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3 years ago

What is the friction factor for fully developed flow in a circular pipe where Reynolds number is 1000

1 answer:

Pie3 years ago

7 0

Shear stress decreases along the flow direction. That is why the pressure drop is highest in the entrance region of a pipe, which increases the average friction factor for the whole pipe. ... In fully developed region the pressure gradient and the shear stress in flow are in balance.

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Who wants fight with me

vladimir1956 [14]

Yea, ‘Who wants to fight with me’

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3 years ago

Is a steam power plant a heat engine?

Novosadov [1.4K]

Answer:

Yes

Explanation:

Yes steam power plant is heat engine.

As we know that a heat engine takes from source heat and produce some amount of work and reject the heat to the sink.

So the steam power plant fulfill the all requirement like it take heat from boiler and produce some amount work and then reject the heat by using condenser.So we can say that steam power plant is an example of heat engine.

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4 years ago

A tensile test uses a test specimen that has a gage length of 50 mm and an area = 206 mm2. During the test, the specimen yields

vampirchik [111]

Answer:

The percent elongation in the length of the specimen is 42%

Explanation:

Given that:

The gage length of the original test specimen $L_o$ = 50 mm

The final gage length $L_f$ = 71 mm

The area = 206 mm²

maximum load = 162,699 N

To determine the percent elongation in %, we use the formula:

$\%EL = \dfrac{L_f-L_o}{L_o}\times 100$

$\%EL = \dfrac{71 \ mm-50 \ mm}{50 \ mm}\times 100$

$\%EL = \dfrac{21 mm}{50 \ mm}\times 100$

$\%EL = 0.42 \times 100$

$\mathbf{\%EL = 42 \%}$

The percent elongation in the length of the specimen is 42%

4 0

4 years ago

Asperities are microscopic peaks and valleys on

Rudiy27

Answer:

On a surface.

Explanation:

Asperities are microscopic peaks and valleys on a surface. This simply means that, when asperities contacts, surfaces also make contact. Asperities are the roughness or unevenness of a surface coming in contact during wear or friction.

8 0

3 years ago

Read 2 more answers

An uncovered swimming pool loses 1.0 inch of water off its 1,000 ft^2 surface each week due to evaporation. The heat of vaporiza

soldi70 [24.7K]

Answer:

The affirmation is true, the cover will be worth buying

Explanation:

The equation necessary to use is

E = m*cv,

Where

cv: the heat of vaporization.

Finding the rate at which the water evaporates (m^3/week).

The swimming pool loses water at 1 inch/week off its 1,000 ft^2

Than,

1000 ft² * 1 in/wk * 1 ft/12 in = 83.33 ft³/week

To obtains the rate of mass loss it is necessary to multiply it for the density of water

83.33 ft³/week * 62.4 lb/ft³ = 5200 lb/week

Knowing the vaporization heat it is possible to find the rate of heat which is leaving the swimming pool

5200 lb/week * 1050 BTU/lb = 5460000 btu/week

Over a 15-week period, the pool loses 81.9 million BTU.

Knowing the cost of energy to heat the pool is $10.00 per million btu

The price = $819

This way, the affirmation is true, the cover will be worth buying

3 0

3 years ago