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3 years ago

6

One AA battery in a flashlight stores 9400 J. The three LED flashlight bulbs consume 0.5 W. How many hours will the flashlight l

ast?

1 answer:

Effectus [21]3 years ago

7 0

Answer:

time = 5.22 hr

Explanation:

Given data:

Energy of battery = 9400 J

Power consumed by three led bulb is 0.5 watt

we know Power is give as

$Power = \frac{ energy}{time}$

plugging all value and solve for time

$time = \frac{energy}{power}$

$time = \frac{9400}{0.5}$

time = 18,800 sec

in hour

1 hour = 3600 sec

therefore in 18,800 sec

$time = \frac{18800}{3600} = 5.22 hr$

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Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per

kirill [66]

Solution :

Given :

$V_1 = 7 \ m/s$

Operation time, $T_1$ = 3000 hours per year

$V_2 = 10 \ m/s$

Operation time, $T_2$ = 2000 hours per year

The density, ρ = $1.25 \ kg/m^3$

The wind blows steadily. So, the K.E. = $(0.5 \dot{m} V^2)$

$= \dot{m} \times 0.5 V^2$

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = $\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$

Regarding that $\dot m \propto V$ . Then,

Power $\propto V^3$ → Power = constant x $V^3$

Since, $\rho_a$ is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, $P_1= \text{const.} \times V_1^3$

$P_1 = \text{const.} \times 343 \ W$

For the second site,

Power, $P_2 = \text{const.} \times V_2^3 \ W$

$P_2 = \text{const.} \times 1000 \ W$

5 0

3 years ago

An induced-draft cooling tower cools 90,000 gallons per minute of water from 84 to 68oF. Air at 14.61 psia, 70oF dry bulb and 60

belka [17]

Answer:

a. V = 109.64 × 10⁵ ft/min

b. Mw = 654519.54 kg/hr

Explanation:

Given Parameters

mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s

inlet temperature of water, T1 = 84 F = 28.89 C

outlet temperature of water, T2 = 68 F = 20 C

specific heat capacity of water, c = 4.18kJ/kgK

rate of heat remover from water, Qw is given by

Qw = 6607.33[28.89 - 20] * 4.18

Qw = 245529.545kw

For air, inlet condition

DBT = 70 F hi = 43.43 kJ/kg

WBT = 60 F wi = 0.00874 kJ/kg

u1 = 0.8445 m/kg

oulet condition,

DBT = 70 F RH = 100.1

h1 = 83.504kJ/kg

Wo = 0.222kJ/kg

check the attached file for complete solution

3 0

3 years ago

Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure

iris [78.8K]

Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3 = √9 = 3

P5/P6 = P7/P8 = √9 =3

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg

Calculate the net work done

Wnet = (Wturb)out - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

4 0

3 years ago

HOW TO CALCULATE MARGINAL RATE

Basile [38]

Answer:

Divide the difference in tax by the amount of income from the investment, and you'll get the economic marginal tax rate from investing. Most people refer to marginal tax rates as being identical to tax brackets.

hope this helps

have a good day :)

Explanation:

8 0

2 years ago

Which explanation best identifies why the company in the following scenario has decided against investing in solar energy?

pickupchik [31]

Answer:

The company found the cost of the required photovoltaic cells too expensive.

Explanation:

Solar energy can be used as an alternative source of supply for fuel. Solar energy is a renewable source of energy, that is it keeps on replenishing every day. Also solar energy does not require a lot of maintenance.

The cost required is starting a solar system is very high because one needs to buy solar panel, photovoltaic cells for batteries, inverters and so on.

From the question, the company decided against solar energy for the time being. This means that probably in the future they might consider it. Therefore it is as a result of the economic situation of the company that they have not set up a solar system because the cost of the required photovoltaic cells too expensive.

8 0

3 years ago