e., the largest of a group of values) is used frequently in

salantis [7]

Answer:

See Explanation

Explanation:

Required

- Pseudocode to determine the largest of 10 numbers

- C# program to determine the largest of 10 numbers

The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

6.1 if Largest < Number[i] Then

6.2 Largest = Number[i]

6.3 End Loop:

7. Display “The largest input is “+Largest

8. Stop

C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

class Program

{

static void Main(string[] args)

{

int[] Number = new int[10]; // Declare array of 10 elements

//Accept Input

int counter = 0;

while(counter<10)

{

Console.WriteLine("Enter Number " + (counter + 1)+": ");

string var = Console.ReadLine();

Number[counter] = Convert.ToInt32(var);

counter++;

}

//Initialize largest to first element of the array

int Largest = Number[0];

//Determine Largest

for(int i=0;i<10;i++)

{

if(Largest < Number[i])

{

Largest = Number[i];

}

}

//Print Largest

Console.WriteLine("The largest input is "+ Largest);

Console.ReadLine();

}

}

}

8 0

3 years ago

An experiment compares the initial speed of bullets fired from two handguns: a 9 mm and a 0.44 caliber. The guns are fired into

olasank [31]

Answer:

1.176

Explanation:

When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.

v2 * (M + mb) = v1 * mb

Where

v1: muzzle velocity of the bullet

M: mass of the bob

mb: mass of the bullet

v2: mass of the bob with the bullet after being hit

v2 = v1 * mb / (M + mb)

Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.

Ek = 1/2 (M + mb) * v2^2

Ep = (M + mb) * g * h

Ek = Ep

1/2 (M + mb) * v2^2 = (M + mb) * g * h

1/2 * (v1 * mb / (M + mb))^2 = g * h

1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h

v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)

$v2 = \sqrt{\frac{g *h * (M+ mb)^2}{\frac{1}{2} * mb^2}}$

The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.

h = L * (1 - cos(a))

$v2 = \sqrt{\frac{g * L * (1 - cos(a)) * (M+ mb)^2}{\frac{1}{2} * mb^2}}$

For the 9 mm:

$v2 = \sqrt{\frac{9.81 * L * (1 - cos(4.3)) * (10+ 0.006)^2}{\frac{1}{2} * 0.006^2}} = \sqrt{L} * 391$

For the 0.44 caliber:

$v2 = \sqrt{\frac{9.81 * L * (1 - cos(10.1)) * (10+ 0.012)^2}{\frac{1}{2} * 0.012^2}} = \sqrt{L} * 460$

The ratio is 460 / 391 = 1.176

6 0

3 years ago

An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is

devlian [24]

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= $2\pi\sqrt{\frac{L}{g}}$

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= $2\pi\sqrt{\frac{L}{9.81}}$

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = $\frac{\textup{AE}}{\textup{L}}$

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = $\frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}$

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = $\sqrt{\frac{K}{m}}$

or

mass, m = $\sqrt{\frac{K}{\omega_n^2}}$

or

mass, m = $\sqrt{\frac{20.66\times10^8}{0.95^2}}$

mass, m = 22.90 × 10⁸ kg

4 0

3 years ago