Thermodynamic and Mechanical Study

1. Scope and method

This study describes the complete driven cycle of the Dada machine in symbolic form. The model couples prescribed kinematics of the two pistons, well-mixed gas volumes, 0D heat exchangers, passive check valves, and a generic compressible hydraulic closure model.

The cycle origin is set at the top position of the large cylinder, with:

$t=0,\theta (0)=0,V_L(0)=V_{L,max}.$

The four gas volumes are:

• S: small cylinder, directly associated with the cold heat exchanger C;
• L: large cylinder, directly associated with the hot heat exchanger H;
• C: cold heat exchanger, which absorbs heat from the cold reservoir;
• H: hot heat exchanger, which rejects heat to the hot reservoir.

The check valves allow H → S during Phase II and C → L during Phase IV.

The cycle comprises four phases:

• Phase I — check valves closed, nominally adiabatic: compression on the L+H side;
• Phase II — heat exchange, nominally isothermal: transfer L → H → S, with heat rejection on the hot side;
• Phase III — check valves closed, nominally adiabatic: expansion on the S+C side;
• Phase IV — heat exchange, nominally isothermal: transfer S → C → L, with heat absorption on the cold side.

The terms nominally adiabatic and nominally isothermal describe the objective of the cycle. They do not constitute exact thermodynamic constraints: temperatures are calculated from the balances, the heat exchangers remain coupled to their reservoirs, and the low-displacement regions of the pistons are not assumed to be perfectly stationary.

1.1 First-level assumptions

The model is based on the following assumptions:

• single-phase, ideal and calorically perfect gas;
• constant properties $R$, $C_p$, $C_v$, $\gamma$, with $R=C_p-C_v$ and $\gamma =C_p/C_v$;
• each gas volume is uniform and well mixed;
• gas kinetic and potential energies are neglected in the 0D balances;
• fixed heat-exchanger volumes;
• prescribed piston kinematics;
• no mechanical friction in the thermodynamic model;
• heat exchange represented by an overall conductance $UA$;
• passive check valves controlled by the pressure difference.

2. Notation and conventions

2.1 Geometry and kinematics

For $i\in \{S,L\}$:

$V_{i,min}>0,V_{i,max}>V_{i,min}.$

The swept volume is:

$V_{i,\mathrm{s}\mathrm{w}\mathrm{e}\mathrm{p}\mathrm{t}}=V_{i,max}-V_{i,min}.$

The thermodynamic volumes prescribed by the mechanism are:

$V_S=V_S(t),V_L=V_L(t),$

with their signed derivatives:

${\dot{V}}_S=\frac{d{V}_S}{dt},{\dot{V}}_L=\frac{d{V}_L}{dt}.$

If the crank rotates at constant angular speed $\omega$:

$\theta (t)=\omega t,{\dot{V}}_i=\omega \frac{d{V}_i}{d\theta }.$

When only the absolute value of the volumetric speed is useful:

${\nu }_{V,i}=|{\dot{V}}_i|.$

A quasi-stationary region denotes an interval in which the displacement or $|\dot{V}|$ remains small compared with the transfer phases. This region corresponds to the “plateau” of the kinematic optimization, without assuming $\dot{V}=0$ exactly.

2.2 Kinematic closure fraction

To describe the normalized closure of a cylinder:

${\Lambda }_i(t)=\frac{V_{i,max}-V_i(t)}{V_{i,max}-V_{i,min}},i\in \{S,L\}.$

Thus ${\Lambda }_i=0$ corresponds to maximum volume and ${\Lambda }_i=1$ to minimum volume. The values ${\Lambda }_L^{\ast }$ and ${\Lambda }_S^{\ast }$ are nominal kinematic targets at the transitions. The value actually reached at a check-valve event is:

${\Lambda }_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}=\Lambda (t_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}}).$

The physical transitions remain determined by the pressures; ${\Lambda }^{\ast }$ is therefore not an imposed opening condition.

2.3 Thermodynamic variables

For each volume $i\in \{S,L,C,H\}$:

$m_i,U_i,T_i,P_i,V_i.$

The complete state vector is chosen as:

Failed to parse (syntax error): {\displaystyle \boxed{\mathbf X=(m_S,U_S,m_L,U_L,m_C,U_C,m_H,U_H)}.}

Temperatures and pressures are derived from it:

$T_i=\frac{U_i}{m_i{C}_v},P_i=\frac{m_{i}R{T}_i}{V_i}.$

The volumes $V_S(t)$ and $V_L(t)$ are prescribed by the kinematics and are not independent thermodynamic state variables. The volumes $V_C$ and $V_H$ are constant.

2.4 Energy sign convention

Heat is positive when it is received by the gas. Work is positive when the gas delivers work:

$\dot{W}=P\dot{V}.$

For the intended refrigeration operation:

$Q_C>0,Q_H<0,W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}<0.$

3. Thermal closure and validity domain

3.1 Exchange with the thermal reservoirs

For a heat exchanger $HX\in \{C,H\}$:

Failed to parse (syntax error): {\displaystyle \boxed{\dot Q_{HX}=(UA)_{HX}(T_{HX,\mathrm{res}}-T_{HX})}.}

$T_{HX,\mathrm{r}\mathrm{e}\mathrm{s}}$ is the temperature of the external thermal reservoir, prescribed and constant in the base model. $T_{HX}$ is the mean 0D temperature of the gas in the heat exchanger. $(UA{)}_{HX}$ represents the overall thermal conductance, which may combine convection, wall conduction, and contact resistances.

Cold side:

${\dot{Q}}_C=(UA{)}_C(T_{C,\mathrm{r}\mathrm{e}\mathrm{s}}-T_C).$

In refrigeration operation, $T_C<T_{C,\mathrm{r}\mathrm{e}\mathrm{s}}$ gives ${\dot{Q}}_C>0$.

Hot side:

${\dot{Q}}_H=(UA{)}_H(T_{H,\mathrm{r}\mathrm{e}\mathrm{s}}-T_H).$

In refrigeration operation, $T_H>T_{H,\mathrm{r}\mathrm{e}\mathrm{s}}$ gives ${\dot{Q}}_H<0$.

A nominally isothermal phase therefore does not mean $T_{HX}=T_{HX,\mathrm{r}\mathrm{e}\mathrm{s}}$: a finite temperature difference is required to transfer finite thermal power when $UA$ is finite.

An indicator of isothermal quality may be defined over a given phase by:

${\epsilon }_T=\frac{T_{max}-T_{min}}{T_{\mathrm{r}\mathrm{e}\mathrm{f}}}.$

3.2 Thermophysical validity domain

The base model assumes:

$PV=mRT,Z=1,$

$C_p=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t},C_v=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t},\gamma =\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}.$

Validity must be checked a posteriori over the entire cycle, notably through:

$|Z-1|\ll 1,$

and through small variations of the thermophysical properties, for example:

${\epsilon }_{C_p}=\frac{C_{p,max}-C_{p,min}}{C_{p,\mathrm{r}\mathrm{e}\mathrm{f}}}\ll 1.$

The working fluid must remain single-phase and gaseous, and sufficiently far from any condensation or phase transition throughout the $(P,T)$ domain traversed.

If these criteria become insufficient, an extension may use $Z(P,T)$, $C_p(T)$, $C_v(T)$, or a real-gas equation of state without changing the general architecture of the mass and energy balances.

4. Reduced formulation of a quasi pressure-equalized pair

When a cylinder and its heat exchanger are connected by a very low-resistance internal path, the approximation

$P_{\mathrm{c}\mathrm{y}\mathrm{l}}\approx P_{HX}=P_{\mathrm{p}\mathrm{a}\mathrm{i}\mathrm{r}}.$

may be used. It is acceptable if:

Failed to parse (syntax error): {\displaystyle \boxed{\varepsilon_P=\frac{|\Delta P_{\mathrm{int}}|}{P_{\mathrm{pair}}}\ll1},\qquad \Delta P_{\mathrm{int}}=P_{\mathrm{cyl}}-P_{HX}.}

A low internal Mach number provides an additional check:

$M{a}_{\mathrm{i}\mathrm{n}\mathrm{t}}=\frac{|u_{\mathrm{i}\mathrm{n}\mathrm{t}}|}{a}\ll 1,$

but it is not sufficient on its own to guarantee pressure equalization.

4.1 Closed-pair case

For a closed pair, with fixed $V_{HX}$ and $V=V_{\mathrm{c}\mathrm{y}\mathrm{l}}+V_{HX}$:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{dP}{dt}= \frac{(\gamma-1)(UA)_{HX}(T_{HX,\mathrm{res}}-T_{HX})-\gamma P\dot V_{\mathrm{cyl}}} {V_{\mathrm{cyl}}+V_{HX}} }.}

When $(UA{)}_{HX}=0$:

$P(V_{\mathrm{c}\mathrm{y}\mathrm{l}}+V_{HX}{)}^{\gamma }=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}.$

4.2 Internal redistribution flow rate

The internal flow rate is defined as positive from cylinder → heat exchanger. Let:

$N=V_{HX}\dot{P}+(\gamma -1)(UA{)}_{HX}(T_{HX}-T_{HX,\mathrm{r}\mathrm{e}\mathrm{s}}).$

The flow carries the enthalpy of the upstream state:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot m_{\mathrm{int}}= \begin{cases} \dfrac{N}{\gamma R T_{\mathrm{cyl}}}, & N\ge0 \quad (\mathrm{cyl}\to HX),\\[6pt] \dfrac{N}{\gamma R T_{HX}}, & N<0 \quad (HX\to\mathrm{cyl}). \end{cases} }}

The heat-exchanger temperature evolves according to:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot T_{HX}=\frac{T_{HX}}{P}\dot P- \frac{RT_{HX}^2}{PV_{HX}}\dot m_{\mathrm{int}} }.}

4.3 Open receiving pair

During an active phase, the external flow physically enters the receiving cylinder, not directly its associated heat exchanger. If ${\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}>0$ enters the cylinder at temperature $T_{\mathrm{e}\mathrm{x}\mathrm{t}}$:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot P= \frac{\gamma RT_{\mathrm{ext}}\dot m_{\mathrm{ext}} +(\gamma-1)(UA)_{HX}(T_{HX,\mathrm{res}}-T_{HX}) -\gamma P\dot V_{\mathrm{cyl}}} {V_{\mathrm{cyl}}+V_{HX}} }.}

The masses satisfy:

${\dot{m}}_{\mathrm{c}\mathrm{y}\mathrm{l}}={\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}-{\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}},{\dot{m}}_{HX}={\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}},$

and therefore:

${\dot{m}}_{\mathrm{p}\mathrm{a}\mathrm{i}\mathrm{r}}={\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}.$

The closed case is obtained immediately with ${\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}=0$.

5. Hydraulic closure

5.1 Generic formulation

Any hydraulic connection is described by a generic law:

Failed to parse (syntax error): {\displaystyle \boxed{\dot m=\Phi(P_u,P_d,T_u,\mathcal G,\mathcal F)}}

$u$ and $d$ respectively denote the upstream and downstream states. The transported enthalpy is that of the upstream state:

${\dot{H}}_{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}=\dot{m}{C}_p{T}_u.$

For a bidirectional connection, the upstream state is determined by the actual direction of the pressure gradient. For a check valve, reverse flow is prohibited.

5.2 First-level closure using a compressible orifice

A first approximation consists in using an effective hydraulic area:

$(C_{d}A{)}_{\mathrm{e}\mathrm{f}\mathrm{f}},$

which represents the overall ease of gas flow through the actual connection.

With:

$r=\frac{P_d}{P_u},r_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}={\left(\frac{2}{\gamma +1}\right)}^{\gamma /(\gamma -1)},$

the unchoked flow rate, for $r>r_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$, is:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot m=(C_dA)_{\mathrm{eff}}P_u \sqrt{\frac{2\gamma}{RT_u(\gamma-1)} \left(r^{2/\gamma}-r^{(\gamma+1)/\gamma}\right)} }.}

For $r\le r_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot m=(C_dA)_{\mathrm{eff}}P_u \sqrt{\frac{\gamma}{RT_u}} \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}} }.}

This closure is not essential to the model: it may later be replaced by a law that more faithfully represents the pressure losses of a real heat exchanger, pipe, or check valve.

6. Complete thermodynamic cycle

6.1 Phase I — check valves closed, nominally adiabatic: compression on the L side

Both check valves are closed. The L+H and S+C pairs are closed. The L+H side is nominally compressed; the motion of the small piston remains that provided by the actual kinematics.

For L+H:

${\dot{P}}_{LH}=\frac{(\gamma -1)(UA{)}_H(T_{H,\mathrm{r}\mathrm{e}\mathrm{s}}-T_H)-\gamma P_{LH}{\dot{V}}_L}{V_L+V_H}.$

For S+C:

${\dot{P}}_{SC}=\frac{(\gamma -1)(UA{)}_C(T_{C,\mathrm{r}\mathrm{e}\mathrm{s}}-T_C)-\gamma P_{SC}{\dot{V}}_S}{V_S+V_C}.$

The internal redistribution and temperature equations of §4 apply to both pairs.

The transition to Phase II occurs when the H → S check valve satisfies its opening condition.

6.2 Phase II — heat exchange, nominally isothermal: L → H → S

The gas leaves L, passes through the hot heat exchanger H, where it rejects heat, crosses the H → S check valve, and then enters the receiving cylinder S. The S+C pair remains quasi pressure-equalized if the criterion ${\epsilon }_P\ll 1$ is satisfied.

6.2.1 Donor cylinder L

The fundamental balance is:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{d(m_LC_vT_L)}{dt} =-P_L\dot V_L-\dot m_{L,\mathrm{out}}C_pT_L }.}

The kinematics are designed to maintain an outflow from the donor cylinder. In this case, the analytical solution is:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{T_L}{T_{L,\mathrm{ref}}}= \left[\frac{m_L}{m_{L,\mathrm{ref}}}\frac{V_{L,\mathrm{ref}}}{V_L}\right]^{\gamma-1} }}

and:

Failed to parse (syntax error): {\displaystyle \boxed{ P_L=P_{L,\mathrm{ref}} \left[\frac{m_L}{m_{L,\mathrm{ref}}}\frac{V_{L,\mathrm{ref}}}{V_L}\right]^\gamma }.}

One also obtains:

$\frac{T_L}{T_{L,\mathrm{r}\mathrm{e}\mathrm{f}}}={\left(\frac{P_L}{P_{L,\mathrm{r}\mathrm{e}\mathrm{f}}}\right)}^{(\gamma -1)/\gamma },$

and the specific entropy of the remaining gas satisfies $ds=0$ under these assumptions. If reverse flow occurs despite the kinematic design, the analytical solution is no longer applicable and the complete open-system balance in $(m_L,U_L)$ must be used.

6.2.2 Hot heat exchanger H

Mass conservation:

Failed to parse (syntax error): {\displaystyle \boxed{\dot m_H=\dot m_{\mathrm{in}}-\dot m_{\mathrm{out}}}.}

Fundamental energy balance:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{d(m_HC_vT_H)}{dt} =\dot m_{\mathrm{in}}C_pT_L -\dot m_{\mathrm{out}}C_pT_H +(UA)_H(T_{H,\mathrm{res}}-T_H) }.}

In expanded form:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot T_H= \frac{\dot m_{\mathrm{in}}(C_pT_L-C_vT_H) -\dot m_{\mathrm{out}}RT_H +(UA)_H(T_{H,\mathrm{res}}-T_H)} {C_vm_H} }.}

The flow rates are determined by the hydraulic laws:

${\dot{m}}_{\mathrm{i}\mathrm{n}}={\Phi }_{\mathrm{H}\mathrm{X},\mathrm{H}}(P_L,P_H,T_L,\dots ),$

${\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}={\Phi }_{\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{v}\mathrm{e},\mathrm{H}}(P_H,P_{SC},T_H,\dots ).$

6.2.3 Receiving pair S+C

The flow from H enters S. The pair pressure satisfies:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot P_{SC}= \frac{\gamma R\dot m_{\mathrm{out}}T_H +(\gamma-1)(UA)_C(T_{C,\mathrm{res}}-T_C) -\gamma P_{SC}\dot V_S} {V_S+V_C} }.}

The internal redistribution equations of §4 remain unchanged: the external flow acts on heat exchanger C indirectly through the evolution of the pair pressure.

The transition to Phase III is the closing event of the H → S check valve.

6.3 Phase III — check valves closed, nominally adiabatic: expansion on the S side

Both check valves are closed. The S+C pair expands nominally; the L+H pair also remains closed. Neither piston is assumed to be strictly stationary.

For S+C:

${\dot{P}}_{SC}=\frac{(\gamma -1)(UA{)}_C(T_{C,\mathrm{r}\mathrm{e}\mathrm{s}}-T_C)-\gamma P_{SC}{\dot{V}}_S}{V_S+V_C}.$

For L+H:

${\dot{P}}_{LH}=\frac{(\gamma -1)(UA{)}_H(T_{H,\mathrm{r}\mathrm{e}\mathrm{s}}-T_H)-\gamma P_{LH}{\dot{V}}_L}{V_L+V_H}.$

The transition to Phase IV occurs when the C → L check valve satisfies its opening condition.

6.4 Phase IV — heat exchange, nominally isothermal: S → C → L

The gas leaves S, passes through the cold heat exchanger C, where it receives heat from the cold reservoir, crosses the C → L check valve, and then enters the receiving cylinder L. The L+H pair remains quasi pressure-equalized if the criterion ${\epsilon }_P\ll 1$ is satisfied.

6.4.1 Donor cylinder S

The fundamental balance is:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{d(m_SC_vT_S)}{dt} =-P_S\dot V_S-\dot m_{S,\mathrm{out}}C_pT_S }.}

For outflow guaranteed by the kinematic design:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{T_S}{T_{S,\mathrm{ref}}}= \left[\frac{m_S}{m_{S,\mathrm{ref}}}\frac{V_{S,\mathrm{ref}}}{V_S}\right]^{\gamma-1} }}

and:

Failed to parse (syntax error): {\displaystyle \boxed{ P_S=P_{S,\mathrm{ref}} \left[\frac{m_S}{m_{S,\mathrm{ref}}}\frac{V_{S,\mathrm{ref}}}{V_S}\right]^\gamma }.}

Reverse flow requires returning to the complete open-system balance in $(m_S,U_S)$.

6.4.2 Cold heat exchanger C

Failed to parse (syntax error): {\displaystyle \boxed{\dot m_C=\dot m_{\mathrm{in}}-\dot m_{\mathrm{out}}}.}

The fundamental energy balance is:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{d(m_CC_vT_C)}{dt} =\dot m_{\mathrm{in}}C_pT_S -\dot m_{\mathrm{out}}C_pT_C +(UA)_C(T_{C,\mathrm{res}}-T_C) }.}

In expanded form:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot T_C= \frac{\dot m_{\mathrm{in}}(C_pT_S-C_vT_C) -\dot m_{\mathrm{out}}RT_C +(UA)_C(T_{C,\mathrm{res}}-T_C)} {C_vm_C} }.}

The flow rates are determined by:

${\dot{m}}_{\mathrm{i}\mathrm{n}}={\Phi }_{\mathrm{H}\mathrm{X},\mathrm{C}}(P_S,P_C,T_S,\dots ),$

${\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}={\Phi }_{\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{v}\mathrm{e},\mathrm{C}}(P_C,P_{LH},T_C,\dots ).$

6.4.3 Receiving pair L+H

The flow from C enters L. The pair pressure satisfies:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot P_{LH}= \frac{\gamma R\dot m_{\mathrm{out}}T_C +(\gamma-1)(UA)_H(T_{H,\mathrm{res}}-T_H) -\gamma P_{LH}\dot V_L} {V_L+V_H} }.}

Closing the C → L check valve returns the system to Phase I. Geometric closure of the pistons alone is not sufficient to guarantee thermodynamic closure of the cycle.

7. Physical transitions of the check valves

For a check valve oriented from upstream $u$ to downstream $d$:

$P_u-P_d\ge \Delta P_{\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{n}}\Rightarrow \text{opening},$

$P_u-P_d\le \Delta P_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}}\Rightarrow \text{closing},$

with hysteresis, if present:

Failed to parse (syntax error): {\displaystyle \boxed{\Delta P_{\mathrm{close}}\le\Delta P_{\mathrm{open}}}.}

An opening or closing event changes the hydraulic topology and therefore the active equations; it does not cause any instantaneous jump in the thermodynamic state. For each volume:

$m_i^{+}=m_i^{-},U_i^{+}=U_i^{-},V_i^{+}=V_i^{-}.$

For an ideal gas:

$T_i^{+}=T_i^{-},P_i^{+}=P_i^{-}.$

There is therefore no instantaneous pressure equalization when a check valve opens.

The values ${\Lambda }^{\ast }$ serve as kinematic design targets; the values actually observed at the transitions are ${\Lambda }_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}=\Lambda (t_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}})$.

8. Work, heat, and coefficient of performance

The instantaneous work delivered by the gas on the two pistons is calculated during all phases:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot W=P_S^\star\dot V_S+P_L^\star\dot V_L }.}

$P_S^{\star }$ and $P_L^{\star }$ denote the thermodynamic pressure effectively applied to the gas in each cylinder according to the phase topology: pair pressure when the cylinder belongs to a quasi pressure-equalized pair, and its own pressure when it is a hydraulically isolated donor.

The net work over the cycle is:

Failed to parse (syntax error): {\displaystyle \boxed{ W_{\mathrm{cycle}}= \int_0^\tau \left(P_S^\star\dot V_S+P_L^\star\dot V_L\right)dt }.}

The exchanged heats are:

$Q_C={\int }_0^{\tau }(UA{)}_C(T_{C,\mathrm{r}\mathrm{e}\mathrm{s}}-T_C)dt,$

$Q_H={\int }_0^{\tau }(UA{)}_H(T_{H,\mathrm{r}\mathrm{e}\mathrm{s}}-T_H)dt.$

In periodic steady operation:

$\Delta U_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}=0,$

and the first law gives:

Failed to parse (syntax error): {\displaystyle \boxed{Q_C+Q_H=W_{\mathrm{cycle}}}.}

The refrigeration COP is:

Failed to parse (syntax error): {\displaystyle \boxed{COP_c=\frac{Q_C}{-W_{\mathrm{cycle}}}}.}

The heat-pump COP is:

Failed to parse (syntax error): {\displaystyle \boxed{COP_h=\frac{-Q_H}{-W_{\mathrm{cycle}}}=COP_c+1}.}

The signs $Q_C>0$, $Q_H<0$, and $W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}<0$ provide checks of the intended refrigeration regime.

The thermodynamic force exerted by the gas on a piston face may be written $F_{\mathrm{g}\mathrm{a}\mathrm{s}}=PS$. Net mechanical force, inertia, and friction belong to the subsequent mechanical sizing stage.

9. Gas charge and periodic regime

The total amount of enclosed gas is a physical parameter:

Failed to parse (syntax error): {\displaystyle \boxed{M_{\mathrm{tot}}=m_S+m_L+m_C+m_H=\mathrm{const}}.}

It may be imposed directly or defined experimentally by a charging pressure and temperature. The chosen reference configuration is that at $t=0$, with the large cylinder at maximum volume. If all volumes communicate and are in uniform equilibrium during charging:

Failed to parse (syntax error): {\displaystyle \boxed{ M_{\mathrm{tot}}= \frac{P_{\mathrm{charge}} \left[V_S(0)+V_{L,\max}+V_C+V_H\right]} {RT_{\mathrm{charge}}} }.}

$P_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}$ and $T_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}$ define the amount of gas charged; they are not conditions that the periodic cycle must recover.

The established periodic regime is a solution of the system such that, between two successive passages through the top position of the large cylinder with the same kinematic direction:

Failed to parse (syntax error): {\displaystyle \boxed{\mathbf X(t+\tau)=\mathbf X(t)}.}

Geometric periodicity alone:

$V_i(t+\tau )=V_i(t)$

is not sufficient to guarantee thermodynamic periodicity.

The numerical state used to initialize a calculation may be approximate; it must not be confused with a physical parameter of the machine. The future solver may search for the periodic fixed point by successive cycles, a shooting method, or a Newton method.

10. Parameters, design data, and results

10.1 Prescribed data

• working fluid and reference properties $R$, $C_p$, $C_v$, $\gamma$;
• reservoir temperatures $T_{C,\mathrm{r}\mathrm{e}\mathrm{s}}$, $T_{H,\mathrm{r}\mathrm{e}\mathrm{s}}$;
• total charge $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$, or equivalently $(P_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}},T_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}})$ in the charging configuration defined in §9;
• kinematics $V_S(t)$, $V_L(t)$, and, where relevant, $\omega$.

10.2 Design parameters

• $V_{S,min}$, $V_{S,max}$, $V_{L,min}$, $V_{L,max}$;
• $V_C$, $V_H$;
• $(UA{)}_C$, $(UA{)}_H$;
• hydraulic parameters of the heat exchangers, pipes, and check valves, represented at first level by $(C_{d}A{)}_{\mathrm{e}\mathrm{f}\mathrm{f}}$;
• thresholds $\Delta P_{\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{n}}$, $\Delta P_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}}$;
• kinematic targets ${\Lambda }_L^{\ast }$, ${\Lambda }_S^{\ast }$.

10.3 Calculated variables and results

• $m_i$, $U_i$, $T_i$, $P_i$;
• internal and external mass flow rates;
• ${\dot{Q}}_C$, ${\dot{Q}}_H$, $Q_C$, $Q_H$;
• $W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}$, $CO{P}_c$, $CO{P}_h$;
• pressure, temperature, and flow-rate extrema;
• actual check-valve events and ${\Lambda }_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}$;
• isothermal quality ${\epsilon }_T$;
• validity criteria ${\epsilon }_P$, $Ma$, $Z$, and property variations.

11. Global conservation checks

11.1 Mass conservation

The solver must satisfy:

Failed to parse (syntax error): {\displaystyle \boxed{\frac{dM_{\mathrm{tot}}}{dt}=0}.}

A useful numerical residual is:

Failed to parse (syntax error): {\displaystyle \boxed{\varepsilon_M(t)=M_{\mathrm{tot}}(t)-M_{\mathrm{tot}}(0)}.}

11.2 Global energy conservation

Whatever the phase, the internal mass and enthalpy fluxes must cancel when the balances of all volumes are summed. The global balance must reduce to:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{dU_{\mathrm{tot}}}{dt} =\dot Q_C+\dot Q_H -P_S^\star\dot V_S -P_L^\star\dot V_L }.}

A cumulative energy residual may be defined by:

Failed to parse (syntax error): {\displaystyle \boxed{ \varepsilon_E(t)= U_{\mathrm{tot}}(t)-U_{\mathrm{tot}}(0) -Q_C(0,t)-Q_H(0,t)+W(0,t) }.}

The solver must keep ${\epsilon }_M$ and ${\epsilon }_E$ close to zero to the expected numerical accuracy.

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Appendix A — Symbolic derivations and validated checks

A.1 Pressure equation for a closed pair

For a cylinder + heat-exchanger pair at quasi-uniform pressure:

$U=\frac{P(V_{\mathrm{c}\mathrm{y}\mathrm{l}}+V_{HX})}{\gamma -1}.$

The first law gives:

$\frac{dU}{dt}=\dot{Q}-P{\dot{V}}_{\mathrm{c}\mathrm{y}\mathrm{l}},$

with:

$\dot{Q}=(UA{)}_{HX}(T_{HX,\mathrm{r}\mathrm{e}\mathrm{s}}-T_{HX}).$

Differentiating $U$:

$\frac{1}{\gamma -1}[(V_{\mathrm{c}\mathrm{y}\mathrm{l}}+V_{HX})\dot{P}+P{\dot{V}}_{\mathrm{c}\mathrm{y}\mathrm{l}}]=\dot{Q}-P{\dot{V}}_{\mathrm{c}\mathrm{y}\mathrm{l}}.$

Hence:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot P= \frac{(\gamma-1)\dot Q-\gamma P\dot V_{\mathrm{cyl}}} {V_{\mathrm{cyl}}+V_{HX}} }.}

If $\dot{Q}=0$:

$\frac{\dot{P}}{P}=-\gamma \frac{\dot{V}}{V},$

then:

Failed to parse (syntax error): {\displaystyle \boxed{PV^\gamma=\mathrm{const}}.}

A.2 Internal flow rate of the pair

For the heat exchanger alone, at fixed volume:

$U_{HX}=\frac{P{V}_{HX}}{\gamma -1}.$

Therefore:

$\frac{V_{HX}}{\gamma -1}\dot{P}=(UA{)}_{HX}(T_{HX,\mathrm{r}\mathrm{e}\mathrm{s}}-T_{HX})+{\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}}{C}_p{T}_{\mathrm{u}\mathrm{p}}.$

Using $C_p=\gamma R/(\gamma -1)$:

$V_{HX}\dot{P}+(\gamma -1)(UA{)}_{HX}(T_{HX}-T_{HX,\mathrm{r}\mathrm{e}\mathrm{s}})=\gamma R{T}_{\mathrm{u}\mathrm{p}}{\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}}.$

This recovers the definition of the numerator $N$ and the selection of the upstream temperature according to the sign of the flow rate.

A.3 Evolution of the heat-exchanger temperature within a pair

For a fixed volume:

$m_{HX}=\frac{P{V}_{HX}}{R{T}_{HX}}.$

Differentiating:

$\frac{{\dot{m}}_{HX}}{m_{HX}}=\frac{\dot{P}}{P}-\frac{{\dot{T}}_{HX}}{T_{HX}}.$

With ${\dot{m}}_{HX}={\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}}$:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot T_{HX}=\frac{T_{HX}}P\dot P- \frac{RT_{HX}^2}{PV_{HX}}\dot m_{\mathrm{int}} }.}

This relation follows solely from mass conservation and the equation of state; it is valid for both flow directions.

A.4 Open receiving pair

For a pair receiving ${\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}$ into its cylinder:

$\frac{dU}{dt}={\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}{C}_p{T}_{\mathrm{e}\mathrm{x}\mathrm{t}}+\dot{Q}-P{\dot{V}}_{\mathrm{c}\mathrm{y}\mathrm{l}}.$

With $U=P(V_{\mathrm{c}\mathrm{y}\mathrm{l}}+V_{HX})/(\gamma -1)$:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot P= \frac{\gamma RT_{\mathrm{ext}}\dot m_{\mathrm{ext}} +(\gamma-1)\dot Q -\gamma P\dot V_{\mathrm{cyl}}} {V_{\mathrm{cyl}}+V_{HX}} }.}

The local mass balance:

${\dot{m}}_{\mathrm{c}\mathrm{y}\mathrm{l}}={\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}-{\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}},$

${\dot{m}}_{HX}={\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}},$

immediately gives:

Failed to parse (syntax error): {\displaystyle \boxed{\dot m_{\mathrm{pair}}=\dot m_{\mathrm{ext}}}.}

A.5 Analytical solution for the adiabatic donor cylinder

For an adiabatic, well-mixed cylinder with outflow only:

$\frac{d(m{C}_{v}T)}{dt}=-P\dot{V}-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{C}_{p}T,$

and:

$\dot{m}=-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}.$

Expanding:

$C_{v}m\dot{T}+C_{v}T\dot{m}=-P\dot{V}+C_{p}T\dot{m},$

thus:

$C_{v}m\dot{T}=-P\dot{V}+RT\dot{m}.$

With $P=mRT/V$ and $R/C_v=\gamma -1$:

$\frac{dT}{T}=(\gamma -1)(\frac{dm}{m}-\frac{dV}{V}).$

After integration:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{T}{T_0}= \left[\frac{m}{m_0}\frac{V_0}{V}\right]^{\gamma-1} }.}

Then, using $PV=mRT$:

Failed to parse (syntax error): {\displaystyle \boxed{ P=P_0\left[\frac{m}{m_0}\frac{V_0}{V}\right]^\gamma }.}

and:

Failed to parse (syntax error): {\displaystyle \boxed{ \frac{T}{T_0}= \left(\frac{P}{P_0}\right)^{(\gamma-1)/\gamma} }.}

Under these assumptions, the specific entropy of the remaining gas is constant: $ds=0$. The total entropy of the gas contained in the cylinder is not constant because its mass varies.

A.6 Active heat exchanger: expanded balance

Fundamental balance:

$\frac{d(m{C}_{v}T)}{dt}={\dot{m}}_{\mathrm{i}\mathrm{n}}{C}_p{T}_{\mathrm{i}\mathrm{n}}-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{C}_{p}T+\dot{Q}.$

Expanding the left-hand side and using:

$\dot{m}={\dot{m}}_{\mathrm{i}\mathrm{n}}-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}},$

one obtains:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot T= \frac{\dot m_{\mathrm{in}}(C_pT_{\mathrm{in}}-C_vT) -\dot m_{\mathrm{out}}RT+\dot Q} {C_vm} }.}

Limiting checks:

• with no flow, the equation recovers the thermal relaxation of a closed volume;
• with equal steady inlet/outlet flow rates, it recovers $m{C}_v\dot{T}=\dot{m}{C}_p(T_{\mathrm{i}\mathrm{n}}-T)+\dot{Q}$;
• if $T_{\mathrm{i}\mathrm{n}}=T$ and ${\dot{m}}_{\mathrm{i}\mathrm{n}}={\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}$, the net contribution of the flow to $\dot{T}$ vanishes.

A.7 Global mass test during an active phase

For Phase II:

${\dot{m}}_L=-{\dot{m}}_{\mathrm{i}\mathrm{n}},$

${\dot{m}}_H={\dot{m}}_{\mathrm{i}\mathrm{n}}-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}},$

${\dot{m}}_{SC}={\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}.$

Summing:

Failed to parse (syntax error): {\displaystyle \boxed{\dot M_{\mathrm{tot}}=0}.}

Phase IV gives exactly the same result by symmetry.

A.8 Global energy test during an active phase

For Phase II:

${\dot{U}}_L=-P_L{\dot{V}}_L-{\dot{m}}_{\mathrm{i}\mathrm{n}}{h}_L,$

${\dot{U}}_H={\dot{m}}_{\mathrm{i}\mathrm{n}}{h}_L-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{h}_H+{\dot{Q}}_H,$

${\dot{U}}_{SC}={\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{h}_H+{\dot{Q}}_C-P_{SC}{\dot{V}}_S.$

The internal enthalpy fluxes cancel exactly:

$-{\dot{m}}_{\mathrm{i}\mathrm{n}}{h}_L+{\dot{m}}_{\mathrm{i}\mathrm{n}}{h}_L=0,$

$-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{h}_H+{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{h}_H=0.$

What remains is:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot U_{\mathrm{tot}} =\dot Q_C+\dot Q_H -P_L\dot V_L-P_{SC}\dot V_S }.}

Phase IV provides the symmetric relation. In the closed phases, the same structure follows directly from summing the balances of the two pairs. Hence, for any phase:

Failed to parse (syntax error): {\displaystyle \boxed{ \dot U_{\mathrm{tot}} =\dot Q_C+\dot Q_H -P_S^\star\dot V_S-P_L^\star\dot V_L }.}

Integrated over a periodic cycle, this relation gives:

Failed to parse (syntax error): {\displaystyle \boxed{Q_C+Q_H=W_{\mathrm{cycle}}}.}

A.9 Continuity at transitions

At the instant of a check-valve event, no finite mass or energy can be transferred in zero time. The conserved variables and geometry are therefore continuous:

$m_i^{+}=m_i^{-},U_i^{+}=U_i^{-},V_i^{+}=V_i^{-}.$

For an ideal gas:

$T_i=\frac{U_i}{m_i{C}_v},P_i=\frac{m_{i}R{T}_i}{V_i},$

which implies:

Failed to parse (syntax error): {\displaystyle \boxed{T_i^+=T_i^-,\qquad P_i^+=P_i^-}.}

The event only creates a change in hydraulic topology and in the active system of equations.