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kolezko [41]
3 years ago
8

An automobile engine consumes fuel at a rate of 22 L/h and delivers 85 kW of power to the wheels. If the fuel has a heating valu

e of 44,000 kJ/kg and a density of 0.8 g/cm3, determine the efficiency of this engine.
Engineering
2 answers:
attashe74 [19]3 years ago
7 0

Answer:

The efficiency of the engine is 39.53%

Explanation:

data given by the exercise

HV = heating value = 44000 kJ/kg

Vfuel = volume flow rate of fuel = 22 L/h = 0.0061 L/s

W = power output = 55 kW

p = density = 0.8 g/cm³

the mass flow rate of fuel is equal to

m_{fuel} =V*p=0.0061*0.8=4.89x10^{-3} kg/s

the rate of heat supplied is

Q_{in} =m_{fuel} *HV=4.89x10^{-3} *44000=215kW

the efficiency is equal to

n=\frac{W}{Q_{in} } *100=\frac{85}{215} *100=39.53%

zepelin [54]3 years ago
5 0

Answer:

Efficiency of the engine is 39.51 %

Explanation:

Mass of fuel consumed per hour = 22 L/h * 0.08 Kg/L = 17.6 Kg/h

Total Energy Consumed per hour = Mass of fuel consumed per hour * Heating  value of fuel

Total Energy Consumed per hour = 17.6 Kg/h * 44,000 KJ/Kg = 774,400 KJ/h

since, 1 KW = 3600 KJ/h

774,400KJ/h = 774,400/3600 kW = 215.1 kW

Efficiency of the engine = \frac{Power to the wheels}{Total Power}

Efficiency = 85/215.1 = 39.51 %

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2 years ago
Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of
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Answer:

#include<iostream>

#include <iomanip>

using namespace std;

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double getTotal(double [], int);

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double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

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double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

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for ( int month = 1; month <= NUM_MONTHS; month++ ){

                     if ( values[month] > largest ){

                 largest = values[month];

return largest;

          }

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

                     if ( values[month] < smallest ){

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int main()

{

double rainFall[NUM_MONTHS];

 for (int month = 0; month < NUM_MONTHS; month++)

  {

     cout << "Enter the rainfall (in inches) for month #";

     cout << (month + 1) << ": ";

     cin >> rainFall[month];

 

     while (rainFall[month] < 0)

     {

      cout << "Rainfall must be 0 or more.\n"

             << "Please re-enter: ";

      cin >> rainFall[month];

     }

  }

  cout << fixed << showpoint << setprecision(2) << endl;

  cout << "The total rainfall for the year is ";

  cout << getTotal(rainFall, NUM_MONTHS)

      << " inches." << endl;

   cout << "The average rainfall for the year is ";

  cout << getAverage(rainFall, NUM_MONTHS)

      << " inches." << endl;

   int subScript;

cout << "The largest amount of rainfall was ";

  cout << getLargest(rainFall, NUM_MONTHS, subScript)

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  cout << (subScript + 1) << "." << endl;

  cout << "The smallest amount of rainfall was ";

  cout << getSmallest(rainFall, NUM_MONTHS, subScript)

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  cout << (subScript + 1) << "." << endl << endl;

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8 0
3 years ago
A batch of 1000 is split into 10 smaller batches of equal size 100. The processing time of each unit is 2
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The lead time of the actual batch will be in

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This refers to the amount of time which is taken for a processor to run a procedure and return a result.

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In a hydraulic system, accumulator is a device that collects liquid and keeps the liquid under pressure.
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Answer:

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        where R = Resistance

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3 0
2 years ago
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