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adell [148]
3 years ago
9

An Otto cycle engine is analyzed using the cold air standard method. Given the conditions at state 1, compression ratio (r), and

pressure ratio (rp). The specific heats for air are Cp = 1.0045 kJ/kg-K and Cv = 0.7175 kJ/kg-K.
T1 = 300K

P1 = 100 kPa

r = 10

rp = 1.65

a) Determine the temperature (K) at state 2.

b) Determine the pressure (kPa) at state 2.

c) Determine the pressure (kPa) at state 3.

d) Determine the temperature (K) at state 3.

e) Determine the temperature (K) at state 4.

f) Determine the pressure (kPa) at state 4.

g) Determine the net work output (kJ/kg/cycle) for the engine.

h) Determine the heat addition (kJ/kg/cycle) for the engine.

i) Determine the efficiency (%) of the engine.

Points will be given to answer showing all the work done to get right answer.

Engineering
1 answer:
lilavasa [31]3 years ago
4 0

Answer:

Explanation:

The detailed and careful step by step calculation and analysis is as shown with appropriate formula in the attached files

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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;

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                   }

               }

           }

       }

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   if (isDog == true)

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3 0
3 years ago
Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin
Vladimir79 [104]

Answer:

a)

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

b)

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

c)

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

a)

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 =  1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 =  1.013 μm

b)

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 =  0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.243 μm

c)

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 =  0.605 μm

# for an I-line λ = 365 nm

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6 0
3 years ago
Question # 3
tino4ka555 [31]

Answer:

I think it's False.

Apologies if I am wrong.

4 0
3 years ago
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Nancy wants to buy a cooking stove that’s electrically insulated and resistant to heat. What material should she choose for the
astra-53 [7]

Answer:

Ceramics

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3 years ago
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bija089 [108]

Answer:

(a) 20 MHz

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(c) 3.33 ns

(d) 33 pF

Explanation:

(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>

(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>

(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>

(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>

8 0
4 years ago
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