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3 years ago

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A rectangular plate casting has dimensions 200mm x 100mm x 20mm. The riser for this sand casting mold is in the shape of a spher

e. The casting takes 3.5mins to solidify. Calculate the diameter of the riser so that it takes 20% longer for the riser to solidify

1 answer:

9966 [12]3 years ago

3 0

Answer:

Diameter of riser =6.02 mm

Explanation:

Given that

Dimensions of rectangular plate is 200mm x 100mm x 20mm.

Volume of rectangle V= 200 x 100 x 20 $mm^3$

Surface area of rectangle A

A=2(200 x 100+100 x 20 +20 x 200) $mm^2$

So V/A=7.69

We know that

Solidification times given as

$t=K\left(\dfrac{V}{A}\right)^2$ -----1

Lets take diameter of riser is d

Given that riser is in spherical shape so V/A=d/6

And

Time for solidification of rectangle is 3.5 min then time for solidificartion of riser is 4.2 min.

Lets take $\dfrac{V}{A}=M$

$\dfrac{M_{rac}}{M_{riser}}=\dfrac{7.69}{\dfrac{d}{6}}$

Now from equation 1

$\dfrac{3.5}{4.2}=\left(\dfrac{7.69}{\dfrac{d}{6}}\right)^2$

So by solving this d=6.02 mm

So the diameter of riser is 6.02 mm.

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Answer:

a) 2.2 m/s² b) 8 s

Explanation:

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As we don´t know the time needed to accelerate, we can use the following equation:

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At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

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a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

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A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b

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Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

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we use here lever rule that is

Wα = Wβ ...............1

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now we take here left side of equation

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$\frac{C_\beta - Co}{C_\beta - Ca}$ = 0.5

$\frac{87 - 57}{87 - Ca}$ = 0.5

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A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during

astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

$\sigma_c$ = tensile stress = 200 Mpa

$K$ = plane strain fracture toughness= 55 Mpa $\sqrt{m}$

$\lambda$ = length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

$\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}$ (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for $\lambda$

Multiplying both sides of equation (1) by $Y\sqrt{\pi\lambda}$

$\sigma_c Y\sqrt{\pi\lambda}=K$ (2)

Sequaring both sides of equation (2):

$(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2$

$\sigma^2_c Y^2 \pi\lambda=K^2$ (3)

Dividing both sides by $\sigma^2_c Y^2 \pi$ we got:

$\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2$ (4)

Replacing the values into equation (4) we got:

$\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m$

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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