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Question on converting heat to kinetic energy.

  1. Aug 5, 2006 #1
    For propulsion at hypersonic speed I want to convert the heat produced by stopping the hypersonic airstream to propulsive thrust. But I keep getting a result that says the kinetic energy out would be greater than the heat energy input.
    Let's say you heat hydrogen fuel in gas form. You could store the
    hydrogen initially as liquid and then use the gaseous hydrogen formed
    from the liquid hydrogen that vaporizes when used to cool the external
    surfaces and engine nozzles and combustion chambers.
    The reason why I'm suggesting using the hydrogen in gas form is that
    you want to apply the full heating effect from stopping the hypersonic
    air to raise the hydrogen to high temperature. However, if you use
    liquid hydrogen a large portion of the heat energy just goes to
    converting the liquid to gas, the "heat of vaporization".
    Say you have 1 kg of hydrogen gas at 1 bar and 300 K. This page gives
    properties of hydrogen at various temperatures and pressures:

    Hydrogen Properties Package.

    These are the results at 1 bar and 300 K:

    Pressure = 1.000e+00 bar
    Temperature = 3.000e+02 K
    Enthalpy = 4.199e+03 kJ/kg
    Entropy = 6.483e+01 kJ/kg.K
    Vel.of sound = 1.310e+03 m/s
    Density = 8.080e-02 kg/m**3
    Them. cond. = 1.939e-01 W/m.K
    Viscosity = 8.948e-06 N.s/m**2
    Spec. heat = 1.485e+01 kJ/Kg.K
    Gamma = 1.385e+00

    The specific heat given is that at constant pressure, Cp. For my
    scenario, I'll heat the hydrogen in a constant volume vessel, so I need
    the specific heat at constant volume, Cv. Gamma is the ratio of these
    so Cv = Cp/gamma = 14,850/1.385 = 10,700 joules per kilo per degree
    Kelvin temperature increase.
    Let's say you were able to get 10,000,000 joules of heat energy from
    stopping the air flow. Then this would raise the temperature of one
    kilo of hydrogen at 300 K by 10,000,000/10,700 = 934.5 K, so to 1234.5
    This page gives the exhaust velocity for a rocket engine according
    to temperature in the combustion chamber:

    Rocket engine nozzles.

    At very high altitudes the exit pressure in the formula is nearly zero
    and the velocity equation simplifies to:

    V = sqrt[2kRT/((k-1)M)] , k the ratio of specific heats, which is the
    same as the gamma from the Hydrogen Properties Package page, R the
    ideal gas constant, and M the molecular weight.
    We need to use the Hydrogen Properties Package page to find k, or
    gamma, at the new temperature 1234.5 K. This requires knowing the
    pressure also. The pressure can be found from the ideal gas law: P =
    density*R*T/M . The density stays the same at a constant volume so is
    .0808 kg/m^3, and P = .0808*8314*1234.5/2 = 414,651 pascals.
    Now this can be input to the Hydrogen Properties Package as 4.15 bar
    and 1234.5 K. The result is that gamma, the k in the velocity equation,
    is 1.364, about the same as in the 1 bar, 300 K case. Then V =
    sqrt[2*1.384*8314*1234.5/(.384*2)] = 6,082 m/s.
    The kinetic energy of 1 kg of hydrogen at 6,082 m/s is .5*1*6082^2 =
    18,500,000 joules, which is more than the 10,000,000 joules of heat
    energy input.
    The kinetic energy of .41 kg of air at 7000 m/s is about 10,000,000
    joules. I am assuming most of this kinetic energy could be converted
    to heat energy [though the above calculations make me wonder about the validity of this too.] The momentum change of stopping this air would
    be .41*7000 = 2870 N-s. However, the impulse from the 1 kg of hydrogen
    at 6,082 m/s would 6,082 N-s, resulting in a net forward velocity.

    Bob Clark
    Last edited by a moderator: Apr 22, 2017 at 11:59 AM
  2. jcsd
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