Question

A person breathes 10 liters per minute of air that is at 68 degree and 50% relative humidity. [Relative humidity is the ratio of the actual vapor density to the saturation vapor density. The saturation vapor density is the maximum amount of water vapor that the air can hold at a particular temperature and is given in grams/meter cu bed] Given: Saturation vapor density at 20 deg. Celsius = 17.3 grams/meter cubed Saturation vapor density at 37 deg. Celsius = 44.0 grams/meter cubed How much water (grams) per minute must the internal membranes supply to saturate this air at normal body temperature? If all of this moisture is subsequently exhaled, how much (net) water per day (grams/day) is given off by the body in this process (of saturation)? In each gram of water extracts 580 calories (58Kcal or food calories) as it is vaporized, then how much daily heat loss in kilocalories does this represent?

Answer 1

volume of the air, V = 10 L/min

convert L/min into L/day -

V = 14400 L/day

Total volume exchanged per day is given by,

V = 14.4 m³/day

Saturation vapor density at 20 ⁰C = 17.3 g/m³

Saturation vapor density at 37 ⁰C = 44 g/m³

(a) The amount of water per minute by an internal membranes supply to saturate the air at 20⁰C which is given as :

Inhaled air is 50% saturated at 20 ⁰C, that means water content is -

_$\rho$in = (0.5) (17.3 g/m³)

_$\rho$in = 8.65 g/m³

So, the net flow of water into the body is given by -

m_in = _$\rho$in V = (8.65g/m³) (0.01 m³/min)

m_in = 0.0865 g/min

(b) The amount of water per day by an internal membranes supply to saturate the air at 37⁰C which is given as :

Exhaled air is 100% saturated at 37 ⁰C, that means water content is -

_$\rho$ex = 44 g/m³

So, the net flow of water into the body is given by -

m_ex = _$\rho$ex V = (44 g/m³) (14.4 m³/day)

m_ex = 633.6 g/day

(c) If each gram of water extracts 580 calories as it is vaporized, then the amount of daily heat loss in kilocalories which is given as :

mass of water evaporated at an internal membranes is given by,

$\Delta$m = m_ex - m_in = (633.6 g/day) - (124.56 g/day)

$\Delta$m = 509.04 g/day

Now, using an equation :

$\Delta$Q = $\Delta$m . $\Delta$H_vap

$\Delta$Q = (509.04 g/day) (580 cal)

$\Delta$Q = 295.2 kcal/day