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Zina [86]
3 years ago
14

A town is designing a rectangular, 4m deep settling tank for treating surface water intake. The tank will have a flow velocity o

f 5 cm/s. The untreated water has particles with the following properties: d=0.025 mm and density of 2,650 kg m-3 The settler is placed following a coagulation-flocculation step which is able to do one of the following: a) Triple the diameter of the particles but have no impact on the particle density. b) Triple the density of the particles but have no impact on their diameter. 1. Which of the options would you select (a or b)? (Please explain your selection in the next question) 2. According to your selection, calculate the length of the settler (in m) required to remove the particles assuming they reach terminal velocity.
Engineering
1 answer:
Oksana_A [137]3 years ago
7 0

Answer:

333335790089

Explanation:

blh

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Compute the number of kilo- grams of hydrogen that pass per hour through a 6-mm-thick sheet of palladium having an area of 0.25
nydimaria [60]

Answer:

The number of kilo- grams of hydrogen that pass per hour through this sheet of palladium is 4.1 * 10^{-3} \frac{kg}{h}

Explanation:

Given

x1 = 0 mm

x2 = 6 mm = 6 * 10^{-3} m

c1 = 2 kg/m^{3}

c2 = 0.4 kg/m^{3}

T = 600 °C

Area = 0.25 m^{2}

D = 1.7 * 10^{8} m^{2}/s

First equation

J = - D \frac{c1 - c2}{x1 - x2}

Second equation

J = \frac{M}{A*t}

To find the J (flux) use the First equation

J = - 1.7 * 10^{8} m^{2}/s * \frac{2 kg/m^{3}  - 0.4 kg/m^{3}}{0 - 6 * 10^{-3} } = 4.53 * 10^{-6} \frac{kg}{m^{2}s }

To find M use the Second equation

4.53 * 10^{-6} \frac{kg}{m^{2}s} = \frac{M}{0.25 m^{2} * 3600s/h}

M = 4.1 * 10^{-3} \frac{kg}{h}

4 0
3 years ago
Which of the following is not one of the systems required to ensure the safe and correct operation of an engine?
velikii [3]
Brake system

Explanation: the engine doesn’t need to be running to make the brake system work the brake system it’s independent
6 0
3 years ago
Recall the steps of the engineering design process. Compare and contrast the
Marta_Voda [28]

Answer:

hi

Explanation:

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3 0
2 years ago
Block D of the mechanism is confined to move within the slot of member CB. Link AD is rotating at a constant rate of ωAD = 6 rad
svet-max [94.6K]

Answer:

1) 1.71 rad/s

2) -6.22 rad/s²

Explanation:

Choose point C to be the origin.

Using geometry, we can show that the coordinates of point A are:

(a cos 30°, a sin 30° − b)

Therefore, the coordinates of point D at time t are:

(a cos 30° − b sin(ωt), a sin 30° − b + b cos(ωt))

The angle formed by CB with the x-axis is therefore:

tan θ = (a sin 30° − b + b cos(ωt)) / (a cos 30° − b sin(ωt))

1) Taking the derivative with respect to time, we can find the angular velocity:

sec² θ dθ/dt = [(a cos 30° − b sin(ωt)) (-bω sin(ωt)) − (a sin 30° − b + b cos(ωt)) (-bω cos(ωt))] / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω [(a cos 30° − b sin(ωt)) sin(ωt) − (a sin 30° − b + b cos(ωt)) cos(ωt)] / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω [(a cos 30° sin(ωt) − b sin²(ωt)) − (a sin 30° cos(ωt) − b + b cos²(ωt))] / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − b sin²(ωt) − a sin 30° cos(ωt) + b − b cos²(ωt)) / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − a sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -abω (cos 30° sin(ωt) − sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²

We know at the moment shown, a = 350 mm, b = 200 mm, θ = 30°, ω = 6 rad/s, and t = 0 s.

sec² 30° dθ/dt = -(350) (200) (6) (cos 30° sin(0) − sin 30° cos(0)) / (350 cos 30° − 200 sin(0))²

sec² 30° dθ/dt = -(350) (200) (6) (-sin 30°) / (350 cos 30°)²

dθ/dt = (200) (6) (1/2) / 350

dθ/dt = 600 / 350

dθ/dt = 1.71 rad/s

2) Taking the second derivative of θ with respect to time, we can find the angular acceleration.

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30° − b sin(ωt))² (ω cos 30° cos(ωt) + ω sin 30° sin(ωt)) − (cos 30° sin(ωt) − sin 30° cos(ωt)) (2 (a cos 30° − b sin(ωt)) (-bω cos(ωt)))] / (a cos 30° − b sin(ωt))⁴

At t = 0:

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30°)² (ω cos 30°) − (0 − sin 30°) (2 (a cos 30°) (-bω))] / (a cos 30°)⁴

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω (a²ω cos³ 30° − 2abω sin 30° cos 30°) / (a⁴ cos⁴ 30°)

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -bω (aω cos² 30° − 2bω sin 30°) / (a² cos³ 30°)

d²θ/dt² + 2 tan θ dθ/dt = -bω² (a cos² 30° − b) / (a² cos 30°)

Plugging in values:

d²θ/dt² + 2 tan 30° dθ/dt = -(200) (6)² (350 cos² 30° − 200) / (350² cos 30°)

d²θ/dt² + 2 tan 30° dθ/dt = -7200 (262.5 − 200) / (350² cos 30°)

d²θ/dt² + 2 tan 30° (1.71) = -4.24

d²θ/dt² = -6.22 rad/s²

4 0
3 years ago
A. The ragion was colonized by European powers
alex41 [277]

Answer:

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Explanation:

3 0
2 years ago
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