Determine the phase or phases in a system consisting of H2O at the following conditions, sketch p-v and T-v diagrams showing the location of each state and determine the specified property at at the indicated state.

a.) p= 5 bar, T= 151.83C; and find the values of v and u at this state if x=0.6

b.) p= 1.25 MPa, T= 200C; and find the values of v and u at this state (Interpolation may be needed)

c.) p= 5 psia, T=124F; and find the values of v and u at this state (Interpolation may be needed)

d.) p=500 psia, T= 320F; and find the values of v and u at this state (Interpolation may be needed)

a) v = v_f + x*(v_g - v_f) = 0.225377 m^3 / kg

u = u_f + x*(u_g - u_f) = 1792.592 kJ/kg

b) v = 0.2060 + (0.1325 - 0.2060) *(1.25 - 1) / (1.5 - 1) = 0.16925 m^3/kg

u = 2621.9 + (2598.1 - 2621.9) *(1.25 - 1) / (1.5 - 1) = 2610 kJ/kg

C )v = 0.0162 ft^3 / lb and u = 92 Btu/lb

D)v = 0.0176 m^3/kg, u = 289 Btu / lb

Explanation:

a)

From steam properties, at p = 5 bar, we get T_sat = 151.83 deg C which is the same as given T. Hence, phase is liquid-vapor mixture.

Further,

v_f = 0.0010926 m^3/kg, and v_g = 0.3749 m^3/kg

u_f = 639.68 kJ/kg, and u_g = 2561.2 kJ/kg

v = v_f + x*(v_g - v_f) = 0.225377 m^3 / kg

u = u_f + x*(u_g - u_f) = 1792.592 kJ/kg

b)

From steam properties, at T = 200 deg C, we get P_sat = 15.54 bar = 1.554 MPa which is greater than 1.25 MPa. Hence, phase is superheated vapor.

From superheated tables, at 200 degg C and p = 1 MPa, we get v = 0.2060 m^3/kg and u = 2621.9 kJ/kg

From superheated tables, at 200 degg C and p = 1.5 MPa, we get v = 0.1325 m^3/kg and u = 2598.1 kJ/kg

By interpolation, At T = 200 deg C and p = 1.25 MPa we get

v = 0.2060 + (0.1325 - 0.2060) *(1.25 - 1) / (1.5 - 1) = 0.16925 m^3/kg

u = 2621.9 + (2598.1 - 2621.9) *(1.25 - 1) / (1.5 - 1) = 2610 kJ/kg

c)

From steam properties, at 5 psia we get T_sat = 162.2 F which is less than 124 F. Hence, phase is sub-cooled liquid.

we get v = 0.0162 ft^3 / lb and u = 92 Btu/lb

d)

p=500 psia, T= 320Fphase = subcooled liquid

v = 0.0176 m^3/kg, u = 289 Btu / lb

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