Question

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a velocity of 600 m/s. Determine the final temperature if the steam is superheated in the final state, and the quality if it is saturate

Answer 1

x2 = 0.99

Step-by-step explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

`\dot m( h1+(v1^2)/(2)+ gz1 )+ Q = \dot m( h2+(v2^2)/(2)+ gz1) + W`

`\dot m( h1+(v1^2)/(2)) = \dot m( h2+(v2^2)/(2))`

`h1+(v1^2)/(2) = h2+(v2^2)/(2)`

`2850.6 + [(50^2)/(2) * (1 kJ/kg)/(1000 m^2/S^2)] = h2 +[ (600^2)/(2) * (1 kJ/kg)/(1000 m^2/S^2)]`

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99