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ch4aika [34]
4 years ago
12

The air standard efficiency ofan Otto cycle compared to diesel cycle for thie given compression ratio is: (a) same (b) less (c)

more () more (d) unpredictable (d) e) more or less depending on power rating
Engineering
1 answer:
Svetach [21]4 years ago
5 0

Answer:

Correct sentence: (c) more.

Explanation:

Otto cycle:

An ideal Otto cycle models the behavior of an explosion engine. This cycle consists of six steps, as indicated in the figure. Prove that the performance of this cycle is given by the expression

η = 1 - (1/r^(ρ-1))

where r = VA / VB is the compression ratio equal to the ratio between the volume at the beginning of the compression cycle and at the end of it. ρ=1.4

Diesel cycle:

The ideal Diesel cycle is distinguished from the ideal Otto in the combustion phase, which in the Otto cycle is assumed at a constant volume and in the Diesel at constant pressure. Therefore the performance is different.

If we write the performance of a Diesel cycle in the form

:

η = 1 - (1/r^(ρ-1)) × ( (r^ρ)-1)/(ρ^(r-1)) )

we see that the efficiency of a Diesel cycle differs from that of an Otto cycle by the factor in parentheses. This factor is always greater than the unit, therefore, for the same compression reasons r

                   diesel performance is less than otto performance

\ eta_ \ mathrm {Diesel} <\ eta_ \ mathrm {Otto} \,

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More discussion about seriesConnect(Ohm) function In your main(), first, construct the first circuit object, called ckt1, using
STALIN [3.7K]

Answer:

resistor.h

//circuit class template

#ifndef TEST_H

#define TEST_H

#include<iostream>

#include<string>

#include<vector>

#include<cstdlib>

#include<ctime>

#include<cmath>

using namespace std;

//Node for a resistor

struct node {

  string name;

  double resistance;

  double voltage_across;

  double power_across;

};

//Create a class Ohms

class Ohms {

//Attributes of class

private:

  vector<node> resistors;

  double voltage;

  double current;

//Member functions

public:

  //Default constructor

  Ohms();

  //Parameterized constructor

  Ohms(double);

  //Mutator for volatage

  void setVoltage(double);

  //Set a resistance

  bool setOneResistance(string, double);

  //Accessor for voltage

  double getVoltage();

  //Accessor for current

  double getCurrent();

  //Accessor for a resistor

  vector<node> getNode();

  //Sum of resistance

  double sumResist();

  //Calculate current

  bool calcCurrent();

  //Calculate voltage across

  bool calcVoltageAcross();

  //Calculate power across

  bool calcPowerAcross();

  //Calculate total power

  double calcTotalPower();

  //Display total

  void displayTotal();

  //Series connect check

  bool seriesConnect(Ohms);

  //Series connect check

  bool seriesConnect(vector<Ohms>&);

  //Overload operator

  bool operator<(Ohms);

};

#endif // !TEST_H

resistor.cpp

//Implementation of resistor.h

#include "resistor.h"

//Default constructor,set voltage 0

Ohms::Ohms() {

  voltage = 0;

}

//Parameterized constructor, set voltage as passed voltage

Ohms::Ohms(double volt) {

  voltage = volt;

}

//Mutator for volatage,set voltage as passed voltage

void Ohms::setVoltage(double volt) {

  voltage = volt;

}

//Set a resistance

bool Ohms::setOneResistance(string name, double resistance) {

  if (resistance <= 0){

      return false;

  }

  node n;

  n.name = name;

  n.resistance = resistance;

  resistors.push_back(n);

  return true;

}

//Accessor for voltage

double Ohms::getVoltage() {

  return voltage;

}

//Accessor for current

double Ohms::getCurrent() {

  return current;

}

//Accessor for a resistor

vector<node> Ohms::getNode() {

  return resistors;

}

//Sum of resistance

double Ohms::sumResist() {

  double total = 0;

  for (int i = 0; i < resistors.size(); i++) {

      total += resistors[i].resistance;

  }

  return total;

}

//Calculate current

bool Ohms::calcCurrent() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  current = voltage / sumResist();

  return true;

}

//Calculate voltage across

bool Ohms::calcVoltageAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double voltAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      voltAcross += resistors[i].voltage_across;

  }

  return true;

}

//Calculate power across

bool Ohms::calcPowerAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double powerAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      powerAcross += resistors[i].power_across;

  }

  return true;

}

//Calculate total power

double Ohms::calcTotalPower() {

  calcCurrent();

  return voltage * current;

}

//Display total

void Ohms::displayTotal() {

  for (int i = 0; i < resistors.size(); i++) {

      cout << "ResistorName: " << resistors[i].name << ", Resistance: " << resistors[i].resistance

          << ", Voltage_Across: " << resistors[i].voltage_across << ", Power_Across: " << resistors[i].power_across << endl;

  }

}

//Series connect check

bool Ohms::seriesConnect(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  vector<node> temp = ohms.getNode();

  for (int i = 0; i < temp.size(); i++) {

      this->resistors.push_back(temp[i]);

  }

  return true;

}

//Series connect check

bool Ohms::seriesConnect(vector<Ohms>&ohms) {

  if (ohms.size() == 0) {

      return false;

  }

  for (int i = 0; i < ohms.size(); i++) {

      this->seriesConnect(ohms[i]);

  }

  return true;

}

//Overload operator

bool Ohms::operator<(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  if (this->sumResist() < ohms.sumResist()) {

      return true;

  }

  return false;

}

main.cpp

#include "resistor.h"

int main()

{

   //Set circuit voltage

  Ohms ckt1(100);

  //Loop to set resistors in circuit

  int i = 0;

  string name;

  double resistance;

  while (i < 3) {

      cout << "Enter resistor name: ";

      cin >> name;

      cout << "Enter resistance of circuit: ";

      cin >> resistance;

      //Set one resistance

      ckt1.setOneResistance(name, resistance);

      cin.ignore();

      i++;

  }

  //calculate totalpower and power consumption

  cout << "Total power consumption = " << ckt1.calcTotalPower() << endl;

  return 0;

}

Output

Enter resistor name: R1

Enter resistance of circuit: 2.5

Enter resistor name: R2

Enter resistance of circuit: 1.6

Enter resistor name: R3

Enter resistance of circuit: 1.2

Total power consumption = 1886.79

Explanation:

Note

Please add all member function details.Its difficult to figure out what each function meant to be.

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A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-
777dan777 [17]

Answer:

1709.07 ft^3/s

Explanation:

Annual peak streamflow = Log10(Q [ft^3/s] )

mean = 1.835

standard deviation = 0.65

Probability of levee been overtopped in the next 15 years = 1/5

<u>Determine the design flow ins ft^3/s </u>

P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2

                         ∴  T = 67.72 years

Q₁₅ = 1 - 0.2 = 0.8

Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )

K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )

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back to equation 1

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hence:

Log₁₀ ( Qt(ft^3/s) ) = Zt  = 3.23276

hence ; Qt = 10^3.23276

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