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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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3 years ago
A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w
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Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

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A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%  

Δd = 5%

B) would the wood swell or shrink

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C) The new diameter of the wood

D2 = D + D( \frac{M1}{100} )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

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Determine the general lighting load for a two-story office building that measures 125 feet by 150 feet.
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The general lighting load for a two-story office building that measures 125 feet by 150 feet is 112, 500 sq ft.

<h3>What is lighting load?</h3>

Lighting loads are the energy used to power electric lights and they make up nearly a third of US commercial building energy use.

Lighting load = n(LW)

where;

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  • W is width of the building
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For one story building, = 3

For two story building, n = 6

Lighting load = 6 x 125 x 150 = 112, 500 sq ft.

Learn more about lighting load here: brainly.com/question/14070748

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