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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.

Respuesta :

Answer:

Explanation:

  Given

[tex]T_1\left ( DBT\right )=36^{\circ}C[/tex]

[tex]WBT=30^{\circ}C[/tex]

[tex]m_1=8 kg/s [/tex]

[tex]m_2=10 kg/s[/tex]

Now total mass [tex]m_3=sum of m_1+m_2[/tex]

[tex]m_3=8+10=18 kg/sec[/tex]

Now from Steam table at DBT=[tex]36^{\circ}C & WBT=30^{\circ}C[/tex]

[tex]\omega _1=0.025[/tex]kg\kg of dry air

for saturated cool air at [tex]12^{\circ}C  [/tex]

Humidity ratio [tex]\omega _2=0.0085[/tex] kg\kg of dry air

Now

[tex]m_1\omega _1+m_2\omega _2=m_3\omega _3[/tex]

[tex]8\cdot 0.025+10\cdot 0.0085=18\cdot \omega _3[/tex]

[tex]\omega _3=0.01585 [/tex]kg/kg of dry air

Now

[tex]m_1h_1+m_2h_2=m_3h_3[/tex]

[tex]m_1C_pT_1+m_2C_pT_2=m_3C_pT_3[/tex]

[tex]m_1T_1+m_2T_2=m_3T_3[/tex]

[tex]8\cdot 36+10\cdot 12=18\cdot T_3[/tex]

[tex]T_3=22.67^{\circ}C[/tex]

From Psychometric chart

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

relative humidity ratio [tex]\phi =0.89 or 89\% [/tex]

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