Question

Saturated humid air at 1 atm and 10°C is to be mixed with atmospheric air at 1 atm, 32°C, and 80 percent relative humidity to form air of 70 percent relative humidity. Determine the proportion at which these two streams are to be mixed and the temperature of the resulting air.

Answer 1

Answer:

Explanation:

From the information given in the question:

The pressure = 1 atm

The saturated humid  air temperature $T_1 = 10^0 \ C$

The saturated humid air relative humidity $\phi_1$ = 100%

The atmospheric air temperature $T_2$ = 32°C;   &

The atmospheric relative humidity $\phi_2$ = 80%

The data obtained at 1 atm pressure from property psychometric chart at $T_1$ = 10°C

$h_1 = 29.4 \ kJ/kg$ of air ; $\omega _1$ = 0.0077  kg/kg  of air

At $T_2= 12^0 \ C$

$h_2 = 94.6 \ kJ/kg$ of air;  $\omega _2 = 0.024 \ kg/kg \ of \ air$

If we take a look at the expression used in combining the conservation of energy and mass for adiabatic mixing of two streams; we have:

$\dfrac{m_1}{m_2}= \dfrac{\omega_2 -\omega _3}{\omega _3-\omega _1}= \dfrac{h_2-h_3}{h_3-h_1}$

$\dfrac{m_1}{m_2}= \dfrac{0.024 -\omega _3}{\omega _3-0.0077}= \dfrac{94.6-h_3}{h_3-29.4}$

The mixture temperature $T_3$ is determined through a trial and error method.

At trial and error method  $T_3$ = 24°C

From the relative humidity of 70%;

From the psychometric chart;

The specific humidity $\omega _3$ = 0.0143 kg/kg of air

The enthalpy $h_3$ = 57.6  kJ/kg  of air

Then;

$\dfrac{m_1}{m_2}=1.3$

Thus, 1.3 is the proportion at which the two streams are being mixed.