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Natali5045456 [20]
3 years ago
5

Water is flowing steadily through a 3-m long, 20 mm innerdiameter cast iron pipe. The water enters at a uniform velocity U and e

xits with a velocity profile that is a function of radial position within the pipe, U (r ). The inlet flow velocity U is 5 m/s and the outlet flow velocity is
Engineering
1 answer:
devlian [24]3 years ago
8 0

Answer:

Hello your question is incomplete attached below is the complete question

answer :

U/r = R  = U_{max} [ 1 - (\frac{R}{R} )^2 ] = 0

∴ No slip condition is satisfied

Explanation:

Given that

Outlet velocity  U(r)  = U_{max} [ 1 - (\frac{r}{R} )^2 ]

<u>prove that the output velocity profile satisfies the no slip condition</u>

at no slip u = 0 ( i.e. at the pipe's inner surface )

at , r = R  ( inference  is at center )

hence  U/r = R  = U_{max} [ 1 - (\frac{R}{R} )^2 ] = 0

∴ No slip condition is satisfied

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Answer:

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7 0
3 years ago
A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are w
adelina 88 [10]

Answer:

Maximum allowable chip power is 0.35 W

Explanation:

This question is incomplete. The complete question is

A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at t[infinity] = 15°c. from reliability considerations, the chip temperature must not exceed t = 85°c. f the coolant is air and the corresponding convection 200 w/m2 k, what is the maximum allowable chip power?

<u>ANSWER:</u>

The heat transfer through convection, we have the equation:

q = hA(T - T∞)

where,

q = power transfer through convection = ?

h = convection coefficient = 200 W/m²K

A = Area of convection surface = (0.005 m)² = 0.000025 m²

T = Chip surface temperature = 85° C

T∞ = Fluid temperature = 15° C

Therefore,

q = (200 W/m².K)(0.000025 m²)(85° C - 15° C)

<u>q = 0.35 W</u>

Since, difference in temperature is same on both Celsius and kelvin scale. Therefore, Celsius is written as kelvin for difference and they shall be cancelled.

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4 years ago
A liquid-liquid extraction process consists of two units, a mixer and a separator. One inlet stream to the mixer consists of two
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Answer:

One inlet stream to the mixer flows at 100.0 kg/hr and is 35wt% species-A and 65wt% species-B.

Explanation:

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4 years ago
The natural material in a borrow pit has a mass unit weight of 110.0 pcf and a water content of 6%, and the specific gravity of
enot [183]

Answer:

A. 288,030.91 cy

B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water

Explanation:

The natural material in the barrow properties are;

The mass unit weight, γ = 110.0 pcf

The water content, w = 6%

The specific gravity of the soil solids, G_s = 2.63

The desired dry unit weight, \gamma _d = 122 pcf

The water content, w₁ = 5.5 %

The net section volume, V_T = 245,000 cy = 6,615,000 ft³

A.  \gamma _d = W_s/V_T

∴ W_s = V_T × \gamma _d = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs

w = (W_w/W_s) ×  100

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The weight of solids

W = W_s + W_w = 807030000 lbs + 48421800 lbs = 855451800 lbs

V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy

V = 288,030.91 cy

The amount of cubic yards of borrow required = 288,030.91 cy

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W_{w1} = (w₁/100) × W_s = (5.5/100) × 807030000 lbs = 44386650 lbs

The amount of water that must be added =  W_{w1} - W_w = 44386650 lbs - 48421800 lbs = -4,035,150 lbs

Therefore, 4,035,150 lbs of water must be removed

The density of water, ρ = 8.345 lbs/gal

Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal  of water must be removed from the natural material

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3 years ago
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Margaret [11]

Answer:

what are simple machines?

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