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Over [174]
3 years ago
9

A wine aerator is a small, in-bottle, hand-held pour-through or decantor top device using the venturi effect for aerating the wi

ne. These devices mix air into the wine as it flows through (or over), increasing exposure to oxygen and causing aeration. Apparently, wine which has been aerated tastes better. You want to design a new aerator that will also chill the wine as it is poured. This aerator will be aventuri-shaped tube with an input diameter of 1 cm and an output diameter of 1 mm. Determine the flow rate of the wine exiting theventure tube while pouring as a function of pouring angle and volume of wine in the bottle. The wine bottle can be modeled as a cylinder of 10 cm in diameter and length of 35 cm. You can assume the pressure at the exit is 1 atm and you can assume the head in the bottle while pouring has a pressure of 1 atm.You will need to determine the pressure at the inlet of the aerator as a function of time at various angles. Assume the bottle starts full. Also, calculate the resultant force needed to keep the device attached to the top of your wine bottle as a function of time for the same angles. Comment on maximum forces. Finally, you need to determine the rate that energy needs to be removed to chill the wine from room temperature to 4 C as it flows through the aerator.

Engineering
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

Input area=0.785x10^-4m^2

Output area=0.785x10^-6m^2

P1-p2=0.49x0.99v2

V1 =0.01v2

Explanation:

Please see attachment for step by step guide

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Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at
a_sh-v [17]

Answer:

(a) 62460 kg/hr

(b) 17,572.95 kW

(c) 3,814.57 kW

Explanation:

Volumetric flow rate, G = 30 m³ / 1 min => 90 / 60 => 1.5

Calculate for h₁ , h₂ , h₃

h₁ is h at P = 40 bar, 500°C => 3445.84 KJ/Kg

Specific volume steam, ц = 0.086441 m³kg⁻¹

h₂ is h at P = 20 bar, 400°C => 3248.23 KJ/Kg

h₃ is h at P = 20 bar, 500°C => 3468.09 KJ/Kg

h₄ is hg at P = 0.6 bar from saturated water table => 2652.85 KJ/Kg

a)

Mass flow rate of the steam, m = G / ц

m = 1.5 / 0.086441

m = 17.35 kg/s

mass per hour is m = 62460 kg/hr

b)

Total Power produced by two stages

= m (h₁ - h₂) + m (h₃ - h₁)

= m [(3445.84 - 3248.23) + (3468.09 - 2652.85)]

= m [ 197.61 + 815.24 ]

= 17.35 [1012.85]

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c)

Rate of heat transfer to the steam through reheater

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= 17.35 x 219.86

= 3,814.57 kW

8 0
3 years ago
Find the current Lx in the figure
AleksandrR [38]

Explanation:

\frac{1}{8}  +  \frac{1}{2}   \\ 1.6 + 1.4 = 3 \\  \frac{1}{3}  +  \frac{1}{9}   \\ 2.25 + 2 = 4.25 \: ohm

R total = 4.25 ohm

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I total= 17/4.25= 4 A

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\frac{9}{9 + 3}  \times 4 = 3\\   \frac{2}{2 + 8} \times 3 = 0.6a \\  = 0.6 \: milli \: amper

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How many 10" diameter circles can be cut from a semicircular shape that has a 20"
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Answer:

  1

Explanation:

Only one such circle can be drawn. The diameter of the 10" circle will be a radius of the semicircle. In order for the 10" circle to be wholly contained, the flat side of the semicircle must be tangent to the 10" circle. There is only one position in the figure where that can happen. (see attached).

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