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

Step-by-step 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 * 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 * 3054.29 + m_(B) * 2546.54` =
`(3 + m_(B)) * 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.