Question

A non-mechanical, rigid, and fully insulated mixing tank is used to combine two inflows to produce wet steam. One inflow consists of superheated steam flowing at 3 kg/s, 300+C and 8 bar, while the other inflow is saturated liquid water at 25+C. The combined stream exiting the tank consists of wet steam at 2 bar and 90% quality. What is the mass rate of wet steam that can be produced and what is its temperature? How much mass inflow of saturated liquid water is needed?

Answer 1

Explanation:

As it is given that stream A is super heated. From stream tables, we get that specific enthalpy, ($h_{A}$) is 3054.29 kJ/kg.

For stream B, it is saturated water at 25 degree celsius. It's $h_{B}$ is 2546.54 kJ/kg.

Stream C, data will be as follows.

                P = 200 kPa,         $\chi$ = 0.9

So, $h_{c}$ = $h_{f} + \chi \times h_{fg}$  

        = 504.47 + 0.9 \times 2201.7

        = 2486 kJ/kg

Now, energy balance formula will be as follows.

            $m_{A}h_{A} + m_{B}h_{B}$ = $(m_{A} + m_{B})h_{c}$

              $3 \times 3054.29 + m_{B} \times 2546.54$ = $(3 + m_{B}) \times 2486$

             $m_{B}$ = 28.16 kg/s

Hence, inflow of saturated liquid is 28.16 kg/s. According to steam table, temperature of wet steam is $120.23^{o}C$

Mass flow rate of out flow is 31.16 kg/s.