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xeze [42]
3 years ago
10

The town of Mustang, TX is concerned that waste heat discharged from a new up- stream power plant will decimate the minnow popul

ation in its local stream. The power plant is 35% efficient with a rated output of 350 MW, and 75% of the waste heat is discharged into the stream, which flows at 150 m3/s. Before the power plant starts operations, the stream temperature was 20◦C. If the minnows can handle max- imum stream temperatures of 22◦C, will the water be safe at the point of cooling water discharge? (Assume that the waste heat is discharged to the flow of the whole stream.)
Engineering
1 answer:
vitfil [10]3 years ago
7 0

Answer:

Yes the water will be safe at the point of cooling water discharge

Explanation:

Power losses in plant= 350- 350×0.35=227.5MW

Rate of heat rejection to stream= 0.75× 227.5= 170.625MW

Rate of heat rejection= rate of flow of water× c × ΔT

170625000= 150000000× 4.186 × (Final temperature- 20)

Final temperature= 20.3 ◦C

The final temperature of stream will be 20.3 ◦C. Thechange is very small so the minnows will be able to handle this temperature.

You might be interested in
C++ - Green Crud Fibonacci programThe following program is to be written with a loop. You are to write this program three times
Fynjy0 [20]

Answer:

Below is the required code:

Explanation:

Using for loop

#include <iostream>

using namespace std;

int main()

{

    //Initial crud size

    int init = 0;

    int newCrud;

    int next=0;

    //Number of days to simulate

    int no_days;

    int day;

    cout << "Enter initial amount of green crud: ";

    cin >> newCrud;

    cout << "Enter number of days to simulate: ";

    cin >> no_days;

    for (day = 10; day<=no_days; day++)

    {

         if (day % 10 == 0)

         {

             next = newCrud + init;

         }

             newCrud = init;

             init = next;

    }

    if (no_days < 5)

    cout << "\nCrud reproduce only after 5 days minimum.Hence the current amount is "

    << newCrud << " pounds.";

    else

    cout << "On day " << no_days << " you have " << init

    << " pounds of green crud." << endl;

    cout << "\nWould you like to continue? (y or n): ";

    cin >> ans;

         return 0;

}

Output:

         Enter initial amount of green crud: 5

         Enter number of days to simulate: 220

    On day 220 you have 10485760 pounds of green crud.

Using while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans='y';

    while (ans == 'Y' || ans == 'y')

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud:

         ";

         cin >> newCrud;

         cout << "Enter number of days to simulate:

         ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         cout << "On day " << no_days << " you have "

         << init

         << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or

         n): ";

         cin >> ans;

    }

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

Using do while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans;

    do

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud: ";

         cin >> newCrud;

         cout << "Enter number of days to simulate: ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         

         cout << "On day " << no_days << " you have " <<

         init << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or n):

         ";

         cin >> ans;

    } while (ans == 'Y' || ans == 'y');

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

7 0
3 years ago
An ant starts at one edge of a long strip of paper that is 34.2 cm wide. She travels at 1.3 cm/s at an angle of 61◦ with the lon
ankoles [38]

Answer:

t = 30.1 sec

Explanation:

If the ant is moving at a constant speed, the velocity vector will have the same magnitude at any point, and can be decomposed in two vectors, along directions perpendicular each other.

If we choose these directions coincident with the long edge of the paper, and the other perpendicular to it, the components of the velocity vector, along these axes, can be calculated as the projections of this vector along these axes.

We are only interested in the component of the velocity across the paper, that can be calculated as follows:

vₓ = v* sin θ, where v is the magnitude of the velocity, and θ the angle that forms v with the long edge.

We know that v= 1.3 cm/s, and θ = 61º, so we can find vₓ as follows:

vₓ = 1.3 cm/s * sin 61º = 1.3 cm/s * 0.875 = 1.14 cm/s

Applying the definition of average velocity, we can solve for t:

t =\frac{x}{vx} = \frac{34.2 cm}{1.14 cm/s} =30.1 sec

⇒ t = 30.1 sec

7 0
3 years ago
Two fluids, A and B exchange heat in a counter – current heat exchanger. Fluid A enters at 4200C and has a mass flow rate of 1 k
Volgvan

Answer:

Your question has some missing information below is the missing information

Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )

answer : 300 kW , 95°c

Explanation:

Given data:

Fluid A ;

Temperature of Fluid ( Th1 )  = 420° C

mass flow rate (mh)  = 1 kg/s

Fluid B :

Temperature ( Tc1) = 20° C

mass flow rate ( mc ) = 1 kg/s

effectiveness of heat exchanger = 75% = 0.75

<u>Determine the heat transfer rate and  exit temperature of fluid</u> <u>B</u>

Cph = 1000 J/kgk

Cpc = 4000 J/Kgk

Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B

exit temp of fluid  B = 95°C

heat transfer = 300 kW

attached below is a the detailed solution

5 0
3 years ago
A driver counts 21 other vehicles using 3 EB lanes on one section of I-80 between her rented car and an overpass ahead. It turne
SVEN [57.7K]

Answer:

vehicle density =  28.205 veh/mile

flow rate =  0.909 veh/hr

Explanation:

given data

count n = 21

distance = 0.78 miles

speed = 52 mph

solution

we get here vehicle density that is express as

vehicle density = n ÷ distance    ...............1

vehicle density = ( 21 + 1 )  ÷ 0.78

vehicle density  k =  28.205 veh/mile

and

now we get here flow rate that is express as

flow rate = k × vs    .................2

flow rate = 28.205 × ( 52 × 0.00062 ÷ 1m )

flow rate =  0.909 veh/hr

6 0
2 years ago
How does geothermal work
puteri [66]

Answer:

geothermal heating works by moving temperature-conducting fluid through an underground loop of pipes beneath or near your home. This allows the fluid to collect the thermal energy deposited in the earth from the sun.

Explanation:

5 0
3 years ago
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