WORK-TO-HEAT TRANSDUCTION IN THERMO-FLUID SYSTEMSEnergy-based modeling is built on thermodynamics— the fundamental science of physical processes.Thermodynamics is to physical system dynamics what geometry is to mechanics.Why should we care about thermal phenomena?— they can profoundly influence dynamic behavior.EXAMPLE: A CLOSED BICYCLE PUMP compressible gas (air) in a closed container with variable volumeCompress the gas and it gets hot.When the gas is hotter than its surroundings the temperature gradient induces heat flow. This is another form of energy transduction— mechanical work to compress the gas is converted to heat. — governed by the first law.Energy transduction is bilateral (it works both ways)Heat the gas and its pressure increases.If the pressure moves the piston (to increase gas volume) mechanical work is done.Question:If you do work on the gas, can you get all of it back?(Answer: No — due to the second law.)Under what conditions is this energy “loss” significant?How do we integrate this behavior with our previous models?At least two different phenomena are involved:energy storagepower dissipationENERGY STORAGE IN A COMPRESSIBLE GASEnergy is added to (or taken from) the gas in two forms— mechanical work or heatModel this as a multiport energy storage element— but of what kind?A RELEVANT STATEMENT OF THE FIRST LAW:U = Q – WU: internal energy of the gasQ: heat added to the gasW: work done by expansion of the gas(Notation and sign convention are standard for engineering thermodynamics.)Apply this on an instant-by-instant basis using its differential form.dU = dQ – dWON THE MECHANICAL SIDE:dW = PdVP: pressureV: volumeThe gas behaves as a capacitor on the mechanical side work is a form of energypressure is an effortvolume is a displacementA COMMENT ON SIGN CONVENTIONPositive work compresses the gas— a negative volume change, –dV.dW = P(–dV)Positive work increases gas internal energydUdQ = 0 = dW = –PdVconsistent with the usual convention.ON THE THERMAL SIDE:Gibbs’ relation(i.e., the relevant part of it)dU = TdS – PdVT: absolute temperatureS: entropyThe gas behaves as a capacitor on the thermal side also. heat is a form of energytemperature behaves as an effort variablee.g., temperature gradient induces heat flowelectric potential gradient induces charge motionforce induces mechanical motionentropy behaves as a displacement variableA COMMENT ON ENTROPYA classical definition of entropy isdS = dQ/TRearranging yieldsdQ = TdSOn the thermal side, heat added isQ1 = + Q0ThusdUdW = 0 = TdSconsistent with Gibbs’ relation.DrawbackThis suggests that entropy production requires heat transfer.Not so— adiabatic processes may generate entropy.(adiabatic: no heat transfer)This classical definition is not essential for our treatment.MULTIPORT CAPACITOR MODEL Energy stored in a variable volume of compressible gasis a function of volume and entropy.U = U(S,V)Internal energy also depends on the mass of gas. U = U(S,V,m)As we assume a closed container, the mass is constant. — for now, mass, m, is a parameter. (Later we will allow mass to vary, thereby adding a third port to the capacitor.)Mathematical properties:Internal energy is a scalar.U(S,V) is a scalar potential function (or field) defined on the space of displacements spanned by S and V.q = U = U(q)Efforts in the two domains (thermal and mechanical) may be defined as the gradients of this potential with respect to the two displacements.e = IN TERMS OF COMPONENTS: (gradient “in the direction of” S) (gradient “in the direction of” –V) Using the chain rule, dU = TdS – PdVwhich recapitulates Gibbs’ relation.NOTE:A two-port capacitor requires two constitutive equations, one for each port. T = T(S,V)P = P(S,V)However, both may be derived as gradients of a single energy function.U = U(S,V)SYMMETRY:The curl of the gradient of a potential function is zero. In terms of components: (the order of differentiation doesn’t matter)Regrouping — Maxwell’s “reciprocity” condition.INTRINSIC STABILITY:Define inverse capacitance Maxwell’s reciprocity means this is a symmetric matrix.Stabilitydeterminant C-1 > 0—a sufficient conditionALTERNATIVE CAUSAL ASSIGNMENTSClassical thermodynamics identifies several different functions related to energy:enthalpyHelmholtz free energyGibbs free energyThese are nothing more than the co-energy functions associated with different causal assignments for the two-port capacitor. Internal energy corresponds to integral causality on both ports. flow variable input on each porttime-integrate to find displacementconstitutive equations define output effort variablesEnthalpy is a Legendre transformation of energy with respect to volume.L = U(S,V) – V = H(S,P)i.e.,H = U + PVAlternatively, differentiate the definition of enthalpy:dH = dU + PdV + VdPSubstituting for the differential of internal energy:dH = TdS –PdV + P dV + VdP = TdS + VdPIntegrating yieldsH = H(S,P)as before.Enthalpy is a co-energy function corresponding to differential causality on the mechanical port. Note that and Because enthalpy is a scalar function of S and P, we obtain the symmetry relation Helmholtz free energy is a Legendre transformation of internal energy with respect to entropy.L = U(S,V) – S = F(T,V)i.e.,F = U – TSAgain, an alternative is to differentiate the definition of Helmholtz free energy:dF = dU – TdS – SdTSubstituting for internal energy:dF = TdS –PdV – TdS – SdT = –SdT – PdVIntegrating yieldsF = F(T,V)Helmholtz free energy is a co-energy function corresponding to differential causality on the thermal port. Note that and Because Helmholtz free energy is a scalar function of T and V, we obtain the symmetry relation(S/∂V = ∂P/∂TGibbs free energy is a Legendre transformation of internal energy with respect to both entropy and volume.L = U(S,V) – S – V = G(T,P)i.e.,Gibbs free energyG = U – TS + PVAgain, an alternative is to differentiate the definition of Gibbs free energy:dG = dU – TdS – SdT + PdV + VdPSubstituting for the internal energy:dG = VdP – SdTIntegrating yieldsG = G(T,P)Gibbs free energy corresponds to differential causality on both ports. Note that∂G/∂T = –Sand∂G/∂P = VBecause Gibbs free energy is a scalar function of T and P, we obtain the symmetry relation∂S/∂P = –∂V/∂TREMARKSOnly internal energy is a true energy—it's the only one that's guaranteed conserved. —the others are