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4 years ago

10

Liquid water enters a valve at 300 kPa and exits at 275 kPa. As water flows through the valve, the change in its temperature, st

ray heat transfer with the surroundings, and potential energy effects are negli-gible. Operation is at steady state. Modeling the water as incompress-ible with constant rho= 1000 kg/m3, determine the change in kinetic energy per unit mass of water flowing through the valve, in kJ/kg

1 answer:

Elan Coil [88]4 years ago

3 0

Answer:

The change in kinetic energy per unit mass of water flowing through the valve is - ΔKE = 0.025 KJ/Kg

Explanation:

Knowing

-Fluid is air

-inlet 1: P1 = 300 kPa

-exit 2: P2 = 275 kPa

density - rho= 1000 kg/m3

Using the formula

Δh = cΔT + Δp/rho

as change in temperature is neglected then change in enthalpy becomes

Δh = Δp/rho

energy equation could be defined by

Q - W = m(out) [h(out) $V^{2}$ (out)/2 + g Z(out)] - m(in) [h(in) $V^{2}$ (in)/2 + g Z(in)]

Q - W = m2 [h2 $V^{2}$ 2/2 + g Z2] - m1 [h1 $V^{2}$ 1/2 + g Z1]

as for neglecting potential energy effects

Q - W = m2(h2) - m1(h1)

as the system is adiabatic and has no work done

0 = m2 [h2 $V^{2}$ 2/2] - m1 [h1 $V^{2}$ 1/2]

from mass balance m1 = m2

0 = [h2 $V^{2}$ 2/2] - [h1 $V^{2}$ 1/2]

Change in kinetic energy could be defined by

ΔKE = $V^{2}$ 2/2 - $V^{2}$ 1/2

Change in specific enthalpy could be defined by

Δh = h2 - h1

Then the change in kinetic energy per unit mass of water flowing through the valve could be calculated as following

ΔKE = -Δh = ΔP/rho

-(275 - 300)/1000 = 0.025 KJ/Kg

- ΔKE = 0.025 KJ/Kg

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The state of plane strain on an element is:

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Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

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Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

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To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

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To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

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$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

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ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

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$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

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Below is the required code:

Explanation:

Using for loop

#include <iostream>

using namespace std;

int main()

{

//Initial crud size

int init = 0;

int newCrud;

int next=0;

//Number of days to simulate

int no_days;

int day;

cout << "Enter initial amount of green crud: ";

cin >> newCrud;

cout << "Enter number of days to simulate: ";

cin >> no_days;

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newCrud = init;

init = next;

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Enter initial amount of green crud: 5

Enter number of days to simulate: 220

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Using while loop

Program:

#include <iostream>

using namespace std;

int main()

{

char ans='y';

while (ans == 'Y' || ans == 'y')

{

//Initial crud size

int init = 0;

int newCrud;

int next=0;

//Number of days to simulate

int no_days;

int day;

cout << "Enter initial amount of green crud:

";

cin >> newCrud;

cout << "Enter number of days to simulate:

";

cin >> no_days;

for (day = 10; day<=no_days; day++)

{

if (day % 10 == 0)

{

next = newCrud + init;

}

newCrud = init;

init = next;

}

if (no_days < 5)

cout << "\nCrud reproduce only after 5 days

minimum.Hence the current amount is "

<< newCrud << " pounds.";

else

cout << "On day " << no_days << " you have "

<< init

<< " pounds of green crud." << endl;

cout << "\nWould you like to continue? (y or

n): ";

cin >> ans;

}

return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

Using do while loop

Program:

#include <iostream>

using namespace std;

int main()

{

char ans;

do

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//Initial crud size

int init = 0;

int newCrud;

int next=0;

//Number of days to simulate

int no_days;

int day;

cout << "Enter initial amount of green crud: ";

cin >> newCrud;

cout << "Enter number of days to simulate: ";

cin >> no_days;

for (day = 10; day<=no_days; day++)

{

if (day % 10 == 0)

{

next = newCrud + init;

}

newCrud = init;

init = next;

}

if (no_days < 5)

cout << "\nCrud reproduce only after 5 days

minimum.Hence the current amount is "

<< newCrud << " pounds.";

else

cout << "On day " << no_days << " you have " <<

init << " pounds of green crud." << endl;

cout << "\nWould you like to continue? (y or n):

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cin >> ans;

} while (ans == 'Y' || ans == 'y');

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Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

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Enter initial amount of green crud: 5

Enter number of days to simulate: 225

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