Thermodynamic and Mechanical Study/en

1. Scope and method

This study describes the complete thermodynamic cycle of the Dada engine in symbolic form, in both refrigeration and motor operation. 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 with the large cylinder at maximum volume, with:

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

The four gas volumes are:

• $S$: small cylinder;
• $L$: large cylinder;
• $H_i$: heat-in exchanger, located on the $S\to L$ hydraulic path and transferring heat into the gas;
• $H_o$: heat-out exchanger, located on the $L\to S$ hydraulic path and transferring heat out of the gas.

The hydraulic topology is fixed:

$S\to H_i\to L,$

$L\to H_o\to S.$

The passive check valves therefore always allow:

$H_i\to L,H_o\to S.$

The physical function of each heat exchanger is independent of the operating mode. What changes between refrigeration and motor operation is the external thermal reservoir connected to each exchanger. The cycle comprises four hydraulic/thermodynamic phases:

• Phase I — check valves closed, nominally adiabatic: compression on the $L+H_o$ side;
• Phase II — heat exchange, nominally isothermal: transfer $L\to H_o\to S$, with heat removed from the gas;
• Phase III — check valves closed, nominally adiabatic: expansion on the $S+H_i$ side;
• Phase IV — heat exchange, nominally isothermal: transfer $S\to H_i\to L$, with heat supplied to the gas.

These phase definitions describe the permanent physical topology of the machine. The external reservoir associated with each heat exchanger depends on the operating mode. The phase boundaries and their angular positions are determined independently in each operating mode by the actual check-valve events.

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 $k\in \{S,L\}$:

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

The swept volume is:

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

The thermodynamic volumes prescribed by the mechanism are:

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

with their signed derivatives:

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

The crank angular velocity is signed:

$\theta (t)=\omega t,{\dot{V}}_k=\omega \frac{d{V}_k}{d\theta },k\in \{S,L\}.$

The driven refrigeration direction is chosen as positive:

$\omega >0.$

Motor operation uses the opposite crank direction:

$\omega <0.$

The geometric origin is identical in both modes:

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

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

${\nu }_{V,k}=|{\dot{V}}_k|.$

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:

${\Lambda }_k(t)=\frac{V_{k,max}-V_k(t)}{V_{k,max}-V_{k,min}},k\in \{S,L\}.$

Thus ${\Lambda }_k=0$ corresponds to maximum volume and ${\Lambda }_k=1$ 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 $j\in \{S,L,i,o\}$:

$m_j,U_j,T_j,P_j,V_j.$

The subscripts $i$ and $o$ denote respectively the gas contained in $H_i$ and $H_o$.

The complete state vector is chosen as:

$𝐗=(m_S,U_S,m_L,U_L,m_i,U_i,m_o,U_o).$

Temperatures and pressures are derived from:

$T_j=\frac{U_j}{m_j{C}_v},P_j=\frac{m_{j}R{T}_j}{V_j}.$

The volumes $V_S(t)$ and $V_L(t)$ are prescribed by the kinematics and are not independent thermodynamic state variables. The volumes $V_i$ and $V_o$ 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:

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

By definition of the two physical heat exchangers:

$Q_i>0,Q_o<0$

in the intended operating regime of both refrigeration and motor operation.

The net cycle work distinguishes the two modes:

$W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}<0$

for driven refrigeration operation, whereas:

$W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}>0$

for motor operation.

3. Thermal closure and validity domain

3.1 Exchange with the thermal reservoirs

For the heat-in exchanger:

${\dot{Q}}_i=(UA{)}_i(T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}-T_i).$

In the intended operating regime:

$T_i<T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}\Rightarrow {\dot{Q}}_i>0.$

For the heat-out exchanger:

${\dot{Q}}_o=(UA{)}_o(T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}-T_o).$

In the intended operating regime:

$T_o>T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}\Rightarrow {\dot{Q}}_o<0.$

The reservoir temperatures depend on the operating mode.

For refrigeration operation:

$T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}=T_{\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{d}},T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}=T_{\mathrm{h}\mathrm{o}\mathrm{t}}.$

For motor operation:

$T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}=T_{\mathrm{h}\mathrm{o}\mathrm{t}},T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}=T_{\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{d}}.$

Thus the heat-transfer equations themselves are identical in both modes.

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:

${\epsilon }_P=\frac{|\Delta P_{\mathrm{i}\mathrm{n}\mathrm{t}}|}{P_{\mathrm{p}\mathrm{a}\mathrm{i}\mathrm{r}}}\ll 1,\Delta P_{\mathrm{i}\mathrm{n}\mathrm{t}}=P_{\mathrm{c}\mathrm{y}\mathrm{l}}-P_{HX}.$

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 $V=V_{\mathrm{c}\mathrm{y}\mathrm{l}}+V_{HX}$:

$\frac{dP}{dt}=\frac{(\gamma -1)(UA{)}_{HX}(T_{HX,\mathrm{r}\mathrm{e}\mathrm{s}}-T_{HX})-\gamma P{\dot{V}}_{\mathrm{c}\mathrm{y}\mathrm{l}}}{V_{\mathrm{c}\mathrm{y}\mathrm{l}}+V_{HX}}.$

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 \to 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:

${\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}}=\{\begin{array}{cc}\frac{N}{\gamma R{T}_{\mathrm{c}\mathrm{y}\mathrm{l}}},& N\ge 0(\mathrm{c}\mathrm{y}\mathrm{l}\to HX),\\ \frac{N}{\gamma R{T}_{HX}},& N<0(HX\to \mathrm{c}\mathrm{y}\mathrm{l}).\end{array}$

The heat-exchanger temperature evolves according to:

${\dot{T}}_{HX}=\frac{T_{HX}}{P}\dot{P}-\frac{R{T}_{HX}^2}{P{V}_{HX}}{\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}}.$

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 $T_{\mathrm{e}\mathrm{x}\mathrm{t}}$:

$\dot{P}=\frac{\gamma R{T}_{\mathrm{e}\mathrm{x}\mathrm{t}}{\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}+(\gamma -1)(UA{)}_{HX}(T_{HX,\mathrm{r}\mathrm{e}\mathrm{s}}-T_{HX})-\gamma P{\dot{V}}_{\mathrm{c}\mathrm{y}\mathrm{l}}}{V_{\mathrm{c}\mathrm{y}\mathrm{l}}+V_{HX}}.$

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:

$\dot{m}=\Phi (P_u,P_d,T_u,𝒢,ℱ)$

$u$ 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:

$\dot{m}=(C_{d}A{)}_{\mathrm{e}\mathrm{f}\mathrm{f}}{P}_u\sqrt{\frac{2\gamma }{R{T}_u(\gamma -1)}(r^{2/\gamma }-r^{(\gamma +1)/\gamma })}.$

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

$\dot{m}=(C_{d}A{)}_{\mathrm{e}\mathrm{f}\mathrm{f}}{P}_u\sqrt{\frac{\gamma }{R{T}_u}}{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}.$

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 $L+H_o$ and $S+H_i$ pairs are closed. The $L+H_o$ side is nominally compressed; the motion of the small piston remains that provided by the actual kinematics.

For $L+H_o$:

${\dot{P}}_{Lo}=\frac{(\gamma -1)(UA{)}_o(T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}-T_o)-\gamma P_{Lo}{\dot{V}}_L}{V_L+V_o}.$

For $S+H_i$:

${\dot{P}}_{Si}=\frac{(\gamma -1)(UA{)}_i(T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}-T_i)-\gamma P_{Si}{\dot{V}}_S}{V_S+V_i}.$

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

The transition to Phase II occurs when the $H_o\to S$ check valve satisfies its opening condition.

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

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

6.2.1 Donor cylinder L

The fundamental balance is:

$\frac{d(m_L{C}_v{T}_L)}{dt}=-P_L{\dot{V}}_L-{\dot{m}}_{L,\mathrm{o}\mathrm{u}\mathrm{t}}{C}_p{T}_L.$

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

$\frac{T_L}{T_{L,\mathrm{r}\mathrm{e}\mathrm{f}}}={\left[\frac{m_L}{m_{L,\mathrm{r}\mathrm{e}\mathrm{f}}}\frac{V_{L,\mathrm{r}\mathrm{e}\mathrm{f}}}{V_L}\right]}^{\gamma -1}$

and:

$P_L=P_{L,\mathrm{r}\mathrm{e}\mathrm{f}}{\left[\frac{m_L}{m_{L,\mathrm{r}\mathrm{e}\mathrm{f}}}\frac{V_{L,\mathrm{r}\mathrm{e}\mathrm{f}}}{V_L}\right]}^{\gamma }.$

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 Heat-out exchanger Ho

Mass conservation:

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

Fundamental energy balance:

$\frac{d(m_o{C}_v{T}_o)}{dt}={\dot{m}}_{\mathrm{i}\mathrm{n}}{C}_p{T}_L-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{C}_p{T}_o+(UA{)}_o(T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}-T_o).$

In expanded form:

${\dot{T}}_o=\frac{{\dot{m}}_{\mathrm{i}\mathrm{n}}(C_p{T}_L-C_v{T}_o)-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}R{T}_o+(UA{)}_o(T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}-T_o)}{C_v{m}_o}.$

The flow rates are determined by the hydraulic laws:

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

${\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}={\Phi }_{\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{v}\mathrm{e},\mathrm{o}}(P_o,P_{Si},T_o,\dots ).$

6.2.3 Receiving pair S+Hi

The flow from $H_o$ enters S. The pair pressure satisfies:

${\dot{P}}_{Si}=\frac{\gamma R{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{T}_o+(\gamma -1)(UA{)}_i(T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}-T_i)-\gamma P_{Si}{\dot{V}}_S}{V_S+V_i}.$

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

The transition to Phase III is the closing event of the $H_o\to S$ check valve.

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

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

For $S+H_i$:

${\dot{P}}_{Si}=\frac{(\gamma -1)(UA{)}_i(T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}-T_i)-\gamma P_{Si}{\dot{V}}_S}{V_S+V_i}.$

For $L+H_o$:

${\dot{P}}_{Lo}=\frac{(\gamma -1)(UA{)}_o(T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}-T_o)-\gamma P_{Lo}{\dot{V}}_L}{V_L+V_o}.$

The transition to Phase IV occurs when the $H_i\to L$ check valve satisfies its opening condition.

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

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

6.4.1 Donor cylinder S

The fundamental balance is:

$\frac{d(m_S{C}_v{T}_S)}{dt}=-P_S{\dot{V}}_S-{\dot{m}}_{S,\mathrm{o}\mathrm{u}\mathrm{t}}{C}_p{T}_S.$

For outflow guaranteed by the kinematic design:

$\frac{T_S}{T_{S,\mathrm{r}\mathrm{e}\mathrm{f}}}={\left[\frac{m_S}{m_{S,\mathrm{r}\mathrm{e}\mathrm{f}}}\frac{V_{S,\mathrm{r}\mathrm{e}\mathrm{f}}}{V_S}\right]}^{\gamma -1}$

and:

$P_S=P_{S,\mathrm{r}\mathrm{e}\mathrm{f}}{\left[\frac{m_S}{m_{S,\mathrm{r}\mathrm{e}\mathrm{f}}}\frac{V_{S,\mathrm{r}\mathrm{e}\mathrm{f}}}{V_S}\right]}^{\gamma }.$

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

6.4.2 Heat-in exchanger Hi

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

The fundamental energy balance is:

$\frac{d(m_i{C}_v{T}_i)}{dt}={\dot{m}}_{\mathrm{i}\mathrm{n}}{C}_p{T}_S-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{C}_p{T}_i+(UA{)}_i(T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}-T_i).$

In expanded form:

${\dot{T}}_i=\frac{{\dot{m}}_{\mathrm{i}\mathrm{n}}(C_p{T}_S-C_v{T}_i)-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}R{T}_i+(UA{)}_i(T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}-T_i)}{C_v{m}_i}.$

The flow rates are determined by:

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

${\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}={\Phi }_{\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{v}\mathrm{e},\mathrm{i}}(P_i,P_{Lo},T_i,\dots ).$

6.4.3 Receiving pair L+Ho

The flow from $H_i$ enters L. The pair pressure satisfies:

${\dot{P}}_{Lo}=\frac{\gamma R{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{T}_i+(\gamma -1)(UA{)}_o(T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}-T_o)-\gamma P_{Lo}{\dot{V}}_L}{V_L+V_o}.$

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

6.5 Motor operation

No second set of thermodynamic balance equations is required for motor operation.

The physical hydraulic topology remains:

$L\to H_o\to S,S\to H_i\to L.$

The check-valve directions remain:

$H_o\to S,H_i\to L.$

Motor operation is obtained by:

1. reversing the crank direction,
2. exchanging the external reservoirs connected to $H_i$ and $H_o$.

Thus:

$\omega <0,$

with the same cycle origin:

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

The thermal-reservoir assignment becomes:

$T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}=T_{\mathrm{h}\mathrm{o}\mathrm{t}},T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}=T_{\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{d}}.$

The heat-in exchanger therefore absorbs heat from the hot reservoir:

$Q_i>0,$

while the heat-out exchanger rejects heat to the cold reservoir:

$Q_o<0.$

The pressure histories, mass flow rates, check-valve events, and periodic thermodynamic state must be recalculated with the reversed kinematics.

The motor regime is obtained when:

$W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}>0.$

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:

$\Delta P_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}}\le \Delta P_{\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{n}}.$

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:

$m_j^{+}=m_j^{-},U_j^{+}=U_j^{-},V_j^{+}=V_j^{-}.$

For an ideal gas:

$T_j^{+}=T_j^{-},P_j^{+}=P_j^{-}.$

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

The two passive check valves have permanent physical orientations:

$H_o\to S,H_i\to L.$

Their opening and closing conditions are identical in both operating modes. Only the pressure histories change because the crank kinematics are reversed in motor operation.

8. Work, heat, and performance

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

$\dot{W}=P_S^{\star }{\dot{V}}_S+P_L^{\star }{\dot{V}}_L.$

$P_S^{\star }$ 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:

$W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}={\int }_0^{\tau }(P_S^{\star }{\dot{V}}_S+P_L^{\star }{\dot{V}}_L)dt.$

The exchanged heats are:

$Q_i={\int }_0^{\tau }(UA{)}_i(T_{i,\mathrm{r}\mathrm{e}\mathrm{s}}-T_i)dt,$

$Q_o={\int }_0^{\tau }(UA{)}_o(T_{o,\mathrm{r}\mathrm{e}\mathrm{s}}-T_o)dt.$

In periodic steady operation:

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

and the first law gives:

$Q_i+Q_o=W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}.$

The refrigeration COP is:

$CO{P}_c=\frac{Q_i}{-W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}}.$

The heat-pump COP is:

$CO{P}_h=\frac{-Q_o}{-W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}}=CO{P}_c+1.$

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

For motor operation:

$Q_i>0,Q_o<0,W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}>0.$

The thermal efficiency is:

${\eta }_{\mathrm{t}\mathrm{h}}=\frac{W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}}{Q_i}=1+\frac{Q_o}{Q_i}.$

The mean thermodynamic motor power is:

$\bar{\dot{W}}=\frac{W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}}{\tau }.$

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:

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}=m_S+m_L+m_i+m_o=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}.$

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:

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}=\frac{P_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}[V_S(0)+V_{L,max}+V_i+V_o]}{R{T}_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}}.$

$P_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}$ 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 maximum volume of the large cylinder with the same kinematic direction:

$𝐗(t+\tau )=𝐗(t).$

Geometric periodicity alone:

$V_k(t+\tau )=V_k(t),k\in \{S,L\}.$

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

• operating mode;
• reservoir temperatures $T_{\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{d}}$ and $T_{\mathrm{h}\mathrm{o}\mathrm{t}}$;
• signed crank angular velocity $\omega$.
• working fluid and reference properties $R$, $C_p$, $C_v$, $\gamma$;
• 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 $V_S(t)$, $V_L(t)$.

10.2 Design parameters

• $V_{S,min}$, $V_{S,max}$, $V_{L,min}$, $V_{L,max}$;
• $V_i$, $V_o$;
• $(UA{)}_i$, $(UA{)}_o$;
• 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

• $m_j,U_j,T_j,P_j,j\in \{S,L,i,o\}$;
• internal and external mass flow rates;
• ${\dot{Q}}_i$, ${\dot{Q}}_o$, $Q_i$, $Q_o$;
• $W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}$, $CO{P}_c$, $CO{P}_h$;
• ${\eta }_{\mathrm{t}\mathrm{h}}$ in motor operation.
• 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:

$\frac{d{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{dt}=0.$

A useful numerical residual is:

${\epsilon }_M(t)=M_{\mathrm{t}\mathrm{o}\mathrm{t}}(t)-M_{\mathrm{t}\mathrm{o}\mathrm{t}}(0).$

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:

$\frac{d{U}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{dt}={\dot{Q}}_i+{\dot{Q}}_o-P_S^{\star }{\dot{V}}_S-P_L^{\star }{\dot{V}}_L.$

A cumulative energy residual may be defined by:

${\epsilon }_E(t)=U_{\mathrm{t}\mathrm{o}\mathrm{t}}(t)-U_{\mathrm{t}\mathrm{o}\mathrm{t}}(0)-Q_i(0,t)-Q_o(0,t)+W(0,t).$

Over a periodic cycle:

$Q_i+Q_o=W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}.$

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:

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

If $\dot{Q}=0$:

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

then:

$P{V}^{\gamma }=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}.$

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

${\dot{T}}_{HX}=\frac{T_{HX}}{P}\dot{P}-\frac{R{T}_{HX}^2}{P{V}_{HX}}{\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}}.$

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

$\dot{P}=\frac{\gamma R{T}_{\mathrm{e}\mathrm{x}\mathrm{t}}{\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}+(\gamma -1)\dot{Q}-\gamma P{\dot{V}}_{\mathrm{c}\mathrm{y}\mathrm{l}}}{V_{\mathrm{c}\mathrm{y}\mathrm{l}}+V_{HX}}.$

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:

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

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:

$\frac{T}{T_0}={\left[\frac{m}{m_0}\frac{V_0}{V}\right]}^{\gamma -1}.$

Then, using $PV=mRT$:

$P=P_0{\left[\frac{m}{m_0}\frac{V_0}{V}\right]}^{\gamma }.$

and:

$\frac{T}{T_0}={\left(\frac{P}{P_0}\right)}^{(\gamma -1)/\gamma }.$

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:

$\dot{T}=\frac{{\dot{m}}_{\mathrm{i}\mathrm{n}}(C_p{T}_{\mathrm{i}\mathrm{n}}-C_{v}T)-{\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}RT+\dot{Q}}{C_{v}m}.$

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:

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

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

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

Summing:

${\dot{M}}_{\mathrm{t}\mathrm{o}\mathrm{t}}=0.$

Phase IV gives exactly the same result by symmetry.

A.8 Global energy test during an active phase

For Phase II:

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

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

${\dot{U}}_{Si}={\dot{m}}_{\mathrm{o}\mathrm{u}\mathrm{t}}{h}_o+{\dot{Q}}_i-P_{Si}{\dot{V}}_S.$

The internal enthalpy fluxes cancel exactly:

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

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

What remains is:

${\dot{U}}_{\mathrm{t}\mathrm{o}\mathrm{t}}={\dot{Q}}_i+{\dot{Q}}_o-P_L{\dot{V}}_L-P_{Si}{\dot{V}}_S.$

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:

${\dot{U}}_{\mathrm{t}\mathrm{o}\mathrm{t}}={\dot{Q}}_i+{\dot{Q}}_o-P_S^{\star }{\dot{V}}_S-P_L^{\star }{\dot{V}}_L.$

Integrated over a periodic cycle, this relation gives:

$Q_i+Q_o=W_{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}}.$

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:

$j\in \{S,L,i,o\}$

$m_j^{+}=m_j^{-},U_j^{+}=U_j^{-},V_j^{+}=V_j^{-}.$

For an ideal gas:

$T_j=\frac{U_j}{m_j{C}_v},P_j=\frac{m_{j}R{T}_j}{V_j},$

which implies:

$T_j^{+}=T_j^{-},P_j^{+}=P_j^{-}.$

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