Question

10 kg/s Propane at 10 bar and 20 C is directed to an adiabatic rigid mixer and is mixed with 20 kg/s Propane at 10 bar and 40 C. What is the final volumetric flow rate in (m3/s) of the resulting mixture.

Answer 1

The final volumetric flow rate will be "76.4 m³/s".

Step-by-step explanation:

The given values are:

`\dot{m_(1)}=10 \ kg/s`

`\dot{m_(2)}=20 \ Kg/s`

`T_(1)=293 \ K`

`T_(2)=313 \ K`

`P_(1)=P_(2)=P_(3)=10 \ bar`

As we know,

⇒
`E_(in)=E_(out)`

`\dot{m_(1)}h_(1)+\dot{m_(2)}h_(2)=\dot{m_(3)}h_(3)`

`e_(1)\dot{v_(1)}h_(1)+e_(2)\dot{v_(2)}h_(2)=e_(3)\dot{v_(3)}h_(3)`

`(P_(1))/(RP_(1))\dot{v_(1)} \ C_(p)T_(1)+ (P_(2))/(RP_(2))\dot{v_(2)} \ C_(p)T_(1)=(P_(3))/(RP_(3))\dot{v_(3)} \ C_(p)T_(3)`

⇒
`\dot{v_(3)}=\dot{v_(1)}+\dot{v_(2)}`

`=\frac{\dot{m_(1)}}{e_(1)}+\frac{\dot{m_(2)}}{e_(2)}`

On substituting the values, we get

`=(10)/(10* 10^5)* 8314* 293+(20* 8314* 313)/(10* 10^5)`

`=76.4 \ m^3/s`