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?

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OlaMacgregor OlaMacgregor · Answer 1 · 2019-09-02T13:51:49+02:00

Explanation:

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

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

Stream C, data will be as follows.

P = 200 kPa, [tex]\chi[/tex] = 0.9

So, [tex]h_{c}[/tex] = [tex]h_{f} + \chi \times h_{fg}[/tex]

= 504.47 + 0.9 \times 2201.7

= 2486 kJ/kg

Now, energy balance formula will be as follows.

[tex]m_{A}h_{A} + m_{B}h_{B}[/tex] = [tex](m_{A} + m_{B})h_{c}[/tex]

[tex]3 \times 3054.29 + m_{B} \times 2546.54[/tex] = [tex](3 + m_{B}) \times 2486[/tex]

[tex]m_{B}[/tex] = 28.16 kg/s

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

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