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prisoha [69]
3 years ago
13

For problems 1 and 2, six luminaires, similar to Style E used in the Commercial Building, are to be installed in a room that is

12 ft × 18 ft (∼3.7 m × ∼5.5 m) with a 9 ft (2.8 m) floor-to-ceiling height. The spacing ratio for the luminaire is 1:0. The maximum distance that the luminaires can be separated and achieve uniform illuminance is
Engineering
1 answer:
BlackZzzverrR [31]3 years ago
6 0

Answer:

7ft (2.13 m).

Explanation:

From the question above, the following parameters/data are given; the room dimension = 12 ft × 18 ft (∼3.7 m × ∼5.5 m), the floor-to-ceiling height = 9 ft (2.8 m) and the spacing ratio for the luminaire is 1:0.

Note that, we are not given the value or data for the height of the plane, therefore, we will make an assumption that the height of the plane = 2ft.

Hence, the work plane to luminaire distance = (9 - 2)ft = 7ft (2.13 m).

So, the maximum distance that the luminaires can be separated and achieve uniform illuminance is;

= the work plane to luminaire distance × spacing ratio for the luminaire.

= 7ft (2.13 m) × 1 = 7ft (2.13 m).

Thus, the maximum distance that the luminaires can be separated and achieve uniform illuminance is 7ft (2.13 m).

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A heated long cylindrical rod is placed in a cross flow of air at 20°C (1 atm) with velocity of 10 m/s. The rod has a diameter o
postnew [5]

Answer:

Ts = 413.66 K

Explanation:

given data

temperature = 20°C

velocity = 10 m/s

diameter = 5 mm

surface emissivity = 0.95

surrounding temperature = 20°C

heat flux dissipated = 17000 W/m²

to find out

surface temperature

solution

we know that here properties of air at 70°C

k = 0.02881 W/m.K

v = 1.995 ×10^{-5} m²/s

Pr = 0.7177

we find here reynolds no for air flow that is

Re = \frac{\rho V D }{\mu } = \frac{VD}{v}    

Re = \frac{10*0.005}{1.99*10^{-5}}

Re = 2506

now we use churchill and bernstein relation for nusselt no

Nu = \frac{hD}{k} = 0.3 + \frac{0.62 Re6{0.5}Pr^{0.33}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}

h = \frac{0.02881}{0.005}0.3 + \frac{0.62*2506{0.5}0.7177^{0.33}}{[1+(0.4/0.7177)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}

h = 148.3 W/m².K

so

q conv = h∈(Ts- T∞ )

17000 = 148.3 ( 0.95) ( Ts - (20 + 273 ))

Ts = 413.66 K

5 0
4 years ago
Need answers for these please ​
Step2247 [10]

Answer:

Following are given the answers one-by-one with explanation:

(a) 8 records

As records are the horizontal rows of data and we have 8 of them.

(b) 5 fields

As fields are vertical columns of data and we have 5 of them.

(c)E3000

As by sorting Makers_names in ascending order the last one would be Rany and by sorting Processor_type we have two option with Rany that are B1500 and E3000. We can see that E3000 will come at the end.

(d) Computer_type

As only two choices (Laptop and PC )  are being used under the column computer_type so it can be amended to contain Boolean data.

(e) format check

This will check that the field should contain one alphabet (capital letter) followed by 4 digits each.

i hope it will help you!

7 0
3 years ago
Which of the following statements are true concerning AC circuit that contains both resistance and inductance? A. The current an
maw [93]

The current will lag the voltage in AC circuit that contains both resistance and inductance.

Answer: C

Explanation

There is no inductance only circuits in reality.

The circuits containing inductance has also a lower amount of resistance.

The current flows in both resistance and inductance.

There is a drop in the total voltage in resistance and inductance giving rise to the voltage applied in the coil when connected in a series.

An example being inductance coil an AC circuit connected to both resistance and inductance in series.

From the vector diagram, this conclusion can be drawn.

3 0
3 years ago
The moisture content in air (humidity) is measured by weight and expressed in pounds or ____________________.
VikaD [51]

Moisture content is measured in terms of pounds of water per pound of air (lb water/lb air) or grains of water per pound of air (gr. of water/lb air).

Hope this helps❤

3 0
2 years ago
An add tape of 101 ft is incorrectly recorded as 100 ft for a 200-ft distance. What is
baherus [9]

Answer:

the correct distance is 202 ft

Explanation:

The computation of the correct distance is shown below:

But before that correction to be applied should be determined

= (101 ft - 100 ft) ÷ (100 ft) × 200 ft

= 2 ft

Now the correct distance is

= 200 ft +  2 ft

= 202 ft

Hence, the correct distance is 202 ft

The same would be relevant and considered too

4 0
3 years ago
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