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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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Fill in the empty function so that it returns the sum of all the divisors of a number, without including it. A divisor is a numb
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Answer:

// Program is written in C++

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#include<bits/stdc++.h>

#include<iostream>

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3 years ago
What line separates two lanes traveling in the same direction
soldier1979 [14.2K]

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2 years ago
Read 2 more answers
A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De
andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0
3 years ago
As the asteroid falls closer to the Earth's surface its _______ energy decreases and its _______ energy increases.
Levart [38]

Answer:

As the asteroid falls closer to the Earth's surface its <u>Gravitational</u> <u>Potential</u> energy <em>decreases</em> and its <u>Kinetic</u> energy <em>increases</em>.

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Answer:

Explanation:

B. you would grab the plug closest to the outlet

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