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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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A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
2 years ago
Sketch T-s and p-v diagrams for the Diesel cycle.
labwork [276]

Answer:

Diesel cycle:

    All diesel engine works on diesel cycle.It have four processes .These four processes are as follows

1-2.Reversible adiabatic compression

2-3.Heat addition at constant pressure

3-4.Reversible adiabatic expansion

4-1.Heat addition at constant volume

When air inters in the piston cylinder after that it compresses and gets heated due to compression after that heat addition take place at constant pressure after that power is produces when piston moves to bottom dead center.

From the diagram of P-v And T-s we can understand so easily.

3 0
3 years ago
You work in a furniture store. You receive a
spin [16.1K]

Answer:

18 pieces of furniture

Explanation:

Since you receive $120.93 per furniture piece and a the month's commission is $2,176.74 you divide the commission by the furniture price.

  • 2176.74/120.93
3 0
3 years ago
Read 2 more answers
A rigid tank contains 3 kg of water initially at 43.97% quality and at a temperature of 120°C. The water is heated until it reac
makkiz [27]

Explanation: see attachment below

6 0
3 years ago
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core tem
Llana [10]

Answer:

The average thickness of the blubber is<u> 0.077 m</u>

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

7 0
3 years ago
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