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3 years ago

Chaplets are used to support a sand core inside a sand mold cavity. The design of the

Engineering

1 answer:

ziro4ka [17]3 years ago

4 0

Answer:

a) 2 chaplets

b) 9 chaplets

Explanation:

Before we can determine the minimum number of chaplets that should be placed beneath and above the core, we must know the mass of the sand used to make the surface of the mold cavity as well as the mass of the steel metal poured inside the mold.

Density is defined as the ratio of mass of a substance to its density.

Density = Mass/Volume

For the STEEL:

Density of steel = 7.82g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of steel = 7.82×5450

Mass of steel = 42619g

Mass of steel in kg = 42.619kg

For the SAND:

Density of sand = 1.6g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of sand = 1.6×5450

Mass of sand = 8720g

Mass of sand in kg = 8.72kg

a) Since the the chaplet support the sand from beneath the core, and each chaplet weighs 45N, we need to know the amount of force possessed by the sand.

Since the mass of the sand is 87.2kg

Weight = mass × acceleration due to gravity

Weight = 8.72×9.81

Weight of sand used to mold the core = 85.54N

Since 1 chaplet weighs 45N

This means that (85.54N/45N) i.e 1.9 which is approximately 2 chaplets must be placed beneath the core to sustain it before the steel metal is poured.

b) Since the metal poured in the core is steel, this means that the chaplet placed above the core must be able to withstand the strength of the steel.

Weight of steel = mass of steel × acceleration due to gravity

Weight of steel = 42.619×9.81

Weight of steel = 418.09N

Since one chaplet weigh 45N, the amount of chaplets that must be placed above the core is;

418.09/45

= 9.3

Therefore the minimum number of chaplets that should be placed above the core after pouring steel metal is 9chaplets.

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19. A circuit contains four 100 S2 resistors connected in series. If you test the circuit with a digital VOM,

yanalaym [24]

Answer:

Explanation:

in series circuit the resistance is divided Total resistance is equal to the sum of resistances

6 0

3 years ago

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is

Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α = $\frac{1}{\pi} (\frac{K}{\sigma Y})^2$ ....................1

put here value we get

α = $\frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2$

α = 15.043 mm

so minimum length of a surface crack is 15.043 mm

7 0

3 years ago

C++ - Green Crud Fibonacci programThe following program is to be written with a loop. You are to write this program three times

Fynjy0 [20]

Answer:

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;

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.

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;

{

//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;

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

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.

7 0

3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago

What type of engineer makes sure equipment is safe and operational

zvonat [6]

Answer:

mechanical engineer is the best answer

8 0

3 years ago