**2.2 Modeling of the linear alternator**

The linear alternator consists of two main components, a stator and a translator. The permanent magnets are mounted on the stator and the translator is the moving portion of the machine which is made up of coils. A schematic of a three-phase, "U" shaped linear alternator with permanent magnet (PM) excitation is shown in Fig.3.

The FPLA operates on the same basic physical principles as conventional rotary alternators. The principle that governs the voltage generating operation of the alternator is Faraday's law expressed as [14]:

$$
\varepsilon\_{ind} = -\frac{d\lambda}{dt} = -N\_{coll}\frac{d\phi}{dt} \tag{2}
$$

The permanent magnets create a magneto motive force (MMF) in the air gap between the stator and the winding coils as shown in Fig.4, and it can be described by the following mathematic equation:

Dimensionless Parametric Analysis of Spark Ignited Free-Piston Linear Alternator 275

<sup>2</sup>

<sup>1</sup> cos 0 *<sup>F</sup> <sup>x</sup> a M x dx*

 

> 2 *p*

.

 

> 

(5)

(7)

(8)

 

.

(6)

<sup>4</sup> sin sin 2

> 

  2

 

1 4 sin sin

<sup>4</sup> sin sin

0 0 <sup>4</sup> sin sin sin

*<sup>x</sup> Mx M*

*F p*

2 *p*

2 *p F p m x x Bx M x M <sup>B</sup>*

<sup>0</sup> <sup>4</sup> sin

*<sup>p</sup> B M m p <sup>g</sup>*

Both experimental measurements and numerical calculation by finite element method showed that the flux in the air gap of the PM exited linear alternator in Fig.4 could be

*d B x dA B x Hdx*

Then the total flux contained in the coil of one phase at random position *x* is described by

2

*p*

*<sup>x</sup> x N HB x dx N M dx g*

 

<sup>0</sup> <sup>8</sup> sin sin 2 *p*

*<sup>d</sup> dx HN M <sup>x</sup> dt g dt*

*x x p*

 

*coil coil <sup>p</sup> x x*

*HN M x*

*coil p*

  <sup>1</sup> <sup>0</sup> *<sup>F</sup> a M x dx* 

<sup>0</sup> <sup>0</sup>

<sup>1</sup> <sup>0</sup>

<sup>2</sup>

*F p*

<sup>1</sup> <sup>0</sup>

So the flux density in the air gap due to PM is:

*g g*

assumed to be sinusoidal supporting the above result [16]. Therefore, the flux contained in the differential element *dx* is:

 

<sup>0</sup>

<sup>8</sup> sin cos

*coil p*

0 2

Thus, the induced electromotive force produced in the coil of one phase is:

*g*

 <sup>2</sup>

*<sup>x</sup> b M x dx M*

Where,

Then

Where

the following equation:

Fig. 3. Schematic of a U shaped three-phrase linear alternator

Fig. 4. Model of the linear alternator

$$M\_F(\mathbf{x}) = \begin{cases} 0 & 0 \le \mathbf{x} < \frac{\tau - \tau\_p}{2} \\ M\_p & \frac{\tau - \tau\_p}{2} < \mathbf{x} < \frac{\tau + \tau\_p}{2} \\ 0 & \frac{\tau + \tau\_p}{2} < \mathbf{x} < \frac{3\tau - \tau\_p}{2} \\ -M\_p & \frac{3\tau - \tau\_p}{2} < \mathbf{x} < \frac{3\tau + \tau\_p}{2} \\ 0 & \frac{3\tau + \tau\_p}{2} < \mathbf{x} < 2\tau \end{cases} \tag{3}$$

Where *Mp*= *Hc*·*hm*.

The mean value of the MMF can be obtained from the single-order truncated Fourier series [15]:

$$M\_F\left(\mathbf{x}\right) = \frac{a\_0}{2} + a\_1 \cos\left(\frac{\pi\chi}{\pi}\right) + b\_1 \sin\left(\frac{\pi\chi}{\pi}\right) \tag{4}$$

Where,

274 Thermodynamics – Interaction Studies – Solids, Liquids and Gases

Fig. 3. Schematic of a U shaped three-phrase linear alternator

*F*

*M x*

0 0<x

 

 

*p*

*M*

 

 

 

<sup>0</sup>

2 *<sup>F</sup>*

3 0 <x<2 2

 

The mean value of the MMF can be obtained from the single-order truncated Fourier series

*<sup>a</sup> x x Mx a b* 

*p*

*M*

2

*p p*

 

*p p*

 

*p*

2 2

*p p*

(4)

  (3)

 <x< 2 2 3

 

0 <x< 2 2 3 3 <x<

*p*

1 1 cos sin

Fig. 4. Model of the linear alternator

Where *Mp*= *Hc*·*hm*.

[15]:

$$\begin{aligned} a\_0 &= \frac{1}{\pi} \int\_0^{2\pi} M\_F\left(x\right) dx = 0 \\\\ a\_1 &= \frac{1}{\pi} \int\_0^{2\pi} M\_F\left(x\right) \cos\left(\frac{\pi x}{\pi}\right) dx = 0 \\\\ b\_1 &= \frac{1}{\pi} \int\_0^{2\pi} M\_F\left(x\right) \sin\left(\frac{\pi x}{\pi}\right) dx = \frac{4}{\pi} M\_p \sin\left(\frac{\pi \tau\_p}{2\pi}\right) \end{aligned}$$

Then

h

m

$$M\_F\left(\mathbf{x}\right) = \frac{4}{\pi}M\_p \sin\left(\frac{\pi\tau\_p}{2\tau}\right)\sin\left(\frac{\pi\mathbf{x}}{\tau}\right).$$

So the flux density in the air gap due to PM is:

$$B(\mathbf{x}) = \frac{\mu\_0}{g} M\_F \left(\mathbf{x}\right) = \frac{\mu\_0}{g} \frac{4}{\pi} M\_p \sin\left(\frac{\pi \tau\_p}{2\tau}\right) \sin\left(\frac{\pi \chi}{\tau}\right) = B\_m \sin\left(\frac{\pi \chi}{\tau}\right) \tag{5}$$

Where

$$B\_m = \frac{\mu\_0}{\mathcal{S}} \frac{4}{\pi} M\_p \sin\left(\frac{\pi \tau\_p}{2\tau}\right).$$

Both experimental measurements and numerical calculation by finite element method showed that the flux in the air gap of the PM exited linear alternator in Fig.4 could be assumed to be sinusoidal supporting the above result [16].

Therefore, the flux contained in the differential element *dx* is:

$$d\phi = B(\mathbf{x})dA = B(\mathbf{x})Hdx \tag{6}$$

Then the total flux contained in the coil of one phase at random position *x* is described by the following equation:

$$\begin{split} \mathcal{A}(\mathbf{x}) &= \int\_{\mathbf{x}-\tau}^{\mathbf{x}} N\_{coli} H B(\mathbf{x}) \, d\mathbf{x} = \int\_{\mathbf{x}-\tau}^{\mathbf{x}} N\_{coli} \frac{\mu\_0}{g} \frac{4}{\pi} M\_p \sin\left(\frac{\pi \tau\_p}{2\tau}\right) \sin\left(\frac{\pi \mathbf{x}}{\tau}\right) d\mathbf{x} \\ &= -\tau H N\_{coli} M\_p \frac{\mu\_0}{g} \frac{8}{\pi^2} \sin\left(\frac{\pi \tau\_p}{2\tau}\right) \cos\left(\frac{\pi}{\tau} \mathbf{x}\right) \end{split} \tag{7}$$

Thus, the induced electromotive force produced in the coil of one phase is:

$$\varepsilon = -\frac{d\lambda}{dt} = \text{HN}\_{col} M\_p \frac{\mu\_0}{g} \frac{8}{\pi} \sin\left(\frac{\pi \tau\_p}{2\pi}\right) \sin\left(\frac{\pi}{\tau} x\right) \frac{d\lambda}{dt} \tag{8}$$

Dimensionless Parametric Analysis of Spark Ignited Free-Piston Linear Alternator 277

The thermodynamic model is derived based on the first law of thermodynamics and ideal gas law. It consists of the calculation of the process of scavenging, compression, combustion, expansion and exhaust. The zero-dimensional, single zone model is used to describe the

Appling the first law of thermodynamics and ideal gas law on the cylinder as an open thermodynamic system, shown in Fig.5, and assuming that the specific heat *c*V and the gas

*dU dV dQ <sup>p</sup> H H*

*dV p dt*

For the case of compression and expansion process neglecting the crevice flow and the

*dcT dV dQ m p dt dt dt*

Considering the cylinder content is ideal gas, and then at every instant the ideal gas law is

Substitution and mathematical manipulation yield the following equation which is used to

leakage, the first law of thermodynamics applied to the cylinder content becomes:

*<sup>V</sup>*

*in*

*dt dt dt*

. *Hi*

*dQ dt*

. . *i e*

*dU*

(15)

. *H <sup>e</sup>*

(16)

*in pV m RT* (17)

**2.3 Thermodynamic modeling** 

thermodynamic process.

constant *R* are constant:

*dt*

Fig. 5. Thermodynamic system of FPLA

calculate the in-cylinder pressure at each time step.

satisfied:

The induced current from the load circuit can be derived in the following equations:

$$
\varepsilon \left( t \right) = \left( R\_s + R\_L \right) i\_L \left( t \right) + L \frac{d i\_L \left( t \right)}{dt} \tag{9}
$$

$$i\_L\left(t\right) = \frac{\varepsilon\left(t\right)}{R\_s + R\_L} \left(1 - e^{-\frac{R\_s + R\_L}{L}t}\right) \tag{10}$$

The magnetic force has the opposite direction to the direction of the translator's movement. According to Ampere's law, it is described in the following equation:

$$F\_e = 2N\_{col}B\left(\mathbf{x}\right)i\_L H = 4H^2 N\_{col}^2 B\_m^2 \frac{\left(1 - e^{-\frac{R\_s + R\_L}{L}t}\right)}{R\_s + R\_L} \sin^2\left(\frac{\pi x}{\tau}\right)\frac{d\mathbf{x}}{dt} \tag{11}$$

When it comes to three-phase linear alternator, the third phase is derived from another two according to the following equation [17]:

$$\sin \phi = -\sin \left(\phi + \frac{2}{3}\pi\right) - \sin \left(\phi - \frac{2}{3}\pi\right) \tag{12}$$

So the total electromagnetic force produced by a three-phase linear alternator is:

$$\begin{split} F\_c &= 4H^2 N\_{col}^2 B\_m^2 \frac{\left(1 - e^{-\frac{R\_s + R\_l}{L}t}\right)}{R\_s + R\_l} \frac{dx}{dt} \Big(\sin^2\left(\frac{\pi x}{\tau} - \frac{2}{3}\pi\right) + \sin^2\left(\frac{\pi x}{\tau}\right) + \sin^2\left(\frac{\pi x}{\tau} + \frac{2}{3}\pi\right)\Big) \\ &= 6H^2 N\_{col}^2 B\_m^2 \Big(1 - e^{-\frac{R\_s + R\_l}{L}t}\Big) \frac{1}{R\_s + R\_l} \frac{dx}{dt} \\ &= M \Big(1 - e^{-\frac{R\_s + R\_l}{L}t}\Big) \frac{dx}{dt} \end{split} \tag{13}$$

Where

$$M = 6H^2 N\_{cal}^2 B\_m^2 \frac{1}{R\_s + R\_I}$$

In order to develop a first approach analysis, it's assumed that the alternator circuit works in resonance condition, which means that the current and voltage have the same phase, so that it is possible to consider the circuit as only a resistance [18]. Thus, the electromagnetic force is proportional to the speed of the translator:

$$F\_e = M \frac{d\chi}{dt} \tag{14}$$
