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Power of a car engine

  1. May 28, 2017 #1
    1. The problem statement, all variables and given/known data
    The volume displacement in a car engine is 2000 cm3. During the power stroke, the mean pressure inside the cylinder is 15 bar.
    1. Compute the work performed by the engine in one revolution.
    2. Compute the engine's power assuming it runs at 3000 rpm.

    2. Relevant equations
    $$W=-p\cdot \Delta V=-3000\text{ J}$$
    $$P=\frac{3000\cdot\left|W\right|}{60 \text{ s}}=150\text{ kW}$$
    3. The attempt at a solution
    Above numbers are the solutions given by the book. However I'm not sure if they are correct.
    1. Shouldn't the pressure exerted from the outside by the piston be relevant here, since this is what the gas is working against? Or do we need to assume an equilibrium between inside and outside pressure in a thermodynamic treatment? Is this justified in reality?
    2. Usual car engines are four-stroke, so there's a power stroke only in every second revolution. Shouldn't both numbers need to be divided by 2 because of this?
     
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  3. May 28, 2017 #2

    scottdave

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    You make a good point about the way a four cycle engine works. Perhaps something the author did not consider.
     
  4. May 28, 2017 #3

    haruspex

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    Not sure which point you are making.
    If you are concerned about atmospheric pressure, the 15bar might well refer to gauge pressure, not absolute.
    If you feel there would be a difference between the pressure in the cylinder and the backpressure from the piston, no - action and reaction are equal and opposite.
    If you mean the force transmitted through the piston would be less than pressure x area in the cylinder,yes, but only in respect of the net force required to accelerate the piston, which would be quite small.
     
  5. May 28, 2017 #4
    Let's look at an isolated cylinder with atmospheric pressure in- and outside (1 bar), so the piston is at rest. Now combustion is ignited, almost instantly increasing the inside pressure to 15 bar. If the piston now moves, which one is the relevant pressure to compute the work, the 15 bar inside or the 1 bar outside?
     
  6. May 28, 2017 #5

    haruspex

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    The 15 bar inside. It is doing work against a resistance, which has little to do with the 1 bar ambient. If you were to disconnect the load, it would not reach 15 bar.
     
  7. May 28, 2017 #6
    Why? Isn't this resistance exactly given by the 1 bar times area?

    What about following more symmetric situation: Suppose we have a closed cylinder with a (thermally isolating) piston in its centre that is fixed at first. We fill half ##A## with gas up to a pressure of ##p_A=1\text{ bar}##, half ##B## up to ##p_A=15\text{ bar}##. Then the piston is released.
    For simplicity, let's look at a very small displacement ##\Delta V>0## after the release such that the pressures haven't changed considerably. According to you, half ##B## loses the energy
    $$W_B=-p_B\cdot \Delta V\enspace.$$
    However, the energy of the other half increases by
    $$W_A=-p_A\cdot (-\Delta V)=p_A\cdot \Delta V\enspace$$
    Since ##p_B\gg p_A## this obviously violates conservation of energy.
     
  8. May 28, 2017 #7

    haruspex

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    No, the piston is connected through the transmission system to the road wheels, so the resistance is a lot more than that.
    If you disconnect the load completely, so there is just a piston with nothing restraining it, it will accelerate hugely and go flying through the air. You can apply ΔpA=F=ma to the piston to find its acceleration, but the backpressure is still 15bar.
     
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