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kompoz [17]
3 years ago
13

A railroad runs form city A to city B, a distance of 800km, through mountainous terrain. The present one-way travel time (includ

ing time at intermediate yards) is 20 hours, and the rail freight rate is $20 per ton. There is a truck service that competes with the railroad, running over a roughly parallel road for approximately the same distance, at an average speed of 48km per hour and a rate of $30 per ton. A new highway is planned to replace the existing roads; it is expected that most of the traffic will be trucks (auto usage is expected to be negligible). The performance function of the new facility is t_T = t_0 + bV_T, where V_T is the flow in trucks per hour, t_0 = 10 hours, b = 0.08 hour per truck per hour. The railroad's estimate of the demand function is: V_T/V_R = a_0(t_r/t_R)^a_1 (c_T/c_R)^a_2 where t_T and t_R are the trip times (in hours) by truck and rail, respectively, c_T and c_R are the corresponding rates, V_T and V_R are the corresponding flows and a_0, a_1 and a_2 are parameters. The total demand is likely to remain constant at V_TOT = 200 tons per hour. The rail system is utilized at only a fraction of capacity, so its performance function is flat (travel time is constant, independent of volume). If a_0 = 1, a_1 = -1 and a_2 = -2, find the present flows of freight by truck and rail. Make an estimate of the equilibrium flows if the new highway were built. With the new highway built, what would the equilibrium flow be in each of the following two cases: if the railroad dropped its rate to $15 per ton? if truckers were taxed $5 per ton to help pay for the new highway?

Engineering
1 answer:
Verizon [17]3 years ago
4 0

Answer:

i) VT = 52.16

VR = 147.85

ii) VT = 61

VR = 138.99

Explanation:

The step by step solution is been done, please check the attached file below to see it

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A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo
Natalija [7]

Answer:

For 20^{\circ} - 5.556 lb/s

For 15^{\circ} - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = 20^{\circ}

Temperature rise, T' = 15^{\circ}

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, C_{p} = 0.24 btu/lb ^{\circ}F

Now, for a rise of 20^{\circ} in temeprature:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s

Now, for 15^{\circ}:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s

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For a project in C++ we are supposed toDesign a class named Month. The class should have the following private members:-name: a
mart [117]

Answer:

include <iostream>

using namespace std;

 

class Month

{

public:

 Month (char firstLetter, char secondLetter, char thirdLetter);

 

 Month (int monthNum);

.

 

 Month();

 void outputMonth_num();

 

 

 void outputMonthLetters();

private:

 int month;

};

 

 

int main ()

{

 //

 // Variable declarations

 //

 int monthNum;

 char firstLetter, secondLetter, thirdLetter;    

 char testAgain;              

 

 do {

 

   cout << endl;

   cout << "Testing the default constructor ..." << endl;

   Month defaultMonth;

   defaultMonth.outputMonth_num();

   defaultMonth.outputMonthLetters();

 

   //

   // Construct a month using the constructor with one integer argument

   //

   cout << endl;

   cout << "Testing the constructor with one integer argument..." << endl;

   cout << "Enter a month number: ";

   cin >> monthNum;

 

   Month testMonth1(monthNum);

   testMonth1.outputMonth_num();

   testMonth1.outputMonthLetters();

 

   //

   // Construct a month using the constructor with three letters as arguments

   //

   cout << endl;

   cout << "Testing the constructor with 3 letters as arguments ..." << endl;

   cout << "Enter the first three letters of a month (lowercase): ";

   cin >> firstLetter >> secondLetter >> thirdLetter;

   cout << endl;

 

   Month testMonth2(firstLetter, secondLetter, thirdLetter);

   testMonth2.outputMonth_num();

   testMonth2.outputMonthLetters();

 

   //

   // See if user wants to try another month

   //

   cout << endl;

   cout << "Do you want to test again? (y or n) ";

   cin >> testAgain;

 }

 while (testAgain == 'y' || testAgain == 'Y');

 

 return 0;

}

 

 

Month::Month(char firstLetter, char secondLetter, char thirdLetter)

{

if ((firstLetter == 'j')&&(secondLetter == 'a')&&(thirdLetter == 'n'))

  outputMonth_num = 1;

if ((firstLetter == 'f')&&(secondLetter == 'e')&&(thirdLetter == 'b'))

  outputMonth_num = 2;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'r'))

  outputMonth_num = 3;

if ((firstLetter = 'a')&&(secondLetter == 'p')&&(thirdLetter == 'r'))

  outputMonth_num = 4;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'y'))

  outputMonth_num = 5;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(thirdLetter == 'n'))

  outputMonth_num = 6;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(.thirdLetter == 'l'))

  outputMonth_num = 7;

if ((firstLetter == 'a')&&(secondLetter == 'u')&&(thirdLetter == 'g'))

  outputMonth_num = 8;

if ((firstLetter == 's')&&(secondLetter == 'e')&&(thirdLetter == 'p'))

  outputMonth_num = 9;

if ((firstLetter == 'o')&&(secondLetter == 'c')&&(thirdLetter == 't'))

  outputMonth_num = 10;

if ((firstLetter == 'n')&&(secondLetter == 'o')&&(thirdLetter == 'v'))

 outputMonth_num = 11;

if ((firstLetter == 'd')&&(secondLetter == 'e')&&(thirdLetter == 'c'))

 outputMonth_num = 12;

}

 

Month::inputMonthByNumber

{

if (Month_num > 12 && Month_num < 1)

cout << "Invalid number for Month, please choose 1-12)\n";

}

 

void Month::outputMonth_num()

{

 if (month >= 1 && month <= 12)

   cout ><< "Month: " << month << endl;

 else

   cout << "Error - The month is not a valid!" << endl;

}

 

void Month::outputMonthLetters()

{

 switch (month)

   {

   case 1:

     cout << "Jan" << endl;

     break;

   case 2:

     cout << "Feb" << endl;

     break;

   case 3:

     cout << "Mar" << endl;

     break;

   case 4:

     cout << "Apr" << endl;

     break;

   case 5:

     cout << "May" << endl;

     break;

   case 6:

     cout << "Jun" << endl;

     break;

   case 7:

     cout << "Jul" << endl;

     break;

   case 8:

     cout << "Aug" << endl;

     break;

   case 9:  

     cout << "Sep" << endl;

     break;

   case 10:

     cout << "Oct" << endl;

     break;

   case 11:

     cout << "Nov" << endl;

     break;

   case 12:

     cout << "Dec" << endl;

     break;

   default:

     cout << "Error - the month is not a valid!" << endl;

   }

}

7 0
3 years ago
A world class runner can run long distances at a pace of 15 km/hour. That runner expends 800 kilocalories of energy per hour. a)
maks197457 [2]

Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars

Explanation :

A)

Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}

= 3780kJ

And 1 hour = 3600s

Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W

B)

At 15km/hour a 15km run takes 1 hour.

1 hour is 3600s and the runner burns 1050 joule per second.

Energy used in 1 hour = 3600 x 1050 J/s

= 3780000 J or 3.78MJ

C)

1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km

15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ

Finally,

1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ

This means that the runner needs 5320/1008 = 5.3 bars

7 0
3 years ago
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