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larisa [96]
2 years ago
8

We have a tube with a diameter of 5 inches that is 1 foot long. The tube then reduces the diameter to 3 inches. According to the

Venturi effect, what will happen to the velocity of the fluid flowing through this section? What will happen to the pressure?
Engineering
2 answers:
Ksju [112]2 years ago
6 0
Can I get a picture? Of the question so I can understand it more?
NemiM [27]2 years ago
5 0
Can we get a Picture?
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What do auto mechanics do when they take out the spark plugs
ivanzaharov [21]
Check them to see if the cylinder is running rich or lean. This can be determined by looking at the electrodes on the plugs
Hope this helps.
7 0
3 years ago
A wind turbine rotates at 20 rev/min. If its power output is 2.0 MW, the torque produced on the turbine from the wind is
Hoochie [10]

Answer:

The torque torque produced on the turbine from the wind is approximately 955 kN·m

Explanation:

The number of revolution per minute of the turbine = 20 rev/min

The power output of the turbine = 2.0 MW

The power transmitted to a shaft equation is given as follows;

Power , \ P = \dfrac{2 \cdot \pi \cdot N \cdot T}{60}

Where;

P = The power transmitted to a turbine shaft = 2.0 MW

N = The number or revolutions per minute = 20 rev/min

T = The torque produced on the turbine by the wind

Therefore;

Torque , \ T = \dfrac{P  \cdot 60}{2 \cdot \pi \cdot N } = \dfrac{2.0 \times 60}{2 \times \pi \times 20}  = \dfrac{3}{\pi } \ MN \cdot m

The torque torque produced on the turbine from the wind = 3/π MN·m ≈ 0.955 MN·m = 955 kN·m.

4 0
3 years ago
Can anyone h.e.l.p me in Edge . nuity for Architecture
kakasveta [241]

Answer:

yeah

Explanation:

6 0
3 years ago
Calculate the inner area (cross sectional) of a pipe with an outer diameter of 300mm and a
Sunny_sXe [5.5K]

Answer:

46.3 cm²

216 m

Explanation:

The cross sectional area is the area of the outer circle minus the area of the inner circle.

A = πR² − πr²

R = 300/2 = 150 mm = 15 cm

r = 150 − 5 = 145 mm = 14.5 cm

A = π (15 cm)² − π (14.5 cm)²

A ≈ 46.3 cm²

Volume is area times length:

V = AL

10⁶ cm³ = (46.3 cm²) L

L ≈ 21,580 cm

L ≈ 216 m

6 0
3 years ago
An insulated rigid tank of volume 8 m3 initially contains air at 600 kPa and 400 K. A valve connected to the tank is opened, and
Natali5045456 [20]

Answer:

The work done is 3654.83 J/mol

Explanation:

The process is an isothermal process since the air temperature is kept constant during the release.

W = RT ln(P1/P2)

R is gas constant = 8.314 J/mol.K

T is air temperature = 400 K

P1 is initial pressure of air = 600 kPa

P2 is final temperature of air = 200 kPa

W = 8.314×400 ln(600/200) = 3325.6×ln 3 = 3325.6×1.099 = 3654.83 J/mol

6 0
3 years ago
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