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3 years ago

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A disk‑shaped part is to be cast out of aluminum. The diameter of the disk = 401 mm and its thickness = 25 mm. If the mold const

ant = 1.9 sec/mm2 in Chvorinov's rule, how long will it take the casting to solidify (in the unit of seconds)?

1 answer:

kupik [55]3 years ago

3 0

Answer:

w

Explanation:

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Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c

Nitella [24]

Answer:

(a) The volume rate of flow per meter width = 5.6*10⁻³ m²/s

(b) The shear stress acting on the bottom plate = 157.5 N/m²

(c) The velocity along the centerline of the channel = 0.93 m/s

Explanation:

(a)

Calculating the distance of plate from centre line using the formula;

h = d/2

where h = distance of plate

d = diameter of flow = 9 mm

Substituting, we have;

h = 9/2

= 4.5 mm = 4.5*10^-3 m

Calculating the volume flow rate using the formula;

Q = (2h³/3μ)* (Δp/L)

Where;

Q = volume flow rate

h = distance of plate = 4.5*10^-3 m

μ = dynamic viscosity = 0.38 N.s/m²

(Δp/L) = Pressure drop per unit length = 35 kPa/m = 35000 Pa

Substituting into the equation, we have;

Q = (2*0.0045³/3*0.38) *(35000)

= (1.8225*10⁻⁷/1.14) * (35000)

= 1.60*10⁻⁷ * 35000

= 5.6*10⁻³ m²/s

Therefore, the volume flow rate = 5.6*10⁻³ m³/s

(b) Calculating the shear stress acting at the bottom plate using the formula;

τ = h*(Δp/L)

= 0.0045* 35000

= 157.5 N/m²

(c) Calculating the velocity along the centre of the channel using the formula;

u(max) = h²/2μ)* (Δp/L)

= (0.0045²/2*0.38) * 35000

=2.664*10⁻⁵ *35000

= 0.93 m/s

7 0

3 years ago

What is the relative % change in P if we double the absolute temperature of an ideal gas keeping mass and volume constant?

Contact [7]

Answer: 100% (double)

Explanation:

The question tells us two important things:

Mass remains constant
Volume remains constant

(We can think in a gas enclosed in a closed bottle, which is heated, for instance)

In this case we know that, as always the gas can be considered as ideal, we can apply the general equation for ideal gases, as follows:

State 1 (P1, V1, n1, T1) ⇒ P1*V1 = n1*R*T1
State 2 (P2, V2, n2, T2) ⇒ P2*V2 = n2*R*T2

But we know that V1=V2 and that n1=n2, som dividing both sides, we get:

P1/P2 = T1/T2, i.e, if T2=2 T1, in order to keep both sides equal, we need that P2= 2 P1.

This result is just reasonable, because as temperature measures the kinetic energy of the gas molecules, if temperature increases, the kinetic energy will also increase, and consequently, the frequency of collisions of the molecules (which is the pressure) will also increase in the same proportion.

6 0

3 years ago

At a certain location, wind is blowing steadily at 7 m/s. Determine the mechanical energy of air per unit mass and the power gen

Kaylis [27]

Answer:

Explanation:

From the information given;

The velocity of the wind blow V = 7 m/s

The diameter of the blades (d) = 80 m

Percentage of the overall efficiency $\eta_{overall} = 30\%$

The density of air $\rho = 1.25 kg/km^3$

Then, we can use the concept of the kinetic energy of the wind blowing to estimate the mechanic energy of air per unit mass by using the formula:

$e_{mechanic} = \dfrac{mV^2}{2}$

here;

m = $\rho AV$

= $1.25 \times \dfrac{\pi}{4}(80)^2 \times 7$

= 43982.29 kg/s

∴

$W = e_{mechanic} = \dfrac{mV^2}{2}$

$= \dfrac{43982.29 \times 7^2}{2}$

$= 1077566.105 \ W$

$\mathbf{ =1077.566 \ kW}$

The actual electric power is:

$W_{electric} = \eta_{overall} \times W$

$W_{electric} = 0.3 \times 1077.566$

$\mathbf{W_{electric} =323.26 \ kW}$

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2 years ago

If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical

vampirchik [111]

Answer:

critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 × $10^{9}$ N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$ .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 × $10^{-3}$ m

so now put value in equation 1 we get

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$

( σc ) = $\sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}$

( σc ) = 27.396615 × $10^{6}$ N/m²

so critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

6 0

2 years ago

What are the searching algorithms used by search engines?

Juliette [100K]

Search engines use specific algorithms based on their data size and structure to produce a return value.
Linear Search Algorithm. ...
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2 years ago