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Y_Kistochka [10]
2 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]2 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]2 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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Explanation:

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Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h

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2 years ago
The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k
sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

8 0
2 years ago
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