Ask question

Ask question

All categories

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.

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

You might be interested in

A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar

jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading 114.06 m^3/d/m

Explanation:

calculation for single clarifier

$sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d$

$surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2$

$surface area = 314.16 m^2$

volume of tank $V = A\times side\ water\ depth$

$=314.16\times 2 = 628.32m^3$

$Length\ of\ weir = \pi \times diameter of weir$

$= \pi \times 18 = 56.549 m$

overflow rate = $v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2$

Detention time $t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr$

weir loading $= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m$

6 0

3 years ago

12.28 LAB: Output values in a list below a user defined amount - functions Write a program that first gets a list of integers fr

Anastaziya [24]

Answer:

def output_ints_less_than_or_equal_to_threshold(user_values, upper_threshold):

for value in user_values:

if value < upper_threshold:

print(value)

def get_user_values():

n = int(input())

lst = []

for i in range(n):

lst.append(int(input()))

return lst

if __name__ == '__main__':

userValues = get_user_values()

upperThreshold = int(input())

output_ints_less_than_or_equal_to_threshold(userValues, upperThreshold)

8 0

3 years ago

A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N

soldier1979 [14.2K]

Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, $P_{t} = 8 kW = 8000 W$

Now, the force, F is given by:

$P_{t} = Force(F)\times velocity(v) = Fv$ (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, $F_{c}$ is provided by the gravitational force, $F_{G}$ :

$F_{c} = F_{G}$

$\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}$

Thus from the above, velocity comes out to be:

$v = \sqrt{\frac{GM_{e}}{R}}$

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s$

where

R = $R_{e} + H$

R = $\sqrt{GM_{e}(\frac{T}{2\pi})^{2}}$

where

G = Gravitational constant

T = Time period of rotation of Earth

R is calculated as 42166 km

Now, from eqn (1):

$8000 = F\times 3075.36$

F = 2.60 N

6 0

3 years ago

(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor

serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm = 0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( $\frac{\pi }{4}$ ) * 0.0075^2

= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( $\frac{\pi }{4}$ ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%

6 0

2 years ago

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in

Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0

3 years ago