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

Answer:

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

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,

⇒  $E_{in}=E_{out}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}$

$e_{1}\dot{v_{1}}h_{1}+e_{2}\dot{v_{2}}h_{2}=e_{3}\dot{v_{3}}h_{3}$

$\frac{P_{1}}{RP_{1}}\dot{v_{1}} \ C_{p}T_{1}+ \frac{P_{2}}{RP_{2}}\dot{v_{2}} \ C_{p}T_{1}=\frac{P_{3}}{RP_{3}}\dot{v_{3}} \ C_{p}T_{3}$

⇒  $\dot{v_{3}}=\dot{v_{1}}+\dot{v_{2}}$

         $=\frac{\dot{m_{1}}}{e_{1}}+\frac{\dot{m_{2}}}{e_{2}}$

On substituting the values, we get

         $=\frac{10}{10\times 10^5}\times 8314\times 293+\frac{20\times 8314\times 313}{10\times 10^5}$

         $=76.4 \ m^3/s$