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

Step-by-step 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\%`