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Pressure evolution in an engine intake manifold (thermodynamics) |
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| Oct22-10, 09:53 AM | #1 |
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Pressure evolution in an engine intake manifold (thermodynamics)
Hi,
I am using a model that estimates pressure (P) in an intake manifold. I think there is a mistake in its equations but I cannot find it. To simplify the problem we can make the following assumptions: - Only air fills the manifold: air comes into the manifold through the throttle (mass flow rate = mf_thr) and is pumped by the engine through the inlet valve (mass flow rate = mf_vlv). - Heat transfer is neglected: adiabatic manifold. - Air expansion through throttle is isenthalpic: if upstream throttle air temperature is a constant, then manifold temperature is a constant (T). Ideal gas law states: P = (R.T/V).m, with m = manifold air mass and V = manifold volume. With above assumptions (T = constant) and mass conservation law: dP/dt = (R.T/V).dm/dt = (R.T/V).(mf_thr – mf_vlv) {A} However, model equations are as follow: - First principle of thermo for open systems (no heat transfer and no work): dU/dt = mf_thr.h_thr – mf_vlv.h_ vlv [0], with h = specific enthalpy, U = internal energy of air in intake manifold - Internal energy and enthalpy of an ideal gas: U = m.Cv.T, h = Cp.T, with Cp and Cv assumed to be constants on the considered temperature range. - Derivative of U = m.Cv.T: dU/dt = Cv . (dm/dt . T + dT/dt . m) [1] - Derivative of ideal gas law P = m.R.T/V: dP/dt = R/V . (dm/dt . T + dT/dt . m) [2] - [1] and [2] give: dP/dt = R/(Cv.V) . dU/dt [3] - [3] and [0] give: dP/dt = R/(Cv.V) . (mf_thr.h_thr – mf_vlv.h_ vlv) => dP/dt = R/(Cv.V) . (mf_thr.Cp.T_thr – mf_vlv.Cp.T_vlv), with T_thr = temperature at throttle level, T_vlv = temperature at inlet valve level => dP/dt = R.Gamma/V . (mf_thr.T_thr – mf_vlv.T_vlv), with Gamma = Cp/Cv If T_thr = T_vlv = T (manifold temperature), then: dP/dt = (R.T.Gamma/V).(mf_thr – mf_vlv) {B} Equation [B] contains Gamma coefficient, which is not present in equation [A]. Where is the mistake? Thanks in advance for helping me to understand this tricky problem. |
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| first law, ideal gas, open system, pressure, thermodynamics |
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