Question

4.1A membrane made of 0.1 mm thick soft rubber separates pure O2 at 1 atm and 25oC from air at 1.2 atm pressure. Determine the mass flow rate of O2 through the membrane per unit area and the direction of flow.

Molar flow rate of oxygen is given by: ???? = ???????? / ??? (????1 ? ????2 )

where S is solubility of oxygen in rubber.

The mass diffusivity of oxygen in rubber at 298 K is DAB = 2.1 x 10-10 m2 /s. The solubility of oxygen in rubber at 298 K is 0.00312 kmol/m3 bar. The molar mass of oxygen is 32 kg/kmol. The molar fraction of oxygen in air is 21%.

4.2One way of increasing heat transfer from the head on a hot summer day is to wet it. This is especially effective in windy weather, as you may have noticed. Approximating the head as a 30 cm diameter sphere at 30 degrees with an emissivity of 0.95, determine the total rate of heat loss from the head at ambient air conditions of 1 atm, 25 degrees, 40% relative humidity, and 25 km/h winds if the head is dry. (13.Take the surrounding temperature to be 25 degrees.

The properties of air at the free stream temperature of 25oC and 1 atm are:

k = 0.0255 W/mC

Pr = 0.73

u= 1.85 x 10-5 kg/m.s

v = 1.56 x 10-5 m2 /s Also,

u= u@30 degrees = 1.87 x 10-5 kg/m.s

The mass diffusivity of water vapour in air is given by: ?????? = 1.87??10?10 ?? ^2.072 / ??

where T and P should be in Kelvins and atmospheres to get DAB in m2 /s.

The saturation pressure of water at 25degrees is Psat@25°C = 3.169 kPa. Properties of water at 30degrees are: hfg = 2431 kJ/kg Pv = 4.246 kPa The gas constants of dry air and water are Rair = 0.287 kPa.m3 /kg.K and Rwater = 0.4615 kPa.m3 /kg.K.

The emissivity of the head is given to be 0.95. NB: When the head is dry, heat transfer from the head is by forced convection and radiation only.

Nusselt number is determined using:???? = 2 + [0.4????^1?2 + 0.06????^2?3 ]????0.4 ( ??? ???? ) ^0.4

What would happen to the heat loss and why if the head were wet?

Answer 1