All categories

3 years ago

The intake and exhaust processes are not considered in the p-V diagram of Otto cycle. a) true b) false

Engineering

1 answer:

vovangra [49]3 years ago

6 0

Answer:

b) false

Explanation:

We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.

It consist four processes

1-2:Reversible adiabatic compression

2-3:Constant volume heat addition

3-4:Reversible adiabatic expansion

3-4:Constant volume heat rejection

Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.

But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.

You might be interested in

Why does the auto industry prefer uniform (national) standards for automobile emissions as opposed to regionally varying standar

WINSTONCH [101]

Answer:

Explanation:

For automobile emission, a uniform standard is preferred, because no unnecessary advantage is given by it to any company that is located in particular states where the regional standards are less severe.

Since pollution has its impact across the states and in the whole of the USA, then there should be uniform standards across all the states. It will also invalidate the impact of regional standards as a factor in the selection of plant locations for the automobile company. It means that a state offering less valid emission standards, will attract more companies to herself and it will be against the other states who care more about the natural environment. It can make more states to opt for the permissive emission standards, that will be more harmful to the USA as a country, than the good. So, a uniform standard is preferred to eliminate it as a factor in plant location decisions.

Yes, uniform standards are beneficial to everyone, because it will bring effective control upon the pollution level because there will be no state where the culprit firm can hide. Besides, it is more effective as efforts done towards environment conservation.

3 0

3 years ago

Air enters a tank through an area of 0.2 ft2 with a velocity of 15 ft/s and a density of 0.03 slug/ft3. Air leaves with a veloci

Mademuasel [1]

Answer:

please find attached.

Explanation:

4 0

3 years ago

Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user

Thepotemich [5.8K]

Answer:

import java.util.Scanner;

public class FindMatchValue {

public static void main (String [] args) {

Scanner scnr = new Scanner(System.in);

final int NUM_VALS = 4;

int[] userValues = new int[NUM_VALS];

int i;

int matchValue;

int numMatches = -99; // Assign numMatches with 0 before your for loop

matchValue = scnr.nextInt();

for (i = 0; i < userValues.length; ++i) {

userValues[i] = scnr.nextInt();

}

/* Your solution goes here */

numMatches = 0;

for (i = 0; i < userValues.length; ++i) {

if(userValues[i] == matchValue) {

numMatches++;

}

System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

}

8 0

3 years ago

An engineer lives in Hawaii at a location where the annual rain fall is 300 inches. She decides to use the rain to generate elec

Alex777 [14]

Answer:

80.7lbft/hr

Explanation:

Flow rate of water in the system = 3.6x10^-6

The height h = 100

1s = 1/3600h

This implies that

Q = 3.6x10^-6/[1/3600]

Q = 0.0000036/0.000278

Q = 0.01295

Then the power is given as

P = rQh

The specific weight of water = 62.3 lb/ft³

P = 62.3 x 0.01295 x 100

P = 80.675lbft/h

When approximated

P = 80.7 lbft/h

This is the average power that could be generated in a year.

This answers the question and also corresponds with the answer in the question.

4 0

3 years ago

Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres

natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

$h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}$