Question

30 kg/s of superheated steam at 800 kPa, 200°C is mixed with saturated water at 800 kPa in a desuperheater to produce dry saturated steam at 800 kPa. Determine the flow rate of saturated water required.

Answer 1

`\dot{m}` = 3.8 kg/s

Step-by-step explanation:

From steam table, we can find :

The enthalpy of the super heated steam at 800 kPa or 8 bar is

h₁ = 2838.6 kJ/kg

or H₁ = 2838.6 x 30 85158 kJ

Enthalpy of the saturated water at 8 bar is

h₂ =
`h_(f)` = 33.6 kJ/kg

Now let the flow of saturated water is
`\dot{m}`

then h₂ = H₂ = 33.6
`\dot{m}` kJ

Enthalpy of dry saturated steam at 8 bar is

h₃ =
`h_(g)` = 2516.2 kJ/kg

h₃ = H₃ =2516.2 x ( 30+
`\dot{m}` ) kJ

Now we know from energy balance equation,

H₃ = H₁ + H₂

2516.2 x ( 30+
`\dot{m}` ) = 85158 + 33.6
`\dot{m}`

75486 + 2516.2
`\dot{m}` = 85158 + 33.6
`\dot{m}`

2516.2
`\dot{m}` + 33.6
`\dot{m}` = 85158-75486

2549.8
`\dot{m}` = 9672

`\dot{m}` = 3.7932

`\dot{m}` = 3.8 kg/s