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2 years ago

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

Engineering

1 answer:

AlekseyPX2 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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2 years ago

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Lerok [7]

Answer:

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Explanation:

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2 years ago

Steam enters an adiabatic turbine at 8 MPa and 500C with a mass flow rate of 3

Vinil7 [7]

Answer:

a)temperature=69.1C

b)3054Kw

Explanation:

Hello!

To solve this problem follow the steps below, the complete procedure is in the attached image

1. draw a complete outline of the problem

2. to find the temperature at the turbine exit use termodinamic tables to find the saturation temperature at 30kPa

note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

3. Using thermodynamic tables find the enthalpy and entropy at the turbine inlet, then find the ideal enthalpy using the entropy of state 1 and the outlet pressure = 30kPa

4. The efficiency of the turbine is defined as the ratio between the real power and the ideal power, with this we find the real enthalpy.

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3 years ago

A heat engine receives heat from a heat source at 1453 C and has a thermal efficiency of 43 percent. The heat engine does maximu

xxMikexx [17]

Answer:

a) 1253 kJ

b) 714 kJ

c) 946 C

Explanation:

The thermal efficiency is given by this equation

η = L/Q1

Where

η: thermal efficiency

L: useful work

Q1: heat taken from the heat source

Rearranging:

Q1 = L/η

Replacing

Q1 = 539 / 0.43 = 1253 kJ

The first law of thermodynamics states that:

Q = L + ΔU

For a machine working in cycles ΔU is zero between homologous parts of the cycle.

Also we must remember that we count heat entering the system as positiv and heat leaving as negative.

We split the heat on the part that enters and the part that leaves.

Q1 + Q2 = L + 0

Q2 = L - Q1

Q2 = 539 - 1253 = -714 kJ

TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:

η = 1 - T2/T1

T2/T1 = 1 - η

T2 = (1 - η) * T1

The temperatures must be given in absolute scale (1453 C = 1180 K)

T2 = (1 - 0.43) * 1180 = 673 K

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