Question

An amount of liquid water at 30oC and 1MPa (state 1) is to be mixed with superheated steam at 1MPa and 250oC (state 2) to yield saturated liquid at 1MPa (state 3). If the state 2 mass flow rate is to be 1kg/s, determine the other 2 flow rates.

Answer 1

stream 1:

T = 30 C , P = 1 MPa

specific enthalpy = Cp(T-0)...................base temperature is 0 C in the steam tables

.: H1 = 4.187*(30 - 0) = 125.61 kJ/kg

Stream 2:

phase: Superheated steam

From superheated steam table at T = 250 C and P = 1 MPa,

specific enthalpy H2 = 2943.22 kJ/kg

Stream 3:

phase: staurated liquid

P = 1 MPa

From saturated steam table at P = 1 MPa,

specific enthalpy of satuarted liquid Hl = 762.683 kJ/kg

.: H3 = 762.683 kJ/kg

Now,

By mass balance,

m1 + m2 = m3

Given: m2 = 1 kg/s

.: m1 + 1 = m3 ...........(1)

By heat balance,

Assuming there is no heat loss,

m1H1 + m2H2 = m3H3

.: m1*125.61 + 1*2943.22 = m3*762.683.............(2)

On solving (1) and (2), we get,

m3 = 4.42274 kg/s

m1 = 3.4227 kg/s