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2 years ago

Which of the following procedures best applies to assessing the embodied energy of a building?

Engineering

1 answer:

galina1969 [7]2 years ago

5 0

analyzing building materials???????????? but i can try i think it is analyzing the materials of the building

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A fan draws air from the atmosphere through a 0.30-mdiameter round duct that has a smoothly rounded entrance. A differential man

marusya05 [52]

Answer:

V = 50 ft³/s

H.P = 0.466

Explanation:

Given that

Diameter of the duct, D = 0.3 m

Vacuum Pressure of the duct, Z = 0.025 m

P(w) = pressure of water, 1000 kg/m³

P(a) = pressure of air, 1.22 kg/m³

To find the pressure change we use the formula

ΔP = P(w) * g * Z

ΔP = 1000 * 9.8 * 0.025

ΔP = 245 Pa.

We need the area, do we find that too

A = πd²/4

A = π * 0.3² * 1/4

A = 0.071 m²

Recall the energy equation to be

1/2v² = ΔP/p(a) , so that if we rearrange, we have

v² = 2ΔP/p(a)

v = √(2ΔP/p(a)), on substituting the values, we have

v = √(2 * 245)/1.22

v = √490/1.22

v = √401.64

v = 20.04

The volume flow rate is then equal to

Velocity * Area.

V = 20.04 * 0.071

V = 1.42 m³/s, converting to ft³/s, we have 50 ft³/s

Horsepower output is gotten using

P = ΔP * V

P = 245 * 1.42

P = 347.9 w, converting this to HP, we have 0.466 HP

4 0

3 years ago

To ensure safe footing on penetrable surfaces,use?

Masteriza [31]

D. Spur plates

Hope this helps

7 0

3 years ago

How do we need to prepare for the future?

Paraphin [41]

B. to achieve sustainable development

5 0

3 years ago

g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What

earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity ( $v$ ), measured in meters per second, is determined by the following expression:

$v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}$ (1)

Where:

$\dot V$ - Flow rate, measured in cubic meters per second.

$D$ - Diameter, measured in meters.

If we know that $\dot V = 0.01\,\frac{m^{3}}{s}$ and $D = 0.05\,m$ , then the flow velocity is:

$v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}$

$v \approx 5.093\,\frac{m}{s}$

The density and dinamic viscosity of the glycerin at 20 ºC are $\rho = 1260\,\frac{kg}{m^{3}}$ and $\mu = 1.5\,\frac{kg}{m\cdot s}$ , then the Reynolds number ( $Re$ ), dimensionless, which is used to define the flow regime of the fluid, is used:

$Re = \frac{\rho\cdot v \cdot D}{\mu}$ (2)

If we know that $\rho = 1260\,\frac{kg}{m^{3}}$ , $\mu = 1.519\,\frac{kg}{m\cdot s}$ , $v \approx 5.093\,\frac{m}{s}$ and $D = 0.05\,m$ , then the Reynolds number is:

$Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }$

$Re = 211.230$

A pipeline is in turbulent flow when $Re > 4000$ , otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ( $f$ ), dimensionless, is determined by the following expression: