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ZanzabumX [31]
3 years ago
7

What are the four processes of the Carnot cycle? Sketch the Carnot cycle (a) on T-s (temperature - entropy) and P-V (pressure -

volume) diagrams and calculate the thermal efficiency of the Carnot cycle if it operates between 800°C and 15 °C

Engineering
1 answer:
Tanya [424]3 years ago
5 0

Answer:

\eta = 0.7315

Explanation:

Carnot cycle

Carnot cycle have four process

1. Iso thermal expansion

2.Reversible adiabatic expansion

3. Iso thermal compression

4.Reversible adiabatic compression

 This is ideal cycle for all work producing devices.All devices have efficiency less than Carnot cycle. Because in Carnot cycle all process is reversible process.

Efficiency of Carnot cycle given as

\eta =1-\dfrac{T_1}{T_2}

But temperature must be in Kelvin

So

\eta =1-\dfrac{288}{1073}

\eta = 0.7315

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The cult of personality that surrounded Joseph Stalin in the Soviet Union led soviet citizens to believe that there was undisput
evablogger [386]

Answer:

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Explanation:

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3 years ago
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fiasKO [112]

Answer:

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4) To create and develop the plan's framework:

To set the objective for high level organizational goals

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Explanation:

8 0
3 years ago
Comparison of density values determines whether an item will float or sink in water. For each of the values below, determine the
geniusboy [140]

Answer:

a) the object floats

b) the object floats

c) the object sinks

Explanation:

when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water

a)

volumen for a cube

V=L^3

L=1.53in=0.0388m

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density=m/v

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b)

m=111mg=0.111g

density=m/v

density=0.111g/0.296ml=0.375g/ml

the metal paperclip floats   because it is less dense than water

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4 0
3 years ago
I'll mark brainliest plz help
Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

7 0
3 years ago
The observations of 124.53, 124.55, 142.51, and 124.52 are obtained when taping the length of a line. What should the observer c
Nesterboy [21]

Answer:

attached below

Explanation:

8 0
3 years ago
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