Thermodynamic and Mechanical Study

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 two hydraulic branches have fixed circulation orientations:

$S\to H_i\to L,$

$L\to H_o\to S.$

Each branch contains one passive check valve. The check valve may be installed on either side of its heat exchanger. The two admissible arrangements for the heat-in branch are:

$S\to {\mathrm{C}\mathrm{V}}_i\to H_i\to L,$

or:

$S\to H_i\to {\mathrm{C}\mathrm{V}}_i\to L.$

Likewise, the heat-out branch may be arranged as:

$L\to {\mathrm{C}\mathrm{V}}_o\to H_o\to S,$

or:

$L\to H_o\to {\mathrm{C}\mathrm{V}}_o\to S.$

In all cases the check valve enforces the same overall circulation direction of its branch. Its position relative to the heat exchanger is a design parameter because, when the valve is closed, it determines which cylinder remains hydraulically connected to the exchanger volume.

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 is described by four hydraulic/thermodynamic stages: compression, transfer through $H_o$ from $L$ to $S$, expansion, and transfer through $H_i$ from $S$ to $L$. In refrigeration operation, starting from the reference origin used in this study, these stages are traversed in that order. Motor operation reverses the kinematics and traverses the thermodynamic cycle in the opposite direction; its chronology is detailed in §6.5.

The two transfer stages are intended to be quasi-isobaric: one cylinder empties while the other fills, and the pressure variation is intended to remain small compared with the pressure change during compression and expansion. Compression and expansion may involve simultaneous motion of both pistons in the same volumetric direction, so both cylinders may contribute to the pressure-changing stage.

The stage boundaries describe the kinematics and the dominant thermodynamic regime. They are not check-valve events. Check-valve opening and closing are determined independently by the local pressure difference across each valve; mass transfer and heat transfer may therefore continue during compression or expansion, and a valve event may occur inside a kinematic stage.

For comparison with an ideal thermodynamic cycle, compression and expansion may be idealized as adiabatic transformations; if they are also reversible, they are isentropic. These are reference transformations only. The real machine does not impose zero heat transfer, zero mass transfer, or closed check valves during compression and expansion, and no construction capable of enforcing perfectly adiabatic stages is assumed here.

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 characteristic piston motion over the cycle. 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 ${\Lambda }_S^{\ast }$ may be used as nominal kinematic reference levels at selected stage boundaries. The value actually observed at a check-valve event may be recorded as:

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

Check-valve events are determined by the local pressure difference across the valve and are independent of the selected kinematic stage boundaries; ${\Lambda }^{\ast }$ 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.

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.

The reduced equations below apply only when the cylinder and exchanger considered remain directly connected and sufficiently pressure-equalized. Which exchanger remains connected to which cylinder when a check valve is closed depends on whether that valve is installed upstream or downstream of the exchanger. The complete four-volume formulation in §6 does not require this reduction.

4.1 Closed-pair limiting 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 quasi-pressure-equalized pair

Consider a cylinder + heat-exchanger pair receiving an external mass flow ${\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}>0$ at temperature $T_{\mathrm{e}\mathrm{x}\mathrm{t}}$. At the level of the complete pair, the pressure equation is independent of whether the external stream enters the cylinder first or the heat exchanger first:

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

The pair mass always satisfies:

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

If the external stream enters the cylinder and ${\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ is positive from cylinder to heat exchanger:

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

If the external stream enters the heat exchanger first:

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

The local energy balances must use the actual plumbing order and upstream enthalpy. The complete formulation of §6 handles both check-valve positions without requiring a pair reduction.

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 instantaneous 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. On each of the two hydraulic branches, the check valve may be placed on either side of the heat exchanger; the hydraulic law must therefore use the pressures immediately adjacent to the actual valve position.

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

The complete model uses the four control volumes directly and does not change thermodynamic equations at stage boundaries. Define the signed mass flow rates:

${\dot{m}}_{Si},{\dot{m}}_{iL},{\dot{m}}_{Lo},{\dot{m}}_{oS},$

positive respectively in the directions:

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

Exactly one of the two links on each branch contains its passive check valve. A flow through that link cannot become negative; the other link may be bidirectional according to its hydraulic closure law.

For any signed link flow ${\dot{m}}_{ab}$, define the signed enthalpy flux in the named direction $a\to b$ as:

${\dot{H}}_{ab}=\{\begin{array}{cc}{\dot{m}}_{ab}{C}_p{T}_a,& {\dot{m}}_{ab}\ge 0,\\ {\dot{m}}_{ab}{C}_p{T}_b,& {\dot{m}}_{ab}<0.\end{array}$

This is simply the upwind-enthalpy rule written with a fixed sign convention.

The four mass balances are:

${\dot{m}}_S=-{\dot{m}}_{Si}+{\dot{m}}_{oS},$

${\dot{m}}_L={\dot{m}}_{iL}-{\dot{m}}_{Lo},$

${\dot{m}}_i={\dot{m}}_{Si}-{\dot{m}}_{iL},$

${\dot{m}}_o={\dot{m}}_{Lo}-{\dot{m}}_{oS}.$

The four energy balances are:

${\dot{U}}_S=-P_S{\dot{V}}_S-{\dot{H}}_{Si}+{\dot{H}}_{oS},$

${\dot{U}}_L=-P_L{\dot{V}}_L+{\dot{H}}_{iL}-{\dot{H}}_{Lo},$

${\dot{U}}_i={\dot{H}}_{Si}-{\dot{H}}_{iL}+{\dot{Q}}_i,$

${\dot{U}}_o={\dot{H}}_{Lo}-{\dot{H}}_{oS}+{\dot{Q}}_o.$

These equations remain valid throughout the entire cycle, independently of the kinematic stage, the instantaneous valve states, and the position of each valve relative to its heat exchanger.

6.1 Compression

During compression, the total enclosed cylinder volume decreases and the pressure rises. The two cylinders are not required to alternate: both may contract simultaneously,

${\dot{V}}_S<0,{\dot{V}}_L<0,$

over part or all of the compression stage.

The corresponding piston-work contributions then have the same sign and add in magnitude. Neither mass flow nor heat transfer is assumed to vanish. The check valves continue to respond only to their local pressure differences.

6.2 Transfer through Ho — L → Ho → S

The gas is transferred from $L$ toward $S$ through the heat-out branch. It passes through $H_o$, where heat is rejected by the gas:

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

The check valve on this branch may be located either before or after $H_o$. In both cases its permitted branch direction is $L\to H_o\to S$.

During the intended exchange regime, $L$ predominantly empties while $S$ fills. With sufficiently small hydraulic pressure differences and suitably matched piston motion, the transfer is approximately isobaric. Exact pressure equality is not imposed.

6.3 Expansion

During expansion, the total enclosed cylinder volume increases and the pressure falls. Both cylinders may expand simultaneously,

${\dot{V}}_S>0,{\dot{V}}_L>0,$

over part or all of the expansion stage.

The two piston-work contributions then have the same sign and add. As during compression, mass flow and heat transfer may remain active and the valve states are not prescribed by the stage definition.

6.4 Transfer through Hi — S → Hi → L

The gas is transferred from $S$ toward $L$ through the heat-in branch. It passes through $H_i$, where heat is supplied to the gas:

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

The check valve on this branch may be located either before or after $H_i$. In both cases its permitted branch direction is $S\to H_i\to L$.

During the intended exchange regime, $S$ predominantly empties while $L$ fills. With sufficiently small hydraulic pressure differences and suitably matched piston motion, the transfer is approximately isobaric. Exact pressure equality is not imposed.

6.5 Motor operation

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

The physical hydraulic branch directions remain:

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

Each check valve retains the direction of its branch, independently of whether it is installed before or after the corresponding heat exchanger.

Thus reversing the operating mode does not reverse either check valve.

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:

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

With the same geometric origin, the intended motor chronology is:

• low-pressure heat-out transfer $L\to H_o\to S$;
• compression;
• high-pressure heat-in transfer $S\to H_i\to L$;
• expansion.

The corresponding refrigeration chronology is traversed in the opposite thermodynamic direction:

• compression;
• high-pressure heat-out transfer $L\to H_o\to S$;
• expansion;
• low-pressure heat-in transfer $S\to H_i\to L$.

The qualifiers “low-pressure” and “high-pressure” therefore belong to the operating mode and stage, not permanently to either heat exchanger.

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, stage timings, 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 admissible hydraulic flow on the valve link; it does not define a kinematic stage boundary and 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 ${\Lambda }^{\ast }$ may serve as kinematic reference levels at selected stage boundaries; the values actually observed at check-valve events 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 branch orientations: one permits circulation along $L\to H_o\to S$, the other along $S\to H_i\to L$.

Each valve may be installed upstream or downstream of its heat exchanger. Its opening and closing conditions therefore use the pressures immediately adjacent to its actual position.

The opening and closing laws 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 stages from the actual cylinder pressures:

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

The net work over the cycle is:

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

When the two cylinder pressures are comparable during a pressure-changing stage,

$\dot{W}\approx P({\dot{V}}_S+{\dot{V}}_L).$

If both cylinders contract during compression, or both expand during expansion, their work contributions therefore add instead of partly cancelling. This allows a larger fraction of the available swept volume to participate in compression and expansion for a comparable pressure history. It does not by itself prove a higher thermal efficiency, because the heat transfers and the resulting pressure history change at the same time.

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 $\Delta P_{\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{n}}$, $\Delta P_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}}$;
• position of each check valve upstream or downstream of its heat exchanger;
• kinematic reference levels ${\Lambda }_L^{\ast }$, ${\Lambda }_S^{\ast }$, when used.

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 ${\Lambda }_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}$;
• 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:

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

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 quasi-pressure-equalized pair

For a pair receiving ${\dot{m}}_{\mathrm{e}\mathrm{x}\mathrm{t}}$ across its external boundary:

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

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

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

If the external stream enters the cylinder and ${\dot{m}}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ is positive from cylinder to heat exchanger:

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

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

If instead the external stream enters the heat exchanger first:

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

In both cases:

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

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

Using the signed link flows defined in §6:

${\dot{m}}_S=-{\dot{m}}_{Si}+{\dot{m}}_{oS},$

${\dot{m}}_L={\dot{m}}_{iL}-{\dot{m}}_{Lo},$

${\dot{m}}_i={\dot{m}}_{Si}-{\dot{m}}_{iL},$

${\dot{m}}_o={\dot{m}}_{Lo}-{\dot{m}}_{oS}.$

Summing the four equations cancels every internal mass flow exactly:

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

This result is independent of the active stage, the flow directions on bidirectional links, and the positions of the two check valves.

A.8 Global energy test

Using the signed enthalpy fluxes defined in §6:

${\dot{U}}_S=-P_S{\dot{V}}_S-{\dot{H}}_{Si}+{\dot{H}}_{oS},$

${\dot{U}}_L=-P_L{\dot{V}}_L+{\dot{H}}_{iL}-{\dot{H}}_{Lo},$

${\dot{U}}_i={\dot{H}}_{Si}-{\dot{H}}_{iL}+{\dot{Q}}_i,$

${\dot{U}}_o={\dot{H}}_{Lo}-{\dot{H}}_{oS}+{\dot{Q}}_o.$

All internal enthalpy fluxes cancel exactly when the four balances are summed:

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

Integrated over a periodic cycle:

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

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 changes the admissible hydraulic flow on the valve link; the thermodynamic state remains continuous.