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NARA [144]
4 years ago
8

Sketch T-s and p-v diagrams for the Diesel cycle.

Engineering
1 answer:
labwork [276]4 years ago
3 0

Answer:

Diesel cycle:

    All diesel engine works on diesel cycle.It have four processes .These four processes are as follows

1-2.Reversible adiabatic compression

2-3.Heat addition at constant pressure

3-4.Reversible adiabatic expansion

4-1.Heat addition at constant volume

When air inters in the piston cylinder after that it compresses and gets heated due to compression after that heat addition take place at constant pressure after that power is produces when piston moves to bottom dead center.

From the diagram of P-v And T-s we can understand so easily.

You might be interested in
The human circulatory system consists of a complex branching pipe network ranging in diameter from
Stels [109]

Answer: the average velocity decreases

Explanation:

From the provided data we have:

Vessel    avg. diameter[mm] number

Aorta                 25.0                   1

Arteries             4.0                    159

Arteioles           0.06                 1.4*10^7

Capillaries         0.012               2.9*10^9

from the information, let \hat{m} be the mass flow rate, \rho is density, n number of vessels, and A is the cross-section area for each vessel

the flow rate is constant so it is equal for all vessels,

The average velocity is related to the flow rate by:

\hat{m} = v* \rho * A * n

we clear the side where v is in:

v = \frac{\hat{m}}{\rho A n}

area is π*R^2 where R is the average radius of the vessel (diameter/2)

we get:

v = \frac{\hat{m}}{\rho \pi R^2 n}

you can directly see in the last equation that if we go from the aorta to the capillaries, the number of vessels is going to increase ( n will increase and R is going to decrease ) . From the table, R is significantly smaller in magnitude orders than n, therefore, it wont impact the results as much as n. On the other hand, n will change from 1 to 2.9 giga vessels which will dramatically reduce the average blood velocity

8 0
3 years ago
This program will store roster and rating information for a soccer team. Coaches rate players during tryouts to ensure a balance
kari74 [83]

Answer:

Using C++ language

#include <iostream>

#include <vector>

using namespace std;

int main()

{

//definition

vector<int> jerseyNumber;

vector<int> rating;

int temp;

//loop

for (int i = 1; i <= 5; i++)

{

cout << "Enter player " << i

<< "'s jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter player " << i

<< "'s rating: ";

cin >> temp;

rating.push_back(temp);

cout << endl;

}

 

cout << "ROSTER" << endl;

 

for (int i = 0; i < 5; i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: " << jerseyNumber.at(i)

<< ", Rating: " << rating.at(i) << endl;

char option;

'

while (true)

{

cout << "MENU" << endl;

cout << "a - Add player" << endl;

cout << "d - Remove player" << endl;

cout << "u - Update player rating" << endl;

cout << "r - Output players above a rating"

<< endl;

cout << "o - Output roster" << endl;

cout << "q - Quit" << endl << endl;

 

cout << "Choose an option: ";

cin >> option;

switch (option)

{

 

case 'a':

case 'A':

cout << "Enter a new player's"

<< "jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter the player's rating: ";

cin >> temp;

 

rating.push_back(temp);

break;

case 'd':

case 'D':

cout << "Enter a jersey number: ";

cin >> temp;

int i;

for (i = 0; i < jerseyNumber.size();

i++)

{

if (jerseyNumber.at(i) == temp)

{

jerseyNumber.erase(

jerseyNumber.begin() + i);

rating.erase(rating.begin() + i);

break;

}

}

break;

case 'u':

case 'U':

cout << "Enter a jersey number: ";

cin >> temp;

 

for (int i = 0; i < jerseyNumber.size();

i++)

{

if (jerseyNumber.at(i) == temp)

{

cout << "Enter a new rating "

<< "for player: ";

cin >> temp;

rating.at(i) = temp;

break;

}

}

break;

case 'r':

case 'R':

 

cout << "Enter a rating: ";

cin >> temp;

cout << "\nABOVE " << temp << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

if (rating.at(i) > temp)

cout << "Player " << i + 1

<< " -- "

<< "Jersey number: "

<< jerseyNumber.at(i)

<< ", Rating: "

<< rating.at(i) << endl;

break;

case 'o':

case 'O':

cout << "ROSTER" << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: "

<< jerseyNumber.at(i) << ", Rating: "

<< rating.at(i) << endl;

break;

 

case 'q':

return 0;

default: cout << "Invalid menu option."

<< " Try again." << endl;

}

}

}

6 0
3 years ago
1.<br> Defensive driving involves?
-Dominant- [34]

Answer:

According to the standard Safe Practices for Motor Vehicle Operations, ANSI/ASSE Z15.1, defensive driving skills involves "driving to save lives, time, and money, in spite of the conditions around you and the actions of others."

Explanation:

Defensive driving is a type of training provided to the learners eligible to drive motor vehicles. It involves the basics of driving and rules involved in driving. The learners learn about the methods of driving safely with the minimum expense of time and money. It also helps in reducing the risks of collision in adverse situations. The methods to rescue oneself from dangerous situations are also trained. Preventing oneself from the mistakes of others is also termed under defensive driving.

4 0
3 years ago
thier only motto and goal is to work for society and not make any profits A.small business entreprenuership B.scalable start up
sweet-ann [11.9K]

Answer:

social entrepreneurship

8 0
3 years ago
Read 2 more answers
A rear wheel drive car of mass 1000 kg is accelerating with a constant acceleration without slipping from 0 to 60 m/s in 1 min.
tester [92]

Answer:

500 N

Explanation:

Given;

Mass of the car, M = 1000 kg

initial speed of the car, u = 0 m/s

Final speed of the car, v = 60 m/s

Time, t = 1 min = 60 s

Now,

Force, F is given as:

F = Ma

where,

a is the acceleration

From the Newton's equation of motion, we have

v = u + at

on substituting the values, we get

60 = 0 + a × 60

or

a = 1 m/s²

Thus,

Force = 1000 × 1 = 1000 N

now,

this force will be equal to the friction force provided by the rear wheels

let the friction force on a single rear wheel be 'f'

thus,

2f = 1000 N

or

f = 500 N

5 0
4 years ago
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