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Fudgin [204]
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.
Engineering
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 sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It
topjm [15]

Answer:

a) the moisture content before it was placed in the oven is 18.18%

b) degree of saturation for soil is 72.19%

Explanation:

Given the data in the question;

Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100

so we substitute

Moisture content = [(53.3 - 45.1) / 45.1 ] × 100

= (8.2/45.1) × 100

= 18.18%

Therefore the moisture content before it was placed in the oven is 18.18%

Dry Unit Weight = dry weight / volume

Dry Unit Weight = 45.1 lb / 0.45 ft³

Dry Unit Weight = 100.22 lb/ft³

we know that;

dry unit weight = (Specific gravity × unit weight of water) / (1 + e)

we also know that; unit weight of water is 62.43 lbf/ft³

so we substitute

e = (2.70×62.43 / 100.22) - 1

e = 1.68 - 1

e = 0.68

so void ratio e = 0.68

Now we determine the degree of saturation using the equation;

degree of saturation = (Moisture content × specific gravity) / void ratio

we substitute

degree of saturation = ( 18.18% × 2.7) / 0.68

= 0.49086 / 0.68

= 0.7219 ≈ 72.19%

Therefore degree of saturation for soil is 72.19%

7 0
3 years ago
Can somebody help me with that
skelet666 [1.2K]

Answer:

I think it's 23 ohms.

Explanation:

Not entirely sure about it.

hope this helps

4 0
3 years ago
Glyphicons is mainly used for​
-Dominant- [34]

Glyphicons are icon fonts which you can use in your web projects. Glyphicons Halflings are not free and require licensing, however their creator has made them available for Bootstrap projects free of cost.

7 0
2 years ago
Explain the LWD process why is it important in drilling operations?
barxatty [35]

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible.

Explanation:

pls mark brainliest

6 0
2 years ago
One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp
pentagon [3]

Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}

1x0.723x(T_{eq} -350)=3x0.780x(450-T_{eq} ) ⇒T_{eq} = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

V_{a,i}  = \frac{mRT_{a,i} }{P_{a,i} }=  \frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar} = 0.201m^{3}

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }

P_{eq}= 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

                             (0.201+1.275)

= 246.67 KPa = 2.47 bar

6 0
3 years ago
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