A person breathes 10 liters per minute of air that is at 68

A person breathes 10 liters per minute of air that is at 68 degrees F 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 cubed} 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)? If 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?

Solution

Total volume exchanged per day is ccalculated as follows

V= 10 L/min= 14.4 m^3/day

Water content is

p =50% *17.3 = 8.65 g/m^3

Net flow of water into the body

mb=p*V=124.56 g/day

Water content is

p_ex= 44 g/m^3

Water flow into The body

m_ex = p_ex*V =633.6 g/day = 0.44 g/min

Mass of evaporated water

m=m_ex - m_b= 508.44 g/day = 0.354 g/min

The energy is

Q= m*h_VAP = 508.44*580 cal= 294895.2 cal/day = 294.9 kcal/day