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Bad White [126]

3 years ago

15

Steel balls 10 mm in diameter are annealed by heating to 1150 k and then slowly cooling to 450 k in an air environment for which

T, = 325K and h = 25 W/m2-K. Assuming the properties of the steel to be k = 40 W/m-K, rho = 7800 kg/m3, and c = 600 J/kg-K, estimate the time required for the cooling process.

1 answer:

Darina [25.2K]3 years ago

4 0

Answer:

The answer is below

Explanation:

Given that:

ρ = 800 kg/m3, c = 600 J/kg-K, k = 40 W/m-K, Initial temperature = Ti = 1150,

Environment temperature = T = 450 K, Final temperature = T∞ = 325 K

Diameter = 10 mm = 0.01 m, A = 6

The estimated time for cooling process (t) is given as:

$t=\frac{\rho dc}{hA} ln\frac{T_i-T_\infty}{T-T_infty}=\frac{7800*0.01*600}{25*6} ln\frac{1150-325}{450-325}\\ \\t=589\ s\\t=0.1635\ h$

The estimated cooling time is 589 s

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A 65% efficient turbine receives 2 m^3/s of water from a reservoir. The reservoir water surface is 45 m above the centerline of

laiz [17]

Answer:

$P_{out} = 508.071 kW$

Given:

efficiency of the turbine, $\eta$ = 65% = 0.65

available gross head, $H_{G}$ = 45 m

head loss, $H_{loss}$ = 5 m

Discharge, Q = $2 m^{3}$

Solution:

The nozzle is 100% (say)

Available power at the inlet of the turbine, $P_{inlet}$ is given by:

$P_{inlet} = \rho Qg(H_{G} - H_{loss})$ (1)

where

$\rho$ = density of water = 997 $kg/m^{3}$

acceleration due to gravity, g = $9.8 m^{2}$

Using eqn (1):

$P_{inlet} = 997\times 2\times 9.8(45 - 5) = 781.65 kW$

Also, efficency, $\eta$ is given by:

$\eta = \frac{P_{out}}{P_{inlet}}$

$0.65 = \frac{P_{out}}{781.648\times 1000}$

$P_{out} = 0.65\times 781.648\times 1000 = 508071 W = 508.071 kW$

$P_{out} = 508.071 kW$

3 0

3 years ago

While supporting load P, the maximum contraction permitted in a 4.75-m long pipe column is 7 mm. Given that E =180 GPa for the c

Alexeev081 [22]

Answer:

$\mathbf{stress \ \sigma = 264.6 \ Mpa}$

Explanation:

From the concept of Hooke's Law,

$E =\dfrac{ stress \ \sigma}{ strain \ \varepsilon}$

where;

$strain \ \varepsilon = \dfrac{change \ in \ dimension }{original \ dimension}$

$strain \ \varepsilon = \dfrac{7 \ mm }{4.75 \times 10^{3} \ mm}$

$strain \ \varepsilon =0.00147368$

Recall:

$E =\dfrac{ stress \ \sigma}{ strain \ \varepsilon}$

$stress \ \sigma = E \times { strain \ \varepsilon}$

$stress \ \sigma = 180 \times 10^{3} \ Mpa \times 0.00147$

$\mathbf{stress \ \sigma = 264.6 \ Mpa}$

Thus, the stress in the pipe at the maximum allowable contraction = 264.6 Mpa

7 0

3 years ago

1 // Lab 2 tryIt2A 2 #include 3 using namespace std; 4 5 int main() 6 { int x = 1, y = 3; 7 int X = 2, Y = 4; 8 9 cout <<

padilas [110]

Answer:

Here is the complete program:

#include <iostream>

using namespace std;

int main()

{ int x = 1, y = 3;

int X = 2, Y = 4;

cout << "tryIt 2A" <<endl;

cout << x << y << endl;

cout << "x" << "y" << endl;

cout << X << " " << Y << endl;

cout << 2 * x + y << endl;

cout << 2 * X + Y << endl;

//cout << x + 2*y << endl;

cout << "x = ";

cout << x;

cout << " y = ";

cout << y;

return 0;

}

Explanation:

I will explain the code line by line in the comment with each line of code and the output of each cout statement.

int x = 1, y = 3;

This statement assigns value 1 to integer variable x and 3 to int variable y

int X = 2, Y = 4;

This statement assigns value 2 to integer variable X and 4 to int variable Y As C++ is a case sensitive language so variable x and y are different from variables X and Y.

cout << "tryIt 2A" <<endl;

This statement has cout which is used to display output on the screen. So the output displayed by this cout statement is:

tryIt2A

cout << x << y << endl;

This statement will print the values stored in x and y variables. So output displayed by cout statement here is 1 and 3. As there is not space or next line specified in the statement so output displayed will look like this:

13

cout << "x" << "y" << endl;

This statement will display x and y but these are not the variable x and y. They are enclosed in double quotation marks so they are treated as strings not variables so the output displayed is:

xy

cout << X << " " << Y << endl;

This statement will print the values stored in X and Y variables. So output displayed by cout statement here is 2 and 4. As there is space " " specified in the statement so 2 and 4 are displayed with a space between them so the output displayed will look like this:

2 4

cout << 2 * x + y << endl;

This statement has an arithmetic operation in which 2 is multiplied by the values stored in variable x and then the result is added by value of y. So 2*1 = 2 and 2 + 3 = 5. So the result produced by this cout statement is:

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cout << 2 * X + Y << endl;

This will work same as above cout statement but the only difference is that the values of capital X and Y variables are calculated here. So 2 * 2 = 4 and then 4 + 4 = 8. The result produced by this cout statement is:

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//cout << x + 2*y << endl;

This is a comment because before this statement // is written which is used for single line comment. So compiler ignores comments and will not compile this statement.

cout << "x = ";

This will display "x = " as it is not variable but it is treated as a line to be displayed on the screen. So cout statement displays:

x =

cout << x;

This will print the value stored in x variable as there are no double quotes around x so it is a variable which contains value 1. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

x = 1

cout << " y = ";

This will display "y = " as it is not variable but it is treated as a line to be displayed on the screen. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen

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cout << y;

This will print the value stored in y variable as there are no double quotes around y so it is a variable which contains value 3. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

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6 0

4 years ago

An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change i

Harman [31]

Answer:2058.992KJ

Explanation:

Given data

Mass of object $\left ( m\right )$ =521kg

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change in kinectic energy is given by substracting final kinetic energy from initial kinetic energy of body.

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7 0

3 years ago

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Rufina [12.5K]

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Weave lanes are known to be lanes that acts as an entrance and exits for a lot of cars in highways.

Hence, one can say that A Weave lane is a single lane used by drivers to enter and exit a freeway.

Learn more about Weave lane from

brainly.com/question/10828527

#SPJ1

6 0

2 years ago