SAHARA<br>Jeep<br>WRANGLES<br>G

A hollow steel composite door with an 18-guage metal facing, hung on butt hinges with nonremovable pins, often with ventilation

Nadya [2.5K]

Answer:

Average industrial personnel door

Explanation:

Going by the description in which the door has ventilation louvers or glass panel, this shows that it can never be any type of security door. This is because with louvers or glass panel, it can be easier for any intruder to look or break into the house or space that is intended to be protected fully.

Hence, an hollow steel composite door with an 18-guage metal facing, hung on butt hinges with nonremovable pins, often with ventilation louvers or glass panels is likely an Average industrial personnel door

5 0

3 years ago

Which welding processes normally use a constant current power source? (Select all that apply) PLEASE HELP

Furkat [3]

Answer:

Explanation:

GTAW AND GMAW

5 0

3 years ago

If you touch a downed power line, covered or bare, what's the likely outcome?

olya-2409 [2.1K]

Answer:

you get electrocuted...........

5 0

3 years ago

Read 2 more answers

Which of the following are examples of engineering controls? Select all that apply.

Neporo4naja [7]

The examples of engineering controls is Biohazard waste containers and Spill clean up kits.

What is engineering controls?

An engineering controls is a workplace process that protect workers by removing hazardous conditions or by placing a barrier between the worker and the hazard.

An example of engineering controls is installation of exhaust ventilation to remove airborne emissions to shield the worker.

Hence, the examples of engineering controls is Biohazard waste containers and Spill clean up kits.

Therefore, the Option C and D is correct.

8 0

2 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$