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AveGali [126]
3 years ago
7

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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name the process by which mild steel can be converted into high carbon steel and explain it briefly ?​
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Answer:

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Explanation:

Mild steel can be converted into high carbons steel by which of the following heat treatment process? Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.

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Multimeter and the LCD is showing Hz. What's she measuring?
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An approach to a signalized intersection has a saturation flow rate of 1800 veh/h. At the beginning of an effective red, there a
Tcecarenko [31]

Answer: The total vehicle delay is

39sec/veh

Explanation: we shall define only the values that are important to this question, so that the solution will be very clear for your understanding.

Effective red time (r) = 25sec

Arrival rate (A) = 900veh/h = 0.25veh/sec

Departure rate (D) = 1800veh/h = 0.5veh/sec

STEP1: FIND THE TRAFFIC INTENSITY (p)

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p = 0.25 ÷ 0.5 = 0.5

STEP 2: FIND THE TOTAL VEHICLE DELAY AFTER ONE CYCLE

The total vehicle delay is how long it will take a vehicle to wait on the queue, before passing.

Dt = (A × r^2) ÷ 2(1 - p)

Dt = (0.25 × 25^2) ÷ 2(1 - 0.5)

Dt = 156.25 ÷ 4 = 39.0625

Therefore the total vehicle delay after one cycle is;

Dt = 39

4 0
3 years ago
Add the following vector given in rectangular form and illustrated the process graphically A = 16+j12, B= 6+j10.4
MariettaO [177]

Answer:

A=16+j12…'B=6+j10.4

Explanation:

add the following vector given in

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vekshin1

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D

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Masonry uses stone work, making a stone wall requires perfect masonry.

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