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3 years ago

1. Pump extracts energy from a flowing fluid. _______

Engineering

1 answer:

blagie [28]3 years ago

3 0

Answer:

FALSE
FALSE
FALSE
FALSE
FALSE

Explanation:

1) Pump does not extract energy from a flowing fluid instead is converts mechanical energy to fluid energy ( i.e. initial statement is FALSE )

2) Flow within a boundary layer cannot be considered an inviscid flow because flow within a boundary layer is considered a viscid flow ( i.e. initial statement is FALSE )

3) Laminar boundary layer velocity profile does not have more mixing than the velocity profile in a turbulent boundary layer because in a turbulent boundary layer the mixing of velocity profile is higher/more ( i.e. initial statement is FALSE )

4 ) Flow within the log-law region is considered Turbulent and not Laminar( i.e. initial statement is FALSE )

5) In a physical modeling of air flow over an aircraft the Reynold number is more important than the Froude number ( i.e. initial statement is FALSE )

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Terri Duncan recommends following up after an interview in which way?

Varvara68 [4.7K]

Answer:

make contact in a way which fits the industry and is comfortable to you

Explanation:

Terri Duncan was a person that helped people going for interviews by offering them professional advice on how to compose and conduct themselves during an interview.

Terri Duncan also helps people going for interviews but providing them with the aid they need concerning what to say during an interview.

Terri Duncan recommends following up after an interview by making contact in a way which fits the industry and is also comfortable to you.

3 0

3 years ago

Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street r

irakobra [83]

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area = 1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000 × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

4 0

3 years ago

A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos

Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

$T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }$

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\$

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\$