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Mila [183]
3 years ago
7

A heat engine operating between energy reservoirs at 20?c and 600?c has 30% of the maximum possible efficiency.

Physics
1 answer:
Anarel [89]3 years ago
8 0
 <span>The maximum possible efficiency, i.e the efficiency of a Carnot engine , is give by the ratio of the absolute temperatures of hot and cold reservoir. 
η_max = 1 - (T_c/T_h) 

For this engine: 
η_max = 1 - [ (20 +273)K/(600 + 273)K ] = 0.66 = 66% 

The actual efficiency of the engine is 30%, i.e. 
η = 0.3 ∙ 0.664 = 0.20 = 20 % 

On the other hand thermal efficiency is defined as the ratio of work done to the amount of heat absorbed from hot reservoir: 
η = W/Q_h 

So the heat required from hot reservoir is: 
Q_h = W/η = 1000J / 0.20 = 5000J</span>
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Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it
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Answer:

Q = 165.95 cm³ / s,  1)    v = \sqrt{0.55^2 + 19.6 y},  2)  v = 2.05 m / s,

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1) the continuity equation is

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where Q is the flow rate, A is area and v is the velocity

         

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substituting

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let's reduce the magnitudes

       v = 0.55 m / s = 55 cm / s

let's calculate

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       Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

        v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

        v = \sqrt{0.55^2 + 2  \ 9.8\  y}

        v = \sqrt{0.55^2 + 19.6 y}

2) ask to calculate the velocity for y = 0.2 m

        v = \sqrt{0.55^2 + 19.6 \ 0.2}

        v = 2.05 m / s

3) We write the continuous equation for this point 2

        Q = v₂ A₂

        A₂ = Q / v₂

let us reduce to the same units of the SI system

        Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

        A₂ = 165.95 10⁻⁶ / 2.05

        A₂ = 80,759 10⁻⁶ m²

area is

        A₂ = π/4   d₂²

        d₂ = \sqrt{4  A_2 / \pi }

        d₂ = \sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }

        d₂ = 10.14 10⁻³ m

        d₂ = 1.014 cm

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