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bonufazy [111]
3 years ago
15

Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600 K. Calculate the

entropy change of the two reservoirs and determine if the increase of entropy principle is satisfied.
Engineering
1 answer:
Fantom [35]3 years ago
8 0

Solution:

Given:

T_{H} = 1200 K

T_{L} = 600 K

Q = 100 kJ

The Entropy change of the two reservoirs is given by the sum of entropy change of each reservoir system and is given by the formula:

\Delta s = \frac{-Q}{T_{H}}+\frac{Q}{T_{L}}

\Delta s = \frac{Q(T_{L}-T_{_{H}})}{T_{H}T_{L}}

\Delta s = \frac{-100(600-1200)}{1200\times 600}

[tex]\Delta s = 0.0833kJ/K

Since, the change in entropy is positive and according to the Increase in entropy principle, for any process the total change in entropy of a system is always greater than or equal to zero (with its enclosing adiabatic surrounding).

Therefore, the entropy principle is satisfied.

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A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg
lakkis [162]

Answer:

L=107.6m

Explanation:

Cold water in: m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C

Hot water in: m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C

D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m

Step 1: Determine the rate of heat transfer in the heat exchanger

Q=m_{c}C_{c}(T_{c,out}-T_{c,in})

Q=1.2*4.18*(80-20)

Q=1.2*4.18*(80-20)

Q=300.96kW

Step 2: Determine outlet temperature of hot water

Q=m_{h}C_{h}(T_{h,in}-T_{h,out})

300.96=2*4.18*(160-T_{h,out})

T_{h,out}=124\°C

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

dT_{1}=T_{h,in}-T_{c,out}

dT_{1}=160-80

dT_{1}=80\°C

dT_{2}=T_{h,out}-T_{c,in}

dT_{2}=124-20

dT_{2}=104\°C

LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}

LMTD = \frac{104-80}{ln(\frac{104}{80})}

LMTD = \frac{24}{ln(1.3)}

LMTD = 91.48\°C

Step 4: Determine required surface area of heat exchanger

Q=UA_{s}LMTD

300.96*10^{3}=649*A_{s}*91.48

A_{s}=5.07m^{2}

Step 5: Determine length of heat exchanger

A_{s}=piDL

5.07=pi*0.015*L

L=107.57m

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Explanation:

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3 years ago
(Gas Mileage) Drivers are concerned with the mileage their automobiles get. One driver has kept track of several trips by record
Effectus [21]

Answer:

import java.util.*;

public class Main {

   

   public static void main(String[] args) {

     

       double milesPerGallon = 0;

       int totalMiles = 0;

       int totalGallons = 0;

       double totalMPG = 0;

       

       Scanner input = new Scanner(System.in);

 

       while(true){

           System.out.print("Enter the miles driven: ");

           int miles = input.nextInt();

           if(miles <= 0)

               break;

           else{

               System.out.print("Enter the gallons used: ");

               int gallons = input.nextInt();

               totalMiles += miles;

               totalGallons += gallons;

               milesPerGallon = (double) miles/gallons;

               totalMPG = (double) totalMiles / totalGallons;

               System.out.printf("Miles per gallon for this trip is: %.1f\n", milesPerGallon);

               System.out.printf("Total miles per gallon is: %.1f\n", totalMPG);

           }

       }

   }  

}

Explanation:

Initialize the variables

Create a while loop that iterates until the specified condition is met inside the loop

Inside the loop, ask the user to enter the miles. If the miles is less than or equal to 0, stop the loop. Otherwise, for each trip do the following: Ask the user to enter the gallons. Add the miles and gallons to totalMiles and totalGallons respectively. Calculate the milesPerGallon (divide miles by gallons). Calculate the totalMPG (divide totalMiles by totalGallons). Print the miles per gallon and total miles per gallon.

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3 years ago
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kati45 [8]

Answer:

attached below

Explanation:

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3 years ago
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