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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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According to the video, what are some of the effects of the increase in the use of automated tools for Loading Machine Operators

Rudik [331]

sixteen is the answer to the question

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4 years ago

Read 2 more answers

There is a proposal in Brooklyn to construct a new mid-rise apartment building on a vacant lot at the intersection of Avenue A a

Soloha48 [4]

Answer:

a. Park should be adjacent to 48th Street, b. Difference in noise level = 707dBa, c. Yes

Explanation:

Data given for Avenue A

Cars = 496, Medium Truck = 52, Heavy Truck = 19, Buses = 10

Data given for 48th Street

Cars = 822, Medium Truck = 22, Heavy Truck = 8, Buses = 3

Consider the PCEs to be Cars = 1, Medium Truck = 13, Heavy Truck = 47, Buses = 18

a. Noise Level = Number of vehicles x PCE

For Avenue A

Noise level = (496 x 1) + (52 x 13) + (19 x 47) + (10 x 18) = 2245dBa

For 48th Street

Noise level = (822 x 1) + (22 x 13 + (8 x 47) + (3 x 18) = 1538dBa

The park should adjacent to 48th street as it is quieter than Avenue A

b. Let the Setback be 50ft. We know that the reduction of noise for 100ft = 5-8 dBa, hence

For Avenue A Noise Reduction due to 50 ft = (8/100) x 50 = 4dBa

Noise at Setback distance = 2245 - 4 = 2241dBa

Considering the same setback the noise at 48th street would be = 1538 - 4 = 1534 dBa

The difference is noise level between the two sides would be = 2241 - 1534 = 707 dBa

c. Yes the developer concerns are valid because there is a clear difference in noise levels of the two sites. This can be seen even after setting the same Setback. Locating the park next to Avenue A will cause serious noise problems.

8 0

4 years ago

Linus is using a calculator to multiply 5,426 and 30. He enters 5,426 x 300 by mistake. What can Linus do to correct his mistake

Alexxx [7]

Answer:

<em>Linus needs to take one of the zero out and it should be 30 instead 300.</em>

Explanation:

It is because Linus put three zero instead two zero.

4 0

3 years ago

Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

$\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%$

Where:

$V_{gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{fe}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

$\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%$

Where:

$m_{fe}$ , $m_{gr}$ - Masses of the ferrite and graphite phases, measured in grams.

$\rho_{fe}, \rho_{gr}$ - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

$\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%$

$\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%$

If $\rho_{gr} = 2.3\,\frac{g}{cm^{3}}$ , $\rho_{fe} = 7.9\,\frac{g}{cm^{3}}$ , $m_{gr} = 3.2\,g$ and $m_{fe} = 96.8\,g$ , the volume percentage of graphite is:

$\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%$

$\%V_{gr} = 10.197\,\%V$

The volume percentage of graphite is 10.197 per cent.

5 0

3 years ago

Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c

Nitella [24]

Answer:

(a) The volume rate of flow per meter width = 5.6*10⁻³ m²/s

(b) The shear stress acting on the bottom plate = 157.5 N/m²

Explanation:

(a)

Calculating the distance of plate from centre line using the formula;

h = d/2

where h = distance of plate

d = diameter of flow = 9 mm

Substituting, we have;

h = 9/2

= 4.5 mm = 4.5*10^-3 m

Calculating the volume flow rate using the formula;

Q = (2h³/3μ)* (Δp/L)

Where;

Q = volume flow rate

h = distance of plate = 4.5*10^-3 m

μ = dynamic viscosity = 0.38 N.s/m²

(Δp/L) = Pressure drop per unit length = 35 kPa/m = 35000 Pa

Substituting into the equation, we have;

Q = (2*0.0045³/3*0.38) *(35000)

= (1.8225*10⁻⁷/1.14) * (35000)

= 1.60*10⁻⁷ * 35000

= 5.6*10⁻³ m²/s

Therefore, the volume flow rate = 5.6*10⁻³ m³/s

(b) Calculating the shear stress acting at the bottom plate using the formula;

τ = h*(Δp/L)

= 0.0045* 35000

= 157.5 N/m²

u(max) = h²/2μ)* (Δp/L)

= (0.0045²/2*0.38) * 35000

=2.664*10⁻⁵ *35000

= 0.93 m/s

7 0

4 years ago