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Andreyy89
3 years ago
10

a refrigerator uses refrigerant-134a as the working fluid and operates on the vapor-compression refrigeration cycle. the evapora

tor and condenser pressures are 200 kpa and 1400 kpa, respectively. the isentropic efficiency of the compressor is 88 percent. If the refrigerant enters the compressor at a rate of 0.025 kg/s, determine the rate of cooling provided by the evaporator, the power input required to the compressor, and the coefficient of performance of the refrigerator.
Engineering
2 answers:
timurjin [86]3 years ago
8 0

Answer:

I. 3.316 kW

II. 1.218 kW

III. 2.72

Explanation:

At state 1, the enthalpy and entropy are determined using the given data from A-13.

At P1 = 200kpa and T1 = 0,

h1 = 253.07 kJ/kg

s1 = 0.9699 kJ/kgK

At state 2, the isentropic enthalpy is determined at P2 = 1400kpa and s1 = s2 by interpolation. Thus

h2(s) = 295.95 kJ/kg

The actual enthalpy is then gotten by

h2 = h1 + [h2(s) - h1]/n

h2 = 253.07 + [295.95 - 253.07]/0.88

h2 = 253.07 + 48.73

h2 = 301.8 kJ/kg

h3 = h4 = 120.43 kJ/kg

Heating load is determined from energy balance, thus,

Q'l = m'(h1 - h4)

Q'l = 0.025(253.07 - 120.43)

Q'l = 0.025 * 132.64

Q'l = 3.316 kW

Power is determined by using

W' = m'(h2 - h1)

W'= 0.025(301.8 - 253.07)

W'= 0.025 * 48.73

W'= 1.218 kW

The Coefficient Of Performance is Q'l / W'

COP = 3.316/1.218

COP = 2.72

erastovalidia [21]3 years ago
8 0

Answer/Explanation:

The enthalpy and entropy of state 1 is obtained from the property table(A-13).

Entropy is denoted as s, and enthalpy as h. P1 and P2 are pressures of state 1 and state 2 respectively.

For state 1

P1 = 200kpa

The efficiency of the compressor, n = 0.88%

Refrigerant mass rate = 0.025kg/s

h1 = 253.07 kJ/kg

s1 = 0.9699 kJ/kgK

At state 2

P2 = 1400kpa

s1 = s2 by interpolation.

Isentropic enthalpy h2(s) = 295.95 kJ/kg.

The actual enthalpy and the isentropic enthalpy is related by:

h2 = h1 + [h2(s) - h1]/n

h2 = 253.07 + [295.95 - 253.07]/0.88

h2 = 253.07 + 48.73

h2 = 301.8 kJ/kg

Power = mass rate(h2 - h1)

Power = 0.025(301.8 - 253.07)

Power = 0.025 x 48.73

Power = 1.218 kW

The Coefficient Of Performance is given by :

Heating load/Power

To calculate the heating load.

Note:

h3 = h4 = 120.43 kJ/kg

Heating load = mass rate(h1 - h4)

Heating load = 0.025(253.07 - 120.43)

Heating load = 0.025 x 132.64

Heating load = 3.316 kW

Coefficient of Performance = 3.316/1.218

Coefficient of performance = 2.7225

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PIT_PIT [208]

Answer:

γ_{xy} =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²;    1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ_{xy} = displacement / total height

     = 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also,  τ = Gγ_{xy}

By comapring both equations, we get

P/A = Gγ_{xy}   ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

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2 years ago
Which measuring tool will be used to determine the diameter of a crankshaft journal?
guapka [62]

Answer:

The dial bore gauge measures the inside of round holes, such as the bearing journals. This one tool can measure 2″ up to 6″ diameter holes. Both tools are needed in order to check the interior and exterior dimensions of the crankshaft, rods and engine block journals, as well as the thickness of the bearings themselves.

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4 0
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A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an
Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

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Total weight of particulate matter = 1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}

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So particulate remaining = 0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136

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Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

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2 years ago
When does the vc-turbo engine use lower compression ratios?.
Veronika [31]

Answer:

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2 years ago
You will create an array manipulation program that allows the user to do pretty much whatever they want to an array. When launch
enyata [817]

Answer:

Check the explanation

Explanation:

#include <iostream>

using namespace std;

void insert(int* arr, int* size, int value, int position){

if(position<0 || position>=*size){

cout<<"position is greater than size of the array"<<endl;

return ;

}

*size = *size + 1 ;

for(int i=*size;i>position;i--){

arr[i] = arr[i-1];

}

arr[position] = value ;

}

void print(int arr[], int size){

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

cout<< arr[i] <<" ";

}

cout<<" "<<endl;

}

void remove(int* arr, int* size, int position){

* size = * size - 1 ;

for(int i=position;i<*size;i++){

arr[i] = arr[i+1];

}

}

int count(int arr[], int size, int target){

int total = 0 ;

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

if(arr[i] == target)

total += 1 ;

}

return total ;

}

int main()

{

int size;

cout<<"Enter the initial size of the array:";

cin>>size;

int arr[size],val;

cout<<"Enter the values to fill the array:"<<endl;

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

cin>>val;

arr[i] = val ;

}

int choice = 5,value,position,target ;

do{

cout<<"Make a selection:"<<endl;

cout<<"1) Insert"<<endl;

cout<<"2) Remove"<<endl;

cout<<"3) Count"<<endl;

cout<<"4) Print"<<endl;

cout<<"5) Exit"<<endl;

cout<<"Choice:";

cin>>choice;

switch(choice){

case 1:

cout << "Enter the value:";

cin>>value;

cout << "Enter the position:";

cin>>position;

insert(arr,&size,value,position);

break;

case 2:

cout << "Enter the position:";

cin>>position;

remove(arr,&size,position);

break;

case 3:

cout<<"Enter the target value:";

cin>>target;

cout <<"The number of times "<<target<<" occured in your array is:" <<count(arr,size,target)<<endl;

break;

case 4:

print(arr,size);

break;

case 5:

cout <<"Thank you..."<<endl;

break;

default:

cout << "Invalid choice..."<<endl;

}

}while(choice!=5);

return 0;

}

Kindly check the attached images below for the code output.

3 0
3 years ago
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