Problem definition

A step-down transformer (turns ratio = 1:7) is used with an electric train to reduce the voltage from the wall receptacle to a v

SOVA2 [1]

Answer:

wait lemme check it out

Explanation:

6 0

3 years ago

Think about how could you design, build, and test a light maze. What specific behavior of light will be essential to the success

Mrac [35]

Answer:

Reflection

Explanation:

The specific behavior of light that will be essential to ensure the success of your design is "Reflection". This is because light maze makes use of a mirror and it's the light that is reflected that we see with our eyes. Also, the manner in which light is reflected off objects will affect the colors that are reflected as well.

4 0

3 years ago

A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per millio

Mama L [17]

Answer:

a) 570 kWh of electricity will be saved

b) the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

$P_{incandescent}$ = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find the total power consumption by the fluorescent lamp

$P_{fluorescent}$ = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

$P_{saved}$ = $P_{incandescent}$ - $P_{fluorescent}$

$P_{saved}$ = 750 - 180

$P_{saved}$ = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved = 570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved = 570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ / million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

3 0

3 years ago

Select the correct answer.

Elodia [21]

I think balance

Can I get Brainlyist

3 0

3 years ago

The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate

Sonja [21]

Answer:

a) Mechanical efficiency ( $\varepsilon$ )=63.15% b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, $Wmec=0.95*6kW=5.7 kW$ .

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump $\Delta P$ . So $P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W$ .

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is $\varepsilon=3.6kW/5.7kW=0.6315=63.15\%$

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and $\rho$ the density of the water:

$[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T$ [/tex]

Finally, the temperature rise is:

$\Delta T=2100/75348 \ºC=0.028 \ºC$

7 0

3 years ago