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Y_Kistochka [10]
3 years ago
13

a sprue is 12 in long and has a diameter of 5 in at the top. The molten metal level in the pouring basing is taken to be 3 in fr

om the top of the sprue for design purposes. If a flow rate of 40 in3/s is to be achieved, what should be the diameter at the bottom of the sprue

Engineering
2 answers:
vampirchik [111]3 years ago
8 0

Answer:

See explaination

Explanation:

We can describe Aspiration Effect as a phenomenon of providing an allowance for the release of air from the mold cavity during the metal pouring.

See the attached file for detailed solution of the given problem.

prohojiy [21]3 years ago
3 0

Answer:

The diameter at the bottom of the sprue is =0.725 in, and the sprue will not occur when 0.021<0.447.

Explanation:

Solution

The first step to take is to define the Bernoulli's eqaution

h effective = v²top/2g + ptop /ρg = hbottom + v² bottom/2g + p bottom/ ρg

h effective  + 0 +0= 0 +v² bottom/2g + 0

Thus,

v bottom = √ 2gh total

=√ 2 (32. 6 ft/ s²) + (12/12 ft)

Which is = 8.074 ft/s

We now, express the relation for flow rate.

Q =π/4 D² bottom v bottom

= 40 in 3/s = π/4 D²₃ ( 8.074 ft/s) (12  in/ ft)

so,

D bottom = 0.725 in.

Then,

We express the relation to avoid aspiration

A₃/A₂ < √ h top /h total

= π/4 D²₃/ π/4 D²₂ < √3/15

= 0.725²/5² < √3/15

=0.021<0.447

Therefore, the aspiration will not happen or occur

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Annette [7]

Answer:

The heat transfer q = 18.32W

Explanation:

In this question, we are asked to calculate the heat entering the tube and also evaluate properties at T =400K

Please check attachment for complete solution and step by step explanation

6 0
3 years ago
An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are
Vika [28.1K]

This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle W_{net = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ^{Y-1 = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = ( T₂ / T₁ )^{\frac{1}{Y-1}

so we substitute

⇒ V₁ / V₂ = (  973 K / 303 K  )^{\frac{1}{1.4-1}

= (  3.21122  )^{\frac{1}{0.4}

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c)  the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂  = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

8 0
3 years ago
Many BLANK apply trial and error to develop a product. Please Help! I have one hour to finish. 30 points
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Answer:

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Explanation:

8 0
3 years ago
Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th
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Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

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3 0
4 years ago
The 10-lb block is pressed against the spring so as to compress it 2 ft when it is at a point A. If the plane is smooth, determi
leva [86]

Answer:

The distance measure from the wall = 36ft

Explanation:

Given Data:

w = 10

g =32.2ft/s²

x = 2

Using the principle of work and energy,

T₁ +∑U₁-₂ = T₂

0 + 1/2kx² -wh = 1/2 w/g V²

Substituting, we have

0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²

170 = 0.15528V²

V² = 170/0.15528

V²     = 1094.796

V = √1094.796

V = 33.09 ft/s

But tan ∅ = 3/4

∅ = tan⁻¹3/4

   = 36.87°

From uniform acceleration,

S = S₀ + ut + 1/2gt²

It can be written as

S = S₀ + Vsin∅*t + 1/2gt²

Substituting, we have

0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)

19.85t - 16.1t² + 3 = 0

16.1t² - 19.85t - 3 = 0

Solving it quadratically, we obtain t = 1.36s

The distance measure from the wall is given by the formula

d = VCos∅*t

Substituting, we have

d = 33.09 * cos 36. 87 * 1.36

d = 36ft

5 0
3 years ago
Read 2 more answers
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