Question 7 of 10

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

BlackZzzverrR [31]

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).

6 0

3 years ago

What does it mean when the service engine light is on?

IRINA_888 [86]

I’m pretty sure the answer is the engine needs to be checked right away

7 0

4 years ago

25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an

Gnesinka [82]

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

$N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)$

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is $0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}$ .

Total pressure P = 1 atm = 101.325 KPa

$P_{A_{1} }$ = 23.76 mm Hg

$P_{A_{1} }$ = $\frac{23.76}{760}$

$P_{A_{1} }$ = 0.03126 atm

$P_{A_{1} }$ = 3.167 K Pa

When air surrounded is dry air, then $PA_{2}$ = 0 mm Hg

R = 8.314 $\frac{K P a \cdot m^{3}}{mol \cdot k}$

$P_{B/M}$ = $\frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}$

$P_{B_{1}}=P_{T}-P_{A1_{}}$

= 101.325 - 3.167

$P_{B_{1} }$ = 98.158 K Pa

$P_{B_{2}}=P_{T}-P_{A_{2}}$

$P_{B_{2} }$ = 101.325 - 0

$P_{B_{2} }$ = 101.325 K Pa

$P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}$

$P_{B/M}$ = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

$N_{A}$ = $\frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }$

$N_{A}$ = 0.11077 × $10^{-6}$ mol/ $m^{2}$ .s

$N_{A}$ = 0.11077 × $10^{-6}$ × 18 × (60×60×24)

$N_{A}$ = 0.1723 $lb_{m}$ / $m^{2}$ .day

$\tilde{N}_{A}=N_{A} \times Area$

Area of individual pipe is

$A=\frac{\pi}{4}(0.0254)^{2}$

A = 0.00051 $m^{2}$

$\bar{N}_{\boldsymbol{A}}$ = 0.1723 × 0.00051

$\bar{N}_{\boldsymbol{A}}$ = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

5 0

3 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

I am sending an email to my teacher. Is this mature enough?

nirvana33 [79]

I dislike how you wrote the 0’s. But honestly Teachers don’t really care as long as they can read it. So sure send it I guess

7 0

3 years ago

Read 2 more answers