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Over [174]
2 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]2 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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Imagine the arc of a football as it flies through the air. How does this motion illustrate classical mechanics?
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Name 3 ways in which robots have improved since the Ebola outbreak.
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2 years ago
simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
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Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

6 0
3 years ago
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