Question

A stream of warm air with a dry-bulb temperature of 36°C and a wet-bulb temperature of 30°C is mixed adiabatically with a stream of saturated cool air at 12°C. The dry air mass flow rates of the warm and cool airstreams are 8 and 10 kg/s, respectively. Assuming a total pressure of 1 atm, determine (a) the temperature, (b) the specific humidity, and (c) the relative humidity of the mixture.

Answer 1

Answer:

Explanation:

  Given

$T_1\left ( DBT\right )=36^{\circ}C$

$WBT=30^{\circ}C$

$m_1=8 kg/s$

$m_2=10 kg/s$

Now total mass $m_3=sum of m_1+m_2$

$m_3=8+10=18 kg/sec$

Now from Steam table at DBT=$36^{\circ}C & WBT=30^{\circ}C$

$\omega _1=0.025$kg\kg of dry air

for saturated cool air at $12^{\circ}C$

Humidity ratio $\omega _2=0.0085$ kg\kg of dry air

Now

$m_1\omega _1+m_2\omega _2=m_3\omega _3$

$8\cdot 0.025+10\cdot 0.0085=18\cdot \omega _3$

$\omega _3=0.01585$kg/kg of dry air

Now

$m_1h_1+m_2h_2=m_3h_3$

$m_1C_pT_1+m_2C_pT_2=m_3C_pT_3$

$m_1T_1+m_2T_2=m_3T_3$

$8\cdot 36+10\cdot 12=18\cdot T_3$

$T_3=22.67^{\circ}C$

From Psychometric chart

$at t=22.67^{\circ}C & \omega _3=0.01585$ kg/kg of dry air

relative humidity ratio $\phi =0.89 or 89\%$