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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, B and C

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A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L
zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

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4 years ago
If you are setting up a race car. What is the cross weight? Does it matter?
lara31 [8.8K]

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4 years ago
An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va
Ksju [112]

Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}

State 2:

\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}

State 3:

\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}

State 4:

Throttling process  h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}

(a)

Magnitude of compressor power input

\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}

w_{c}=4 \cdot 013 \mathrm{kw}

(b)

Refrigerator capacity

Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}

Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega

\ Q_{in} =6 \cdot 583 \text { tons }

(c)

Cop:

\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}

\beta=5 \cdot 758

3 0
3 years ago
A car accelerates from O to 60. miles per hour in 5.2 seconds. Calculate acceleration in m/s seconds. Calculate acceleration in
DochEvi [55]

Answer:

a=5.515\frac{m}{s^{2} }

Explanation:

The first thing we will do is convert the units. Miles per hour to meters per second.

1 mile=1609.34 mts.

1 hora=3600 segundos

Performing the operations

60\frac{mile}{h}=\frac{(60*1609.34)}{3600}\frac{m}{s}=26.822\frac{m}{s}

Now, we will use the acceleration formula

a=\frac{v}{t}

Where v = speed and t = time

Substituting the values ​​of t=5.2s

a=\frac{v}{t} =\frac{26.822\frac{m}{s} }{5.2s} =5.15\frac{m}{s^{2} }

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