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enyata [817]
4 years ago
14

A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons

tant. The pressure developed in the cylinder during 30% and 70% of the compression stroke is 1.5 bar and 2.75 bar, respectively. The relative efficiency of the engine with respect to the air standard efficiency is 50%. Determine the (i) compression ratio (ii) fuel consumption per hour per kW of power generated
Engineering
1 answer:
antoniya [11.8K]4 years ago
5 0

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given  :

PV^{1.3}=C

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}

where r_{c} is compression ratio = \frac{v_{c}+v_{s}}{v_{c}}

           r_{c} = 1+\frac{v_{s}}{v_{c}}

where v_{c} is compression volume

           v_{s} is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, v_{30}=v_{c}+0.7v_{s}

and at 70% compression, 30% of the swept volume remains

    ∴    v_{70}=v_{c}+0.3v_{s}  

We know,

\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}

\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}

\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\

1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}

v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}

0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}

0.2218v_{s} = 0.594v_{c}

v_{c}=0.3734 v_{s}

∴   r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}

Therefore, compression ratio is r_{c} = 3.678

Now efficiency, \eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}

 \eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}

 \eta =0.32342 , this is the ideal efficiency

Therefore actual efficiency, \eta_{act} =0.5\times \eta _{ideal}

           \eta_{act} =0.5\times 0.32342

           \eta_{act} =0.1617

Therefore total power required = 1 kW x 3600 J

                                                    = 3600 kJ

∴ we know efficiency, \eta=\frac{W_{net}}{Q_{supply}}

Q_{supply}=\frac{W_{net}}{\eta _{act}}

Q_{supply}=\frac{3600}{0.1617}

Q_{supply}=22261.78 kJ

Therefore fuel required = \frac{22261.78}{45000}

                                        = 0.4947 kg/hr      

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Argon is compressed in a polytropic process with n = 1.2 from 120 kPa and 10°C to 850 kPa in a piston–cylinder device. Determine
Ivenika [448]

Answer:

W=-109.12 kJ/kg

Q=-76.34 kJ/kg

Explanation:

The needed work W we will calculate by using the work equation for polytropic process and the heat transfer Q we will calculate by using the energy balance equation.

Before the calculations we first need to determine the final temperature T2. We will do that by using the given initial temperature T1 = 10°C, the given initial p_1 = 120 kPa and final p_2 = 800 kPa pressure and the polytropic index n = 1.2. Before the calculation we need to express the temperature in K units.  

T1 = 10°C + 273 K = 283 K  

T2 = ((p_2/p_1)^(n-1)/n)* T1

T2 = 388 K

Now we can use the heat capacity C_v, = 0.3122 kJ /kg K and the temperatures T1 and T2 to determine the change in internal energy ΔU.  

ΔU = C_v*(T2-T1)

ΔU = 32.78 kJ/kg

to determine the work we will also need the initial v1 and final v2 specific volume. The initial specific volume v1 we can determine from the ideal gas equation. For the calculation we will need the initial pressure p_1, temperature T1 and the specific gas constant R = 0.2081 kJ /kg K.  

v1=R*T1/p_1

v1=0.4908 m^3/kg

For the final specific volume we need to replace the initial temperature and pressure with the final.  

v2=R*T2/p_2

v2=0.1009 m^3/kg

The work W is then:  

W=p_2*v2-p_1*v1/n-1

W=-109.12 kJ/kg

The heat transfer Q we can calculate form the energy balance equation. For the calculation we will need the calculated work W and the change in internal energy ΔU.  

Q=W+ΔU

Q=-76.34 kJ/kg

7 0
4 years ago
For this lab you will be creating a auto-expanding dynamic-array. This array will explicitly hold std library strings. As with a
Inessa05 [86]

Answer:

Explanation:

Program is divided into 3 files. stringVector.h , stringVectory.cpp (implementation file) and main.cpp (testing file)

stringVector.h

#include<iostream>

using namespace std;

class stringVector {

private:

string * data;

unsigned length;

unsigned allocated_length;

public:

stringVector();

virtual ~stringVector();

unsigned size();

unsigned capacity();

void reserve(unsigned new_size);

bool empty();

void append(string new_data);

void swap(unsigned pos1, unsigned pos2);

stringVector &operator = (stringVector const &rhs);

string& operator[](unsigned position);

void sort();

};

stringVector.cpp

#include<iostream>

#include<stdexcept>

#include"stringVector.h"

stringVector::stringVector()

{

data = NULL;

length = 0;

allocated_length = 0;

}

stringVector::~stringVector()

{

delete [] data;

}

unsigned stringVector::size()

{

return length;

}

unsigned stringVector::capacity()

{

return allocated_length;

}

void stringVector::reserve(unsigned new_size)

{

string *temp = new string[new_size]; // Create a new array

 

/*Copy the contents of the array*/

for(int i =0; i < new_size; i++){

if(i < length){

temp[i] = data[i];

}

else

break;

}

 

delete []data ;// Delete previous array

data = temp;

allocated_length = new_size;

if(length > new_size){

length = new_size;

}

}

bool stringVector::empty(){

return (length == 0) ? true : false;

}

void stringVector::append(string new_data)

{

string *temp = NULL;

if(length == allocated_length){

if(allocated_length == 0){

data = new string[10];

allocated_length = 10;    

}

else{

temp = new string[2 * allocated_length]; // Create a new array with double the size

for(int i = 0; i < length ; i++){

temp[i] = data[i];

}

allocated_length = 2 * allocated_length;

 

if(data != NULL)

delete []data;

data = temp;

}

}

data[length] = new_data;

length++;

}

void stringVector::swap(unsigned pos1, unsigned pos2)

{

string str;

if((pos1 >= length) || (pos2 >= length)){

cout << "Index Out of bounds" << endl;

return;

}

str = data[pos1];

data[pos1] = data[pos2];

data[pos2] = str;

}

stringVector& stringVector::operator = (stringVector const &rhs)

{

delete [] data;

length = rhs.length;

allocated_length = rhs.allocated_length;

this->data = new string[allocated_length];

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

this->data[i] = rhs.data[i];

}

string& stringVector::operator[](unsigned position)

{

if(position > length){

throw std::out_of_range("Position out of range");

}

return data[position - 1];

}

void stringVector::sort()

{

string str;

for(int i= (length - 1) ; i > 0; i--){

for(int j = 0; j < i; j++){

if(data[j].compare(data[j+1]) > 0){

//Swap

str = data[j];

data[j] = data[j+1];

data[j+1] = str;

}

}

}

}

main.cpp

#include<iostream>

#include<stdexcept>

#include"stringVector.h"

void printVector(stringVector &);

int main()

{

stringVector vector;

string str1("California");

string str2("Alabama");

string str3("Oklahoma");

string str4("Texas");

vector.append(str1);

vector.append(str2);

vector.append(str3);

vector.append(str4);

cout << vector.size() << " - " << vector.capacity() << endl;

printVector(vector);

cout << endl;

vector.reserve(3);

cout << vector.size() << " - " << vector.capacity() << endl;

printVector(vector);

cout << endl;

vector.append(str4);

cout << vector.size() << " - " << vector.capacity() << endl;

printVector(vector);

cout << endl << "Copied vector : " << endl;

stringVector vector2 = vector;

printVector(vector2);

cout << endl;

vector.sort();

cout << "Vector 1 after sorting" << endl;

printVector(vector);

return 0;

}

void printVector(stringVector &vec)

{

cout << "Vector : ";

for(int i = 1 ; i <= vec.size(); i++){

cout << vec[i] << " ";

}

cout << endl;

}

OUTPUT:

[[email protected] cppProg3]$ ./main

4 - 10

Vector: California Alabama Oklahoma Texas

3 -3

Vector: California Alabama Oklahoma

4 - 6

Vector: California Alabama Oklahoma Texas

Copied Vector:

Vector: California Alabama Oklahoma Texas

Vector 1 after sorting

Vector: Alabama California Oklahoma Texas

8 0
3 years ago
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Reil [10]

Answer:

A.

Explanation:

Able to return to it's original shape after distortion.

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3 years ago
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A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of
larisa86 [58]

Answer:

The field strength needed is 0.625 T

Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

area of the rectangular coil, A =  L x B = 0.0611 x 0.05 = 0.003055 m²

number of tuns of the coil, N = 300 turns

peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

B = (emf₀ ) / (NA ω)

B = (24) / (300 x 0.003055 x 41.893)

B = 0.625 T

Therefore, the field strength needed is 0.625 T

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