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Kipish [7]
3 years ago
15

Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows the pouring

cup and flows into the downsprue. The cross section of the sprue is round, with a diameter at the top = 3.4 cm. If the sprue is 20 cm long, determine the proper diameter at its base so as to main- tain the same volume flow rate.
Engineering
1 answer:
umka2103 [35]3 years ago
8 0

Answer:

the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

Explanation:

Given the data in the question ;

flowrate Q = 400 cm³/s

cross section of the sprue is round

Diameter of sprue at the top d_top = 3.4 cm

Height of sprue = 20 cm = 0.2 m³

the proper diameter at its base so as to maintain the same volume flow rate = ?

first we determine the velocity at the sprue base

V_base = √2gh = √( 2×9.81×0.2) = √3.924 = 1.980908 m = 198.0908 cm

so, diameter of the sprue at the bottom  will be

Q = AV = [ (( πd²_bottom)/4) × V_bottom ]

d_bottom =  √(4Q/πV_bottom)

we substitute

d_bottom =  √((4×400)/(π×198.0908 ))

d_bottom =  √( 1600/622.3206)

d_bottom =  √2.571022

d_bottom =  1.6034 cm

Therefore, the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

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Answer:

I = 0.625 A

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Given that,

Power of the light bulb, P = 75 W

Voltage of the circuit, V = 120 V

We need to find the current flowing through it. We know that, Power is given by :

P=V\times I

I is the electric current

I=\dfrac{P}{V}\\\\I=\dfrac{75\ W}{120\ V}\\\\I=0.625\ A

So, the current is 0.625 A.

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3 years ago
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3 years ago
What is the rated power output in ( kw) of a 8 pole motor designed to an IEC 180L motor frame ?<br>​
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Answer:

P=11 kW

Explanation:

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4 0
3 years ago
Typically each development platform consists of the following components, except:Select one:a.Operating systemb.System softwarec
damaskus [11]

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4 0
2 years ago
The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8
il63 [147K]

Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×\frac{100}{w}

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = \frac{D60}{D10}

⇒ Cu = \frac{0.4}{0.12} = 3.33

d.)

Coefficient of Graduation = Cc = \frac{D30^{2}}{D10 . D60}

⇒ Cc = \frac{0.22^{2}}{(0.4) . (0.12)} = 1

3 0
3 years ago
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