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3 years ago

Can a real refrigerator have higher COP than the COP of the Carnot refrigerator?

Engineering

2 answers:

astra-53 [7]3 years ago

5 0

Answer:

Explanation:

As we know that Carnot refrigerator is an ideal refrigerator in which all processes are reversible and COP of refrigerator depends only on temperature limit of evaporator and condenser.

On the other hand a real refrigerator consist irreversibility that is why the COP of real refrigerator is less that Carnot refrigerator.

So we can say that COP of Carnot refrigerator is always more that COP of real refrigerator.

Flauer [41]3 years ago

4 0

Answer: NO

Explanation:

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A barometer reads a height of 78 cmHg. Express this atmospheric pressure to:

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A barometer reads a height of 78 cmHg. Express this atmospheric pressure to $\large\mathsf{\red{\underline{Pascal(pa)}}}$

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2 years ago

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Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe

Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe $\left ( D\right )=0.15m$

Velocity of water in pipe $\left ( V\right )=15cm/s$

We know viscosity of water is $\left (\mu\right )=8.90\times10^{-4}pa-s$

Pressure drop is given by hagen poiseuille equation

$\Delta P=\frac{128\mu \L Q}{\pi D^4}$

We have asked pressure Drop per unit length i.e.

$\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}$

Substituting Values

$\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}$

$\frac{\Delta P}{L}$ =0.1898 Pa/m

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3 years ago

The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin

IceJOKER [234]

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel = 4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

$\dot{m}=\rho AV$

Putting all the value to get the velocity of the flow

$\frac{\dot{m}}{\rho A} = V$

$V = \frac{4000}{1000*4*2}$

V = 0.5 m/s

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3 years ago

In part A you are asked to write the pseudocode for the program. In part B you are asked to write the syntax of the code for the

Naya [18.7K]

Answer:

C++.

Explanation:

#include <iostream>

#include <string>

using namespace std;

///////////////////////////////////////////////////////////////

int main() {

string quote, book;

int page;

cout<<"What is your favorite quote from a book?"<<endl;

getline(cin, quote);

cout<<endl;

/////////////////////////////////////////////

cout<<"What book was that quote from?"<<endl;

getline(cin, book);

cout<<endl;

/////////////////////////////////////////////

cout<<"What page was that quote from?"<<endl;

cin>>page;

cout<<endl;

/////////////////////////////////////////////

int no_of_upper_characters = 0;

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

if (isupper(quote[i]))

no_of_upper_characters++;

}

cout<<"No. of upper case characters: "<<no_of_upper_characters<<endl;

/////////////////////////////////////////////

int no_of_characters = quote.length();

cout<<"No. of characters: "<<no_of_characters<<endl;

/////////////////////////////////////////////

bool isDog = false;

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

if (isDog == true)

break;

else if (quote[i] == 'd') {

for (int j=i+1; j<quote.length(); j++) {

if (isDog == true)

break;

else if (quote[j] == 'o') {

for (int z=j+1; z<quote.length(); z++) {

if (quote[z] == 'g') {

isDog = true;

break;

}

if (isDog == true)

cout<<"This includes 'd' 'o' 'g' in the quote";

//////////////////////////////////////////////

return 0;

}

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3 years ago

"A communication between two devices is over the maximum limit of an ethernet frame size. The Transmission Control Protocol (TCP

Sliva [168]

COmmunication Devices

Explanation:

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The sequence number in a header is used to keep track of which segment out of many this particular segment might be. The next field, the acknowledgment number, is a lot like the sequence number. The acknowledgment number is the number of the next expected segment.

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