Exhaled air is ordinarily at body temperature and 100 humidi

Exhaled air is ordinarily at body temperature and 100% humidity (i.e. it is fully saturated with water). Estimate the mass of H₂O that is lost each day because of normally breathing. The saturated vapour pressure of water at room temperature (25 ^oC) and body temperature (37 ^oC) and 25 mmHg and 47 mmHg respectively. You may assume the temperature outside you body is RT with a relative humidity of 90% and the density of water vapour is 0.8 g/L.

Table 22.2 Effects of Breathing Rate and Depth on Alveolar Ventilation of Three Hypothetical Patients EFFECTIVE DEAD SPACE TIDAL ALVEOLAR MINUTE RESPIRATORY VENTILATION VENTILATION VENTILATION BREATHING PATTERN OF VOLUME VOLUME (TV) (MVR) (AVR) (DSV) (AVR/MVR) HYPOTHETICAL PATIENT RATE 150 ml 7000 ml/min 500 ml I Normal rate and depth 2 /min 10,000 ml/min 70% 10min l slow, deep breathing 150 ml 1000 ml 10,000 ml/min 8500 ml/min 85% 10,000 ml/min 4000 ml/min 40% Ill Rapid, shallow breathing 150 ml 40/min 250 m

Solution

Answer:

The equations required for the calculation are as under:

ACC = dMA/dt (P_H2Oair/P_T) - dMA/dt (P_H2Obody/P_T)

ACC = dVA/dt (_air) (P_H2Oair/P_T) - dVA/dt (_body)(P_H2Obody/P_T)

dVA/dt = 7.5 L/min (1440min/day)

= 10800L/day

P_T = 760 mmHg

P_O2air = 25 mmHg

P_O2body = 47 mmHg, (Considering that _air and _body can be ignored)

Thus, putting these values in the equation, ACC can be calculated.