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3 years ago

What limits the practical realization of higher efficiencies in the Otto cycle?

Engineering

1 answer:

AlekseyPX3 years ago

5 0

Answer:

We know that all petrol engines are works on Otto cycle.Otto cycle have four process out of four two are constant volume process and others two are isentropic processes.

There are lots of limitations for practical Otto cycle these are as follows

1.In practical cycle heat can not add at constant volume.

2.In practical cycle there is a gap between combustion of two set of fuel.

3.Lots of heat is dissipated by cylinders.

4.Valve opening and closing is not a sudden process it requires some time.

5.There is a limitations of cylinder material ,it means that temperature of cycle can not rise after a specified limit of material.

Due to these above limitations practical cycles have low efficiency as compare to ideal cycle.

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3 years ago

A surveyor knows an elevation at Catch Basin to be elev=2156.77 ft. The surveyor takes a BS=2.67 ft on a rod at BM Catch Basin a

fenix001 [56]

Answer:

the elevation at point X is 2152.72 ft

Explanation:

given data

elev = 2156.77 ft

BS = 2.67 ft

FS = 6.72 ft

solution

first we get here height of instrument that is

H.I = elev + BS ..............1

put here value

H.I = 2156.77 ft + 2.67 ft

H.I = 2159.44 ft

and

Elevation at point (x) will be

point (x) = H.I - FS .............2

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3 years ago

What is the thermal efficiency of this regeneration cycle in terms of enthalpies and fractions of total flow?

irga5000 [103]

Answer:

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Explanation:

generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency increase.

Now from T-S diagram

$W_{net}=W_{out}-W_{in}$

$W_{net}=(h_3-h_4)-(h_2-h_1)$

$Q_{in}=h_3-h_5$

Due to generation $(h_5-h_2)$ amount of energy has been saved.

$Q_{generation}=Q_{saved}$

So efficiency of cycle $\eta =\frac{W_{net}}{Q_{in}}$

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Effectiveness of re-generator

$\varepsilon =\dfrac{(h_5-h_2)}{(h_4-h_2)}$

So the efficiency of regenerative cycle

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

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nekit [7.7K]

Answer:

The correct option is;

A. proper protection

Explanation:

Motorcycle riders ride the motorcycle while at some level of speed while having the entire body exposed to be a major part of any collision.

Injuries sustained from motorcycle accidents are several times more severe than injuries sustained by occupants of a car that is fully protected by the metallic panel in the same and even more serious accident scenarios

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