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OleMash [197]
3 years ago
15

Consider a simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater. The two cycles are

very much alike, except the feedwater in the regenerative cycle is heated by extracting some steam just before it enters the turbine. Hence, the simple ideal Rankine cycle is more efficient than the ideal regenerative Rankine cycle. How would you compare the efficiencies of of these two cycles?
Engineering
1 answer:
borishaifa [10]3 years ago
4 0

Answer:

They both have the same efficiency.

Explanation:

The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.

Therefore in comparison, the efficiency is the same for both.

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As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth
telo118 [61]

Answer:

165 mm

Explanation:

The mass on the piston will apply a pressure on the oil. This is:

p = f / A

The force is the weight of the mass

f = m * a

Where a in the acceleration of gravity

A is the area of the piston

A = π/4 * D1^2

Then:

p = m * a / (π/4 * D1^2)

The height the oil will raise is the heignt of a colum that would create that same pressure at its base:

p = f / A

The weight of the column is:

f = m * a

The mass of the column is its volume multiplied by its specific gravity

m  = V * S

The volume is the base are by the height

V = A * h

Then:

p = A * h * S * a / A

We cancel the areas:

p = h * S * a

Now we equate the pressures form the piston and the pil column:

m * a / (π/4 * D1^2) = h * S * a

We simplify the acceleration of gravity

m / (π/4 * D1^2) = h * S

Rearranging:

h = m / (π/4 * D1^2 * S)

Now, h is the heigth above the interface between the piston and the oil, this is at h1 = 42 mm. The total height is

h2 = h + h1

h2 = h1 + m / (π/4 * D1^2 * S)

h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm

7 0
3 years ago
1) Each of the following would be considered company-confidential except
AysviL [449]
Answer is your company’s address
8 0
3 years ago
A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar
jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading  114.06 m^3/d/m

Explanation:

calculation for single clarifier

sewag\  flow Q = \frac{12900}{2} = 6450 m^2/d

surface\  area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2

surface area = 314.16 m^2

volume of tankV  = A\times side\ water\ depth

                             =314.16\times 2 = 628.32m^3

Length\ of\  weir = \pi \times diameter of weir

                       = \pi \times 18 = 56.549 m

overflow rate =v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2

Detention timet_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr

weir loading= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m

6 0
3 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
GenaCL600 [577]

Answer:

62.14\ \text{miles}

6213727.37\ \text{miles}

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = 10^5\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}

1\ \text{mile}=1609.34\ \text{m}

\dfrac{10^5}{1609.34}=62.14\ \text{miles}

The chain would extend 62.14\ \text{miles}

Dislocation density = 10^{10}\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}

\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}

The chain would extend 6213727.37\ \text{miles}

3 0
3 years ago
Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u
Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × 10^{-2} N/m

so average radius = \frac{diameter}{2} =  100 µm = 100 ×10^{-6} m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r²    .....................1

here r is radius and  Δp pressure difference and σ surface tension

Δp = \frac{2 \sigma }{r}    

put here value

Δp = \frac{2*2.69*10^{-2}}{100*10^{-6}}  

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0
3 years ago
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