2. Determine the quality of a two-phase liquid-vapor mixture of two-phase liquid

Question

2. Determine the quality of a two-phase liquid-vapor mixture of two-phase liquid- vapor mixture of Propane at 14.0 bar, v = 0.01 5 m'/kg R-134a at-50F, v = 1.0 ftnbm c. Ammonia at 35 psi, h- 564.22 Btu/lb 3. A closed rigid tank of volume 3 m contains a two-phase liquid-vapor mixture of H,0 at 110°C with a quality of 34.5%. The tank then receives energy by heat transfer until the water becomes a saturated vapor. Determine: a· b. the final pressure and temperature in the tank the mass of water in the tank c. the amount of heat transfer to the tank during this process

Explanation / Answer

2)

a)

Write the saturated properties at the given pressure P = 14 bar

vf = 0.002148 and vg = 0.03231

Now

v = (1-x)*vf + x*vg

x =0.426

b) similarly note properties at -5 F and find the value of x

x = 0.411

c)

At P = 35 psi

Note saturated properties of hf and hg and use the same expression to find x

x = 0.914

3)

Initial state

V = 3 m^3

T1 = 110 C

x = 0.345

Final state

Saturated vapor

Note the saturated properties at T = 110 C

Psat = 143 kPa

vf = 0.001052 and vg = 1.21

uf = 461.1 and ug = 2518

now at x = 0.345

v1 = (1-x)*vf + x*vg = 0.4182 m^3/kg

u1 = (1-x)*uf + x*ug = 1171 kJ/kg

a)

Since it is a closed rigid tank, the volume remains constant

v2 = v1 = 0.4182 m^3/kg = vg at state 2

and saturated vapor

From the properties tables

T = 148 C

P = 446 kPa

u2 = 2558 kJ/kg

b)

Mass of water in the tank is

m = V/v1 = 3/0.4182 = 7.174 kg

c)

The amount of heat transferred is

Q = m*(u2-u1)

There is no Pdv work because the volume is constant

Q = 7.174*(2558-1171) kJ

Q = 9950.4 kJ

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