Ask question

Ask question

All categories

3 years ago

11

In casting experiments performed using a certain alloy and type of sand mold, it took 170 sec for a cube-shaped casting to solid

ify. The cube was 50 mm on a side.
a. Determine the value of the mold constant in Chvorinov's rule.
b. If the same alloy and mold type were used, find the total solidification time for a cylindrical casting in which the diameter = 50 mm and length = 50 mm.

2 answers:

poizon [28]3 years ago

7 0

Answer:

Answer for the question is : Solidification time will be same i.e. 170. See attached file for explanation.

Explanation:

Dimas [21]3 years ago

7 0

Answer:

a) the value of the mold constant in Chvorinov`s rule is 2.45 s/mm²

b) the total solidification time is 170 s

Explanation:

a) The volume is 50³ = 125000 mm³

The area is

A = 6 * 50² = 15000 mm²

The ratio V/A is

125000/15000 = 8.33 mm

The mold constant in Chvorinov`s rule is

$C_{ch } =\frac{T}{(V/A)^{2} } =\frac{170}{8.33^{2} } =2.45s/mm^{2}$

b) The volume is

$V=\frac{\pi D^{2}L }{4} =\frac{\pi 50^{2}*50 }{4} =98174.8mm^{3}$

The area is

$A=\frac{2\pi D}{4} +\pi DL=\frac{2\pi 50^{2} }{4} +\pi *50*50=11781mm^{3}$

V/A = 98174.8/11781 = 8.33

The total solidification is

T =2.45 * (8.33²) = 170 s

You might be interested in

The enforcement of OSHA standards is provided by federal and state

Gnoma [55]

Answer:

Explanation:

Enforcing OSHA, Occupational Safety and Health Administration, standards is not a job for electricians, lawmakers or tax collectors. The right answer is safety inspectors.

3 0

2 years ago

Not sure which one....

Airida [17]

I think downwards as that's how most saw's work.

4 0

3 years ago

Reception of signals from a radio facility, located off the airway being flown, may be inadequate at the designated mea to ident

lakkis [162]

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

<h3>What is altitude?</h3>

Altitude or height exists as distance measurement, usually in the vertical or "up" approach, between a reference datum and a point or object. The exact meaning and reference datum change according to the context.

The MOCA exists in the lower published altitude in effect between fixes on VOR airways, off-airway routes, or route segments that satisfy obstacle support conditions for the whole route segment. This altitude also ensures acceptable navigational signal coverage only within 22 NM of a VOR.

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

Therefore, the correct answer is 22 NM of a VOR.

To learn more about altitudes refer to:

brainly.com/question/1159693

#SPJ4

3 0

2 years ago

H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate

Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

$\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0$ (1)

Where:

$\dot m$ - Mass flow, in kilograms per second.

$h_{in}$ , $h_{out}$ - Specific enthalpies at inlet and outlet, in kilojoules per second.

$v_{in}, v_{out}$ - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

$p = 2000\,kPa$

$T = 300\,^{\circ}C$

$h_{in} = 3024.2\,\frac{kJ}{kg}$

$\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$

Where $\nu_{in}$ is the specific volume of water at inlet, in cubic meters per kilogram.

Outlet (Superheated steam)

$p = 600\,kPa$

$T = 160\,^{\circ}C$

$h_{out} = 2758.9\,\frac{kJ}{kg}$

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $h_{in} = 3024.2\,\frac{kJ}{kg}$ , $h_{out} = 2758.9\,\frac{kJ}{kg}$ and $v_{in} = 30\,\frac{m}{s}$ , then the flow speed at outlet is:

$35765-25\cdot v_{out}^{2} = 0$ (2)

$v_{out} \approx 37.823\,\frac{m}{s}$

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius ( $r_{in}$ ), in meters, by this expression:

$\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}}$ (3)

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $v_{in} = 30\,\frac{m}{s}$ and $\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$ , then the inlet radius is:

$r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}$

$r_{in}\approx 0.258\,m$

The inlet radius of the nozzle is approximately 0.258 meters.

7 0

2 years ago

(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows

maxonik [38]

Answer:

diameter of the sprue at the bottom is 1.603 cm

Explanation:

Given data;

Flow rate, Q = 400 cm³/s

cross section of sprue: Round

Diameter of sprue at the top $d_{top}$ = 3.4 cm

Height of sprue, h = 20 cm = 0.2 m

acceleration due to gravity g = 9.81 m/s²

Calculate the velocity at the sprue base

$V_{base}$ = √2gh

we substitute

$V_{base}$ = √(2 × 9.81 m/s² × 0.2 m )

$V_{base}$ = 1.98091 m/s

$V_{base}$ = 198.091 cm/s

diameter of the sprue at the bottom will be;

Q = AV = (π $d_{bottom}^2$ /4) × $V_{base}$

$d_{bottom}$ = √(4Q/π $V_{base}$ )

we substitute our values into the equation;

$d_{bottom}$ = √(4(400 cm³/s) / (π×198.091 cm/s))

$d_{bottom}$ = 1.603 cm

Therefore, diameter of the sprue at the bottom is 1.603 cm

6 0

2 years ago