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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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Calculate the equivalent capacitance of the three series capacitors in Figure 12-1 A) 0.060 uF B) 0.8 uF C) 0.58 uF D) 0.01 uF
Naddik [55]

Answer:

The correct answer is option (A) 0.060 uF

Note: Kindly find an attached image of the complete question below

Sources: The complete question was well researched from Quizlet.

Explanation:

Solution

Given that:

C₁ = 0.1 μF

C₂ =0.22 μF

C₃ = 0.47 μF

In this case, C₁, C₂ and C₃ are in series

Thus,

Their equivalent becomes:

1/Ceq = (1/C₁ + 1/C₂ +1/C₃

1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]

1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]

1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)

1/Ceq =[(0.1724)/(0.01034)]

1/Ceq = [(16.67)]

1/Ceq =(1/16.67) = 0.059μf

Ceq = 0.059μf ≈ 0.060μf

Therefore the equivalent capacitance of the three series capacitors is 0.060μf

4 0
3 years ago
The device whose operation closely matches the way the clamp-on ammeter works is
Ivanshal [37]

Answer:

The answer is

C. Split phase motor

Explanation:

Clamp meters rely on the principle of magnetic induction to make non contact AC current measurements. Electric current flowing through a wire produces a magnetic field.

Which is similar to basic mode of operation of electric motor and split phase motor is a type of electric motor.

What is a a clamp on meter?

Clamp meters are electrical testers which have wide jaws that are able to clamp around an electrical conductor. Originally designed as a single purpose tool for measuring AC current, clamp meters now include inputs for accepting test leads and other probes that support a wide range of electrical measurements, the jaws of a clamp meter permit work in tight spaces and permits current measurements on live conductors without circuit interruption.

6 0
3 years ago
10. What does a profile of a river from its headwaters to its mouth typically show?
irina1246 [14]

Answer:

c. an abrupt increase followed by a gradual decrease

Explanation:

At the headwater, the flow gradient starts high but then slowly decreases as the river moves downstream to its mouth.

3 0
2 years ago
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the mot
Thepotemich [5.8K]

78950W the answer

Explanation:

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

5 0
3 years ago
. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C
MrRissso [65]

Answer:

The inside temperature, T_{in} is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×(T_s - T_{\infty) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

P = \dfrac{K \times A \times \Delta T}{L}

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

2456.25  = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}

T_{in} = 250 - \dfrac{2456.25  \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C

The inside temperature, T_{in} = 247.99 °C  ≈ 248 °C.

3 0
3 years ago
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