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ch4aika [34]
4 years ago
12

The air standard efficiency ofan Otto cycle compared to diesel cycle for thie given compression ratio is: (a) same (b) less (c)

more () more (d) unpredictable (d) e) more or less depending on power rating
Engineering
1 answer:
Svetach [21]4 years ago
5 0

Answer:

Correct sentence: (c) more.

Explanation:

Otto cycle:

An ideal Otto cycle models the behavior of an explosion engine. This cycle consists of six steps, as indicated in the figure. Prove that the performance of this cycle is given by the expression

η = 1 - (1/r^(ρ-1))

where r = VA / VB is the compression ratio equal to the ratio between the volume at the beginning of the compression cycle and at the end of it. ρ=1.4

Diesel cycle:

The ideal Diesel cycle is distinguished from the ideal Otto in the combustion phase, which in the Otto cycle is assumed at a constant volume and in the Diesel at constant pressure. Therefore the performance is different.

If we write the performance of a Diesel cycle in the form

:

η = 1 - (1/r^(ρ-1)) × ( (r^ρ)-1)/(ρ^(r-1)) )

we see that the efficiency of a Diesel cycle differs from that of an Otto cycle by the factor in parentheses. This factor is always greater than the unit, therefore, for the same compression reasons r

                   diesel performance is less than otto performance

\ eta_ \ mathrm {Diesel} <\ eta_ \ mathrm {Otto} \,

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Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

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v₂ = 9.43 m/s

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