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mixer [17]
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.
Engineering
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:

Download pdf
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

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An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [
crimeas [40]

Answer:

Modulus of resilience will be 3216942.308j/m^3

Explanation:

We have given yield strength \sigma _y=818MPa

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience =\frac{\sigma _y^2}{2E}, here \sigma _y is yield strength and E is elastic modulus

Modulus of resilience =\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3  

5 0
3 years ago
6. Driving with parking lights only (in place of headlights) is against the law. A. True B. False
trasher [3.6K]

Answer:

B false it is illegal to only have got fog lights on though and bright headlights because it can distract other drivers going last and if the y are distracted then that will cause a collision

Hope this helps :)

Explanation:

4 0
3 years ago
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before adjusting drive-belt tension, technician a checks for proper pulley alignment. technician b looks up the specified belt t
Vsevolod [243]

Answer:

Technician b is correct

Explanation:

Before adjusting drive-belt tension, it is very important to check the vehicle workshop manual for specified belt tension, so that you can match your reading against the specification in the vehicle's service manual. If the tension reading you have matches the suggested reading in the vehicle's service manual and the belt is not damaged then you do not need to proceed any further. But if the reading does not match, then you can adjust the belt tension.

Therefore, technician b is correct.

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3 years ago
A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo
Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, e_{min} = 0.51

Void ratio in the loosest state, e_{max} = 0.87

Now,

Dry density, \gamma_d=\frac{\gamma_t}{1+w}

=\frac{18}{1+0.05}

= 17.14 kN/m³

Also,

\gamma_d=\frac{G\gamma_w}{1+e}

here, G = Specific gravity = 2.7 for sand

17.14=\frac{2.7\times9.81}{1+e}

or

e = 0.545

Relative density = \frac{e_{max}-e}{e_{max}-e_{min}}

= \frac{0.87-0.545}{0.87-0.51}

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

or

S = 0.2477

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S = 0.2477 × 100% = 24.77%

7 0
3 years ago
Determine the period of each of the following discrete-time signals (if a signal is not periodic, denote its period by infinity)
sergiy2304 [10]

Answer:

a) it is periodic

N = (20/3)k = 20 { for K =3}

b) it is Non-Periodic.

N = ∞

c) x(n) is periodic

N = LCM ( 5, 20 )

Explanation:

We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.

then the period of the signal is given as

N = ( 2π/w₀)K

k is least integer for which N is also integer

Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2

now

a) cos(2π(0.15)n)

w₀ = 2π(0.15)

Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3

so, it is periodic

N = (20/3)k = 20 { for K =3}

b) cos(2n);

w₀ = 2

Now, 2π/w₀ = 2π/2) = π

so, it is Non-Periodic.

N = ∞

c)  cos(π0.3n) + cos(π0.4n)

x(n) = x1(n) + x2(n)

x1(n) = cos(π0.3n)

x2(n) =  cos(π0.4n)

so

w₀ = π0.3

2π/w₀  = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3

∴ N1 = 20

AND

w₀ = π0.4

2π/w₀  = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5

∴ N² = 5

so, x(n) is periodic

N = LCM ( 5, 20 )

6 0
3 years ago
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