All categories

2 years ago

Disconnecting a circuit while in operation can create a voltage blank

Engineering

1 answer:

zlopas [31]2 years ago

5 0

Answer:

what is the question

Explanation:

confused

You might be interested in

A bathtub faucet has a maximum flow rate of 3 gal/min. The tub is rectangular (3 ft long x 2 ft wide x 1.5 ft deep). Although th

Anvisha [2.4K]

Answer:

Time taken = 136.32 minutes

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

3 0

3 years ago

A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 5 ft from the right edge.

Anon25 [30]

That is too hard but u got that cuz i believe in you!!!

3 0

2 years ago

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

koban [17]

Answer:

lol what

smmsdnndnsnsnnd

8 0

2 years ago

Savabuck University has installed standard pressure-operated flush valves on their water closets. When flushing, these valves de

Dvinal [7]

Answer:

Cost = $2527.2 per month.

Explanation:

Given that

Discharge ,Q = 130 L/min

$Q=0.13\ m^3/min$

Cost = $0.45 per cubic meter

1 month = 30 days

1 days = 24 hr = 24 x 60 min

1 month = 30 x 24 x 60 min

1 month = 43,200 min

Lets x $m^3\ water\ waste\ in\ a\ month$

x = 0.13 x 43,200

$x=5616\ m^3$

So the total cost = 5616 x 045 $

Cost = $2527.2 per month.

7 0

3 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat

ankoles [38]

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

$\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}$

Properties: The density of water $\delta = 1000 kg/m^3$

Conversions:

$165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\$

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

$e_{mech_{in}} - e_{mech_{out}} = gh - 0$

Then;

$gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}$

gh = 0.491 kJ/kg

$\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg$

= 1559 kW

Therefore; the overall efficiency is:

$\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}$

$= \dfrac{1166 \ kW}{1559 \ kW}$

= 0.75

= 75%

b) mechanical efficiency of the turbine:

$\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}$

thus;

$\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%$

6 0

2 years ago