ENGINEERING COLLEGE

Air is compressed in a piston-cylinder device. List three examples of irreversibilities that could occur

Answers

Answer 1

Answer:

Answer:

While air is compressed in a piston cylinder there are following types of irreversibilities

1.Due to finite temperature difference heat transfer take place between cylinder and surrounding.

2.Due friction force between cylinder and piston .

3.Compression process is so fast due to this ,it leads in the irreversibility of system.

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Q1. A truck traveling at 40 mph is approaching a stop sign. At time ????0 and at a distance of 80ft, the truck begins to slow down by decelerating rate of 12 ft/sec2 . Will the truck be able to stop in time?

A 1-m3 tank containing air at 10°C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35°C and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°C. Determine the volume of the second tank and the final equilibrium pressure of air. The gas constant of air is R = 0.287 kPa·m3/kg·K.

Answers

Answer:

V₂=1.76 m³

P=222.03 KPa

Explanation:

Given that

For tank 1

V₁=1 m³

T₁= 10°C = 283 K

P₁=350 KPa

For tank 2

m₂=3 kg

T₂=35°C = 308 K

P₂=150 KPa

We know that for air

P V = m R T

P=pressure ,V= Volume,R= gas constant ,T= temperature ,m =mass

for tank 2

P₂ V₂ = m₂ R T₂

By putting the values

150 x V₂ = 3 x 0.287 x 308

V₂=1.76 m³

Final mass = m₁+m₂

m =m₁+m₂

The final volume V= V₂+V₁

V= 1.76 + 1 m³

V= 2.76 m³

The final temperature T= 19.5°C

T= 292.5 K

$m=(PV)/(RT)$

$m_1=(P_1V_1)/(RT_1)$

$m_1=(350* 1)/(0.287* 283)$

$m_1=4.3\ kg$

m =m₁+m₂

m =4.3 + 3 = 7.3 kg

Now at final state

P V = m R T

P x 2.76 = 7.3 x 0.287 x 292.5

P=222.03 KPa

Currently, system administrators create Ken 7 users in each computer where users need access. In the Active Directory, where will system administrators create Ken 7 users? 2. How will the procedures for making changes to the user accounts, such as password changes, be different in the Active Directory? 3. What action should administrators take for the existing workgroup user accounts after converting to the Active Directory? 4. How will the administrators resolve the differences between the user accounts defined on the different computers? In other words, if user accounts have different settings on different computers, how will the Active Directory address that issue? 5. How will the procedure for defining access controls change after converting to the Active Directory?

Answers

1. First, you would need to open Active Directory Users and Computers. You click on the folder in which you want to add an account, and point to new, and then user. You would fill in the new user's information, such as name and initials.

2. In Active Directory, you input the user logon name, click on the UPN suffix in the drop-down list. It will prompt you to input password and confirm it.

3. Administrators would need to create new user accounts for all users, then join these to the AD domain manually.

4. Administrators will have to manually change the permissions and privileges of the users in order to meet the new established requirements.

5. After converting to the Active Directory, access control will be administered at the object level by setting different levels of access.

Make a copy of the pthreads_skeleton.cpp program and name it pthreads_p2.cpp Modify the main function to implement a loop that reads 10 integers from the console (user input) and stores these numbers in a one-dimensional (1D) array (this code will go right after the comment that says ""Add code to perform any needed initialization or to process user input""). You should use a global array for this.

Answers

Answer:

The solution code is as follows:

#include <iostream>
using namespace std;
int main()
{
int myArray [10] = {};
int i;
for( i = 0; i < 10; i++ )
{
cout <<"Enter an integer: ";
cin>> myArray[i];
}
}

Explanation:

Firstly, we initialize a 10-elements array, myArray (Line 7) with no values.

Next, we create a for-loop (Line 10). Within the loop, we prompt user to enter an integer and assign the input value to the current element of myArray (Line 12-13).

The reservoir pressure of a supersonic wind tunnel is 5 atm. A static pressure probe is moved along the centerline of the nozzle, taking measurements at various stations. For these probe measurements, calculate the local Mach number and area ratio: a. 4 atm; b. 2.64 atm; c. 0.5 atm.

Answers

Answer

Given,

Reservoir pressure of a supersonic wind tunnel = 5 atm

Local Mach number = ?

Area ration = ?

a) 4 atm.

Pressure ratio = $(4)/(5)$

= 0.8

From Isentropic Flow Tables

M = 0.58 A/A* = 1.213

b) 2.64 atm

Pressure ratio = $(2.64)/(5)$

= 0.528

From Isentropic Flow Tables

M = 1 A/A* = 1

c) 0.5 atm

Pressure ratio = $(0.5)/(5)$

= 0.1

From Isentropic Flow Tables

M =2.10 A/A* = 1.8369

A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.113V. Assuming that errors are due to random processes, how many of the readings are expected to be greater than 2.70V?

Answers

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, $\mu = 2.501 V$

standard deviation, $\sigma = 0.113 V$

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, $P(X\leq 2.70)$ :

$z = (x - \mu)/(\sigma) = (2.70 - 2.501)/(0.113) = 1.761$

Now, from the Probability table of standard normal distribution:

$P(z\leq 1.761) = 0.9608$

Now,

$P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%$

Now, for the expected no. of readings greater than 2.70 V:

$P(X\geq 2.70) = (No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V)/(Total\ no.\ of\ readings)$

No. of readings expected to be greater than 2.70 V = $P(X\geq 2.70)* Total\ no.\ of\ readings$

No. of readings expected to be greater than 2.70 V = $0.0392* 60 = 2.352$ ≈ 2

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3 . Due to the wind, 9.4 x 10-5 kg/s of air enters per meter of crack and exits up a chimney. Assume air temperature is the same inside and out and air density is constant at 1.186 kg/m3 . If windows and doors are not opened or closed, estimate the time required for one complete air change in the building.