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3 years ago

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An air-standard Otto cycle has a compression ratio of 6 and the temperature and pressure at the beginning of the compression pro

cess are 520 deg R and 14.2 lbf/in^2, respectively. The heat addition per unit mass of air is 600 Btu/lb. Determine (a) the maximum temperature, in deg R.

1 answer:

tangare [24]3 years ago

3 0

Answer:

The maximum temperature of the cycle is 1065⁰R

Explanation:

The maximum temperature in degree Rankin can be obtained using the formula below;

$\frac{T_2}{T_1} =[\frac{V_1}{V_2}]^{1.4-1}$

Where;

T₂ is the maximum temperature of the cycle

T₁ is the initial temperature of the cycle = 520 deg R = 520 ⁰R

V₁/V₂ is the compression ratio = 6

$T_2 = T_1(\frac{V_1}{V_2})^{0.4}$

$T_2=T_1(6)^{0.4}$

$T_2=520^0R(6)^{0.4}$

T₂ = 1064.96 ⁰R

Therefore, the maximum temperature of the cycle is 1065⁰R

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Answer:

The engine block.

Explanation:

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3 years ago

1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and

belka [17]

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

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3 years ago

Read 2 more answers

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s

C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c

<u>Using effectiveness-NUT method</u>

<em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

.<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

<em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

<em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

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3 years ago

This program will store roster and rating information for a soccer team. Coaches rate players during tryouts to ensure a balance

Flauer [41]

Answer:

#include <iostream>

#include <vector>

using namespace std;

int main() {

vector<int> jerseyNumber;

vector<int> rating;

int temp;

for (int i = 1; i <= 5; i++) {

cout << "Enter player " << i

<< "'s jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter player " << i

<< "'s rating: ";

cin >> temp;

rating.push_back(temp);

cout << endl;

}

cout << "ROSTER" << endl;

for (int i = 0; i < 5; i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: " << jerseyNumber.at(i)

<< ", Rating: " << rating.at(i) << endl;

char option;

'

while (true) {

cout << "MENU" << endl;

cout << "a - Add player" << endl;

cout << "d - Remove player" << endl;

cout << "u - Update player rating" << endl;

cout << "r - Output players above a rating"

<< endl;

cout << "o - Output roster" << endl;

cout << "q - Quit" << endl << endl;

cout << "Choose an option: ";

cin >> option;

switch (option) {

case 'a':

case 'A':

cout << "Enter a new player's"

<< "jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter the player's rating: ";

cin >> temp;

rating.push_back(temp);

break;

case 'd':

case 'D':

cout << "Enter a jersey number: ";

cin >> temp;

int i;

for (i = 0; i < jerseyNumber.size();

i++) {

if (jerseyNumber.at(i) == temp) {

jerseyNumber.erase(

jerseyNumber.begin() + i);

rating.erase(rating.begin() + i);

break;

}

}

break;

case 'u':

case 'U':

cout << "Enter a jersey number: ";

cin >> temp;

for (int i = 0; i < jerseyNumber.size();

i++) {

if (jerseyNumber.at(i) == temp) {

cout << "Enter a new rating "

<< "for player: ";

cin >> temp;

rating.at(i) = temp;

break;

}

}

break;

case 'r':

case 'R':

cout << "Enter a rating: ";

cin >> temp;

cout << "\nABOVE " << temp << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

if (rating.at(i) > temp)

cout << "Player " << i + 1

<< " -- "

<< "Jersey number: "

<< jerseyNumber.at(i)

<< ", Rating: "

<< rating.at(i) << endl;

break;

case 'o':

case 'O':

cout << "ROSTER" << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: "

<< jerseyNumber.at(i) << ", Rating: "

<< rating.at(i) << endl;

break;

case 'q':

return 0;

default:

cout << "Invalid menu option."

<< " Try again." << endl;

}

}

}

Explanation:

4 0

3 years ago

An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$

From the question the, the mass flow rate at the inlet $m_{1}}$ is 2Kg/s $\\$

Since 5% flow is delivered into the feedwater heating, $\\$

$m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s$ $\\$

Also for the outlet 3 the remaining 95% will flow out. Hence

$m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s$ $\\$

Now, from $m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$ we substitute values

$W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}$

$W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)$

$\\$

$W_{out} = 2285.19 kW$ .

Hence the power produced is 2285kW

7 0

3 years ago