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lutik1710 [3]
3 years ago
5

The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200 K and 400 K,

respectively. For each case, evaluate the net power developed by the cycle, in kW, and the thermal efficiency. Also in each case apply the equation below on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
(a) Qh(dot)=600 kW, Qc(dot)=400 kW
(b) Qh(dot)=600 kW, Qc(dot)=0 kW
(c) Qh(dot)=600 kW, Qc(dot)=200 kW

∮ (δQ/T)_b = -σ_cycle
Engineering
1 answer:
melomori [17]3 years ago
3 0

Answer:

(a) Qh(dot)=600 kW, Qc(dot)=400 kW  is an irreversible process.

(b) Qh(dot)=600 kW, Qc(dot)=0 kW  is an impossible process.

(c) Qh(dot)=600 kW, Qc(dot)=200 kW  is a reversible process.

Explanation:

T(hot) = 1200k, T(cold) = 400

efficiency n = (Th - Tc ) / Tc

n = (1200 - 400) / 1200 = 0.667 (this will be the comparison base)

(a)

Qh = 600 kW, Qc = 400 kW

n = (Qh - Qc) / Qh ⇒ (600 - 400) / 600

n = 0.33

0.33 is less than efficiency value from temperature 0.67

∴ it is irreversible process

(b)

Qh = 600 kW, Qc = 0

n = (Qh - Qc) / Qh ⇒ (600 - 0) / 600 = 1

efficiency in any power cycle can never be equal to one.

∴ it is an impossible process.

(c)

Qh = 600 kW, Qc = 200 kW

n = (Qh - Qc) / Qh = (600 - 200) / 600

n = 0.67 (it is equal to efficiency value from temperature)

∴ it is a reversible process

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The inverted U-tube is used to measure the pressure difference between two points A and B in an inclined pipeline through which
JulijaS [17]

Answer:

i) 0.610 m or 610 mm

ii) 0.4 m or 400 mm

Explanation:

The pressure difference between the pipes is

a) Air

Pa + πha +Ha = Pb + πhb +Hb

Pa - Pb = π(hb-ha) + Hb-Ha

Relative density of air = 1.2754 kg /m3

Pa - Pb = 1.2754 * 0.4 + (0.3-0.2) = 0.610 m or 610 mm

b) paraffin of relative density of 0.75

Pa - Pb = π(hb-ha) + Hb-Ha

Pa - Pb = 0.75 * 0.4 + (0.3-0.2) = 0.4 m or 400 mm

8 0
3 years ago
Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What
Tresset [83]

Answer:

the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

Explanation:

Given that:

Initial Temperature T_1 = 15°C

Initial Pressure P_1 = 5 atm

Final Pressure P_2 = 1 atm

Data obtain from steam tables of saturated water at  15°C are as follows:

Specific volume  v = 1.001 cm³/gm

The change in temperature = 2°C

Specific heat of water = 4.19 J/gm.K

volume expansivity β = 1.5 × 10⁻⁴ K⁻¹

The expression to determine the change in temperature can be given as :

\delta \ T = \frac{-V (1- \beta \ T}{C_p} * \delta \ P ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})\delta \ T = \frac{-1.001 \frac{cm^3}{gm} (1- 1.5*10^{-4} \  K^{-1} )*2}{4.19 \ \frac{J}{gm.K}} *(5-1)atm ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})

Δ T = 0.093 K

Now; we can calculate the lost work bt the formula:

W_{lost} = T_{surr} *S

where ;

T_{surr} is the temperature of the surrounding. = 20°C = (20+273.15)K =  293.15 K

From above the change in entropy is:

\delta \  S = C_p \  In (\frac{T+ \delta \ T }{T}) *  \beta V \delta P

\delta \  S = 4.19*  \  In (\frac{288.15+0.093 }{288.15}) -  1.5*10^{-4} * 1.001 (5-1)* (\frac{1}{9.87})

\delta \  S =1.408*10^{-3} \ J/gm.K

W_{lost} = T_{surr} *S

W_{lost} = 293.15* 1.408*10^{-3} \ J/gm.K

W_{lost} = 0.413 \  kJ/kg

Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

6 0
4 years ago
Al of the following are common automotive network configurations EXCEPT:
qwelly [4]

Answer:

Option D

triangle series

Explanation:

Some of the common automotive network configurations include looped series configuration and star parallel. Moreover, bussed parallel network configuration is also used in automotive. However, there's no any network configuration called triangle series.

4 0
3 years ago
For this lab you will be creating a auto-expanding dynamic-array. This array will explicitly hold std library strings. As with a
Inessa05 [86]

Answer:

Explanation:

Program is divided into 3 files. stringVector.h , stringVectory.cpp (implementation file) and main.cpp (testing file)

stringVector.h

#include<iostream>

using namespace std;

class stringVector {

private:

string * data;

unsigned length;

unsigned allocated_length;

public:

stringVector();

virtual ~stringVector();

unsigned size();

unsigned capacity();

void reserve(unsigned new_size);

bool empty();

void append(string new_data);

void swap(unsigned pos1, unsigned pos2);

stringVector &operator = (stringVector const &rhs);

string& operator[](unsigned position);

void sort();

};

stringVector.cpp

#include<iostream>

#include<stdexcept>

#include"stringVector.h"

stringVector::stringVector()

{

data = NULL;

length = 0;

allocated_length = 0;

}

stringVector::~stringVector()

{

delete [] data;

}

unsigned stringVector::size()

{

return length;

}

unsigned stringVector::capacity()

{

return allocated_length;

}

void stringVector::reserve(unsigned new_size)

{

string *temp = new string[new_size]; // Create a new array

 

/*Copy the contents of the array*/

for(int i =0; i < new_size; i++){

if(i < length){

temp[i] = data[i];

}

else

break;

}

 

delete []data ;// Delete previous array

data = temp;

allocated_length = new_size;

if(length > new_size){

length = new_size;

}

}

bool stringVector::empty(){

return (length == 0) ? true : false;

}

void stringVector::append(string new_data)

{

string *temp = NULL;

if(length == allocated_length){

if(allocated_length == 0){

data = new string[10];

allocated_length = 10;    

}

else{

temp = new string[2 * allocated_length]; // Create a new array with double the size

for(int i = 0; i < length ; i++){

temp[i] = data[i];

}

allocated_length = 2 * allocated_length;

 

if(data != NULL)

delete []data;

data = temp;

}

}

data[length] = new_data;

length++;

}

void stringVector::swap(unsigned pos1, unsigned pos2)

{

string str;

if((pos1 >= length) || (pos2 >= length)){

cout << "Index Out of bounds" << endl;

return;

}

str = data[pos1];

data[pos1] = data[pos2];

data[pos2] = str;

}

stringVector& stringVector::operator = (stringVector const &rhs)

{

delete [] data;

length = rhs.length;

allocated_length = rhs.allocated_length;

this->data = new string[allocated_length];

for(int i=0 ; i < length; i++)

this->data[i] = rhs.data[i];

}

string& stringVector::operator[](unsigned position)

{

if(position > length){

throw std::out_of_range("Position out of range");

}

return data[position - 1];

}

void stringVector::sort()

{

string str;

for(int i= (length - 1) ; i > 0; i--){

for(int j = 0; j < i; j++){

if(data[j].compare(data[j+1]) > 0){

//Swap

str = data[j];

data[j] = data[j+1];

data[j+1] = str;

}

}

}

}

main.cpp

#include<iostream>

#include<stdexcept>

#include"stringVector.h"

void printVector(stringVector &);

int main()

{

stringVector vector;

string str1("California");

string str2("Alabama");

string str3("Oklahoma");

string str4("Texas");

vector.append(str1);

vector.append(str2);

vector.append(str3);

vector.append(str4);

cout << vector.size() << " - " << vector.capacity() << endl;

printVector(vector);

cout << endl;

vector.reserve(3);

cout << vector.size() << " - " << vector.capacity() << endl;

printVector(vector);

cout << endl;

vector.append(str4);

cout << vector.size() << " - " << vector.capacity() << endl;

printVector(vector);

cout << endl << "Copied vector : " << endl;

stringVector vector2 = vector;

printVector(vector2);

cout << endl;

vector.sort();

cout << "Vector 1 after sorting" << endl;

printVector(vector);

return 0;

}

void printVector(stringVector &vec)

{

cout << "Vector : ";

for(int i = 1 ; i <= vec.size(); i++){

cout << vec[i] << " ";

}

cout << endl;

}

OUTPUT:

[[email protected] cppProg3]$ ./main

4 - 10

Vector: California Alabama Oklahoma Texas

3 -3

Vector: California Alabama Oklahoma

4 - 6

Vector: California Alabama Oklahoma Texas

Copied Vector:

Vector: California Alabama Oklahoma Texas

Vector 1 after sorting

Vector: Alabama California Oklahoma Texas

8 0
3 years ago
Student A says hazardous waste can take the form of solid, liquid, or gas. Student B says hazardous waste can only take the form
lina2011 [118]

Answer:

Student A

Explanation:

hope this helps have a great day

4 0
3 years ago
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