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3 years ago

6

A 450 MWt combined cycle plant has a Brayton cycle efficiency of 24% and a Rankine cycle efficiency of 29% with no heat augmenta

tion. Determine the net electric output from the plant.

1 answer:

Romashka-Z-Leto [24]3 years ago

4 0

Answer:

Output of plant=207.18 MW

Explanation:

Given input energy Q=450 MW

Brayton cycle efficiency $\eta_1=24$ %

Rankine cycle efficiency $\eta_2=29$ %

Combined efficiency for two cycle given as follow

$\eta _b=\eta _1+\eta _2-\eta _1\eta _2$

$\eta _b=0.24+0.29-0.24\times 0.29$

$\eta_b=0.4604$

We know that efficiency $\eta =\dfrac{out\ put}{in\ put}$

out put of plant= $450\times 0.4604$

So output of plant=207.18 MW

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A steam power plant with a power output of 230 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,0

NARA [144]

Answer:

$\eta =46\%$

Explanation:

Hello!

In this case, we compute the heat output from coal, given its heating value and the mass flow:

$Q_H=60\frac{tons}{h}*\frac{1000kg}{1ton}*\frac{1h}{3600s}*\frac{30,000kJ}{kg}\\\\Q_H=500,000\frac{kJ}{s}*\frac{1MJ}{1000J} =500MW$

Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:

$\eta =\frac{230MW}{500MW}*100\% \\\\\eta =46\%$

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3 years ago

10. Which of these requires a wheel alignment after replacement?

Elden [556K]

C. Both; require a wheel alignment after replacement

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3 years ago

Read 2 more answers

The tropics receive more heat from the sun than is radiated away from the tropics, and polar regions radiate more than they rece

IRINA_888 [86]

Answer:

It is a well known fact that the earth rotates around the sun in an inclined axis which is approximately 23 degree. The inclined nature of earth axis causes variation in the solar heat received at any place on the earth surface. The hemisphere facing the sun due to this axial tilt, gets higher sun energy as compared to the opposite side. The hemisphere which faces the sun will experience summer whereas the hemisphere away from sun will experience winter.

In each of the hemisphere the polar areas will receive higher radiation and longer daytime during the summer season. However it has been observed that there is difference in radiation received at different areas of earth surface and radiated. The tropical areas have lower reflectance and thus a large part of incoming solar radiation have been absorbed along the tropics. The poles though have longer daytime during summer and hence greater solar radiation but due to high reflectance radiate more energy. Thus the tropical areas have surplus energy as compared to deficit energy areas of poles. This difference in energy creates a heat imbalance.

This net heat difference between poles and equator gives rise to a global circulation system leading to flow of heat from the net energy excess areas to deficit areas. This circulation takes place through atmosphere as well as oceans and different process of climate viz. evaporation, transpiration, rainfall, wind, convection, oceanic circulations etc work as tools of this system

4 0

2 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

8 0

3 years ago

What is the resistance of a resistor if the current flowing through it is 3mA and the voltage across it is 5.3V?

Flura [38]

Answer: 1766.667 Ω = 1.767kΩ

Explanation:

V=iR

where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).

3mA = 0.003 A

Rearranging the equation, we get

R=V/i

Now we are solving for resistance. Plug in 0.003 A and 5.3 V.

R = 5.3 / 0.003

= 1766.6667 Ω

= 1.7666667 kΩ

The 6s are repeating so round off to whichever value you need for exactness.

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1 year ago