**9. ICU Evalution: Indicator for lung-status in mechanically ventilated copd patients (using lung ventilation modelling and assessment)**

In chronic-obstructive-pulmonary-disease (COPD), elevated airway resistance and decreased lung compliance (i.e. stiffer lung) make breathing difficult. After these patients are mechanically ventilated, there is a need for accurate predictive indicators of lung-status improvement, for ventilator discontinuation through stepwise reduction in mechanical support, as and when patients are increasingly able to support their own breathing, followed by trials of unassisted breathing preceding extubation, and ending with extubation.

So, in this section, we have provided a biomedical engineering analysis of the lung ventilator volume response to mechanical ventilation of COPD patients, and developed an index to assess the lung status as well as the basis of weaning the patient from ventilator support.

Figure 25 depicts the model for the lung volume (*V*) response to the net driving pressure *PN*=*PL*– *Pe(end-expiratory pressure)*, in which (i) the *driving pressurePL*= *Pm* (*pressure at the mouth*) minus *Pp* (*the pleural pressure*), and (ii) *Pp* is determined by intubating the patient, and assuming that the pressure in the relaxed esophageal tube equals the pressure in the pleural space surrounding it.

Fig. 25. Model of the Lung, depicting Pm ,Pp, Pel (lung elastic pressure recoil) = *Pa* (alveolar pressure) – Pp (pleural pressure) = 2T/(radius of alveolar chamber), and R (resistance to airflow rate) = (Pm - Pa) / (dV/dt ).

The equation representing the lung model response to the net driving pressure in terms of the model parameters lung compliance *(C)* and airflow-resistance *(R)*, is given by:

$$R\stackrel{\circ}{V} + \frac{V}{\underline{C}} = P\_L(t) - P\_e = P\_N(t) \tag{1}$$

wherein:


Fig. 26. Lung ventilatory model data shows air-flow ( *V* ) and volume (*V*) and net pressure (*PN*). Pause pressure (*Ptm*) occurs at *tm*, at which the volume is maximum (TV = tidal volume). Δ*t* is the phase difference between the time of maximum volume and peak pressure (*Pk*). It also the time lag between the peak and pause pressures. *B* is the amplitude of the net pressure waveform *PN* applied by the ventilator. This *PN* oscillates about *Pe* with amplitude of *B*. The difference between peak pressure *Pk* and pause pressure*Ptm* is Δ*p* [12]*.*

*o*

We measure Peak pressure (Pk), Pause pressure (Ptm), tm & ω (or tm ω). Then,

$$\begin{aligned} \mathbf{P\_k} &= \mathbf{P\_L} \begin{pmatrix} \mathbf{t} = \boldsymbol{\pi} \ \boldsymbol{\sigma} \ \boldsymbol{\omega} \boldsymbol{\sigma} \end{pmatrix} = \mathbf{P\_c} + \mathbf{B} \\ \mathbf{P\_m} &= \mathbf{P\_L} \begin{bmatrix} \mathbf{t} = \mathbf{t\_m} = (\boldsymbol{\pi} - \boldsymbol{\theta}) \ \boldsymbol{\sigma} \end{bmatrix} \text{ } \boldsymbol{\alpha} \mathbf{o} \end{aligned} = \mathbf{P\_c} + \mathbf{B} \sin \boldsymbol{\alpha} \mathbf{o} \ \mathbf{t\_m} = \mathbf{P\_c} + \mathbf{B} \sin \left[ \boldsymbol{\alpha} \mathbf{o} \ (\boldsymbol{\pi} \boldsymbol{\theta}) / \boldsymbol{\alpha} \right] \mathbf{o} \\ &= \mathbf{P\_c} + \mathbf{B} \sin(\boldsymbol{\pi} - \boldsymbol{\theta}) = \mathbf{P\_c} + \mathbf{B} \sin \boldsymbol{\theta} \\ \mathbf{r} \cdot \mathbf{B} &= (\mathbf{P\_k} - \mathbf{P\_m}) / (1 - \sin \boldsymbol{\theta}) = \Delta \mathbf{P} / (1 - \sin \boldsymbol{\theta}) \end{aligned}$$

*B* is the amplitude of the net pressure wave form applied by the ventilator. Let *Ca* be the average dynamic lung compliance, *Ra* the average dynamic resistance to airflow, the driving pressure *PL* =*Pe*+ *B* sin *(ωt),* and the net pressure *PN* = *B* sin *(ωt)*. The governing equation (1) then becomes [12]:

$$R\_a \stackrel{\circ}{V} + \frac{V}{C\_a} = P\_N = B \sin(\alpha t) \tag{2}$$

The volume response to *PN,* the solution to eqn (2), is given by:

$$V(t) = \frac{BC\_a \left\{ \sin(\alpha t) - \alpha k\_a \cos(\alpha t) \right\}}{1 + \alpha^2 k\_a^2} + He^{-\frac{\alpha}{k\_a}} \tag{3}$$

wherein:

34 Biomedical Science, Engineering and Technology

The equation representing the lung model response to the net driving pressure in terms of

() () *<sup>o</sup> L eN*

1. the driving pressure = PL ; Pe = the end-expiratory pressure; net pressure

2. the parameters of the governing equation (1) are lung compliance (C) and airflow-

*o*

(*PN*). Pause pressure (*Ptm*) occurs at *tm*, at which the volume is maximum (TV = tidal volume). Δ*t* is the phase difference between the time of maximum volume and peak pressure (*Pk*). It also the time lag between the peak and pause pressures. *B* is the amplitude of the net pressure waveform *PN* applied by the ventilator. This *PN* oscillates about *Pe* with amplitude of *B*. The difference between peak pressure *Pk* and pause pressure*Ptm* is Δ*p* [12]*.*

 *Pe Ptm*

We measure Peak pressure (Pk), Pause pressure (Ptm), tm & ω (or tm ω).

tm L m e m e

= = − =+ =

 ω

θ

P P [t = t ( ) / ] P Bsin t P +B sin [ (π-θ)/ ]

 θ  ω

 θ

*V* ) and volume (*V*) and net pressure

ω

 ω

*<sup>V</sup> RV t P PP <sup>t</sup> <sup>C</sup>* + = −= (1)

 *t (sec)*

 *t (sec)*

the model parameters lung compliance *(C)* and airflow-resistance *(R)*, is given by:

3. resistance(R), with both R & C being instantaneous values 4. V= V(t) – Ve (wherein Ve is the end expiratory lung volume)

V

Fig. 26. Lung ventilatory model data shows air-flow (

k L e

π ω

P P (t = ) = P B

= /2 +

= + − )= ∴ − − =Δ

k tm

e e

B = (P P ) /(1 sin ) P/(1-sin )

πθ

πθ

P Bsin( P +Bsin

wherein:

Then,

PN= PL– Pe


iv. ω is the frequency of the oscillating pressure profile applied by the ventilator An essential condition is that the flow-rate is zero at the beginning of inspiration and end of expiration. Hence, applying this initial condition of *dV/dt* =0 at t=0 to our differential eqn (3), the constant *H* is obtained as:

$$H = \frac{BC\_a a \phi k\_a}{1 + a^2 k\_a^2} \tag{4}$$

Then from eqn (3) & (4), we obtain:

$$V(t) = \frac{BC\_a \left\{ \sin(\alpha t) - \alpha k\_a \cos(\alpha t) \right\}}{1 + \alpha^2 k\_a^2} + \frac{BC\_a \alpha k\_a}{1 + \alpha^2 k\_a^2} e^{-\frac{\gamma}{k\_a}} \tag{5}$$

### **Evaluating parameters Ra & Ca [12,2]:**

For evaluating the parameter *ka(RaCa)*, we will determine the time at which *V (t)* is maximum and equal to the tidal volume (TV), Hence putting *dV/dt* =0 in eqn (5), we obtain:

From equation (6), we obtain (by neglecting *t k*/ *<sup>a</sup> e*<sup>−</sup> ), the following expression for *ka*:

*a*

$$k\_a = -1/\text{aton}(\text{ot}\_m) \tag{7a}$$

$$
\sigma\sigma\_{\prime} \quad \tan^{-1}\left(1/\,\alpha k\_{a}\right) = \pi - \left(\alpha t\_{m}\right) = \Theta \tag{7b}
$$

From eqn (5 & 6):

$$V(t = t\_m) = \text{TV} = \frac{\text{BC}\_a \left\{ \sin(\alpha t\_{\text{III}}) - \alpha k\_a \cos(\alpha t\_{\text{III}}) \right\}}{1 + \alpha^2 k\_a^2} + \frac{\text{BC}\_a \alpha k\_a}{1 + \alpha^2 k\_a^2} e^{-t m/ka} \tag{8}$$

At *t = tm,* the second term,

$$H = \frac{BC\_a \alpha k\_a}{1 + \alpha^2 k\_a^2} e^{-tm/ka} \approx 0\tag{9}$$

Hence, eqn (8) becomes:

$$V(t = t\_m) = \text{TV} = \frac{BC\_a \left\{ \sin(\alpha t\_\text{v}) - \alpha k\_a \cos(\alpha t\_\text{v}) \right\}}{1 + o^2 k\_a^2} \tag{10}$$

In eqn (10), if we put:

$$N = \sin(\alpha t\_\text{v}) - \alpha k\_a \cos(\alpha t\_\text{v}) \tag{11}$$

Then, based on equations (6 and 7), we get:

$$N = \frac{1}{\sqrt{1 + \alpha^2 k\_a^2}} + \frac{\alpha^2 k\_a^2}{\sqrt{1 + \alpha^2 k\_a^2}} = \sqrt{1 + \alpha^2 k\_a^2} \tag{12}$$

Now then, based on equation (12), equation (10) becomes:

Fig. 26. (reproduced)

36 Biomedical Science, Engineering and Technology

1 tan( ) *k t a m* = − ω

( ) () <sup>1</sup> *or*, tan 1 /ω

From eqn (5 & 6):

At *t = tm,* the second term,

Hence, eqn (8) becomes:

In eqn (10), if we put:

Then, based on equations (6 and 7), we get:

*N*

*m*

 ω

 πω

/ {sin( ) cos( )} ( ) TV <sup>1</sup> <sup>1</sup>

*a a m a*

*BC t t m m <sup>k</sup> BC k t k Vt t <sup>e</sup>*

<sup>−</sup> <sup>−</sup> === <sup>+</sup>

ω

ωω

2 2

*BC k t k H e k* ω

ω

*a*

{sin( ) cos( )} ( ) TV

sin( ) cos( ) *m m N tk t* = − ωω

22 22 1 1

+ +

ω

*k k a a*

ω

 ω

2 2 <sup>1</sup> 2 2 <sup>1</sup>

*BC t k t Vt t*

1

<sup>−</sup> = ≈

1

*m*

+

<sup>−</sup> ===

 θ*k t a m* <sup>−</sup> =− = (7b)

2 2 2 2

*k k*

*a a*

 ω

*a a a a m a*

 ω

+ +

/ 0

2 2

*k*

*a*

*m m a a*

ω

 ω

*ka ka*

ω

+

=+= (12)

ωω

+

(7a)

ω

(8)

(9)

(10)

ω

*<sup>a</sup>* (11)

$$V(\ t = t\_{\eta \eta}) = \text{TV} = \frac{BC\_{\bar{a}}}{\sqrt{1 + \alpha^2 \kappa\_a^2}} \tag{13}$$

### **Determining Lung-Compliance (Ca) and Air-Flow Resistance (Ra):**

From equations (13 and 7), we get:

$$\mathbf{C}\_{a} = \frac{\mathbf{T}\mathbf{V}\sqrt{1+\alpha^{2}}k\_{a}^{2}}{B} = \frac{\mathbf{T}\mathbf{V}}{B\sin\theta} = \frac{\mathbf{T}\mathbf{V}(1-\sin\theta)}{\Delta P \sin\theta} \tag{14}$$

Hence, from eqns (7 & 13), the average value of airflow-resistance (*Ra*) is:

$$\mathcal{R}\_a = k\_a \;/\,\mathcal{C}\_a = \frac{\Delta P \sin \theta (1/\,\partial \tan \theta)}{\text{TV}(1 - \sin \theta)} = \frac{\Delta P \cos \theta}{\text{TV} \, o(1 - \sin \theta)}\tag{15}$$

For our patients, the computed ranges of the parameters are:

$$\text{C}\_{\text{a}} = 0.020 - 0.080 \text{ L} / \text{cm} \text{H}\_{2}\text{0} \tag{16}$$

R 9 43 cmH 0 s /L a 2 =− ⋅

Now, that we have determined the expressions for *Ra* and *Ca*, the next step is to develop an integrated index incorporating these parameters.

**Formulating a Lung Ventilatory Index (LVI) incorporating Ra and Ca:** We now formulate a Lung Ventilatory Index (LVI), incorporating Ra and Ca, as:

$$LVI = \frac{R\_d (RF) P\_k}{C\_d (TV)} \tag{17}$$

Let us now obtain order-of-magnitude values of *LVI,* for a mechanically ventilated COPD patient in acute respiratory failure:

$$\begin{aligned} \text{LVI(Intulated COPD)} &= \frac{[15 cm H\_2 \text{Os/L}][0.33 \text{s}^{-1}][20 cm H\_2 \text{O}]}{[0.035 \text{L}/cm H\_2 \text{O}][0.5 \text{L}]} \\ &= 5654 \text{ (cm} H\_2 \text{O/L)}^3 \end{aligned}$$

wherein

$$\begin{aligned} \text{CR}\_{\text{d}} &= 15 \,\text{cm} \,\text{H}\_{2}\text{Os} / \,\text{L} & \qquad \text{C}\_{\text{d}} = 0.035 \,\text{L} / \,\text{cm} \,\text{H}\_{2}\text{O} & \qquad \text{RF} = 0.33 \,\text{s}^{-1} \\ \text{TV} &= 0.5 \,\text{L} & \qquad \text{P}\_{\text{k}} = 20 \,\text{cm} \,\text{H}\_{2}\text{O} \end{aligned}$$

Now, let us obtain an order-of-magnitude of *LVI* (by using representative computed values of *Ra, Ca, RF,* TV*,* and*Pk*) as abovefor a COPD patient with improving lung-status just before successful discontinuation.

$$\begin{aligned} \text{LVI(Output COPD)} &= \frac{[10cmH\_2\text{Os}/L][0.33s^{-1}][12cmH\_2O]}{[0.05L/cmH\_2O][0.35L]} \\ &= 2263 \text{(cm}H\_2O/L)^3 \end{aligned}$$

wherein

$$\begin{aligned} P\_{\text{fl}} &= 10 \, cmH\_{\text{2}} \, \text{Os} \,/\, L & \quad & C\_{\text{a}} = 0.050L \,/\, cmH\_{\text{2}}O & \quad & RF = 0.33 \,\text{s}^{-1} \\ \text{TV} &= 0.35L & \quad & P\_{\text{k}} = 12 \, cmH\_{\text{2}}O \end{aligned}$$

Hence, for *LVI* to reflect lung-status improvement in a mechanically ventilated COPD patient in acute respiratory failure, it has to decrease to the range of *LVI* for an outpatient COPD patient at the time of discontinuation.

In fig. 27, it is shown that for the 6 successfully-discontinued cases, the *LVI* was (2900) ± (567) (*cmH*2*O/L*)3; for the 7 failed-discontinuation cases the *LVI* was (11400) ± (1433) (*cmH*2*O/L*)3. It can be also observed that *LVI* enables clear separation between failed and successful discontinuation, which again points to the efficacy of *LVI.* 

(6) Successful Extubation Case (7) Failed Extubation Case

Fig. 27. Distribution of *LVI* at discontinuation for patients with failed and successful discontinuation. For the 6 successfully-discontinued cases, the *LVI* was (2900) ± (567) (*cmH*2*O/L*)3; for the 7 failed-discontinuation cases the *LVI* was (11400) ± (1433) (*cmH*2*O/L*)3. It is observed that *LVI* enables clear separation between failed and successful discontinuation [12].
