Question

For each case specify the state of the substance and provide the requested values. A) (8 points) Refrigerant 134a at T 10 C and v 0.025 m/kg, find P in kPa and u in B) (7 points) Air at P-1 bar and v-1 m3/kg, find u in kJ/kg. C) (8 points) Water at T 820 C and P 0.35 MPa, find h in kJ/kg. D) (7 points) Refrigerant 134a at T=4°C and u = 55.1 kJ/kg, find v in m3/kg. 1) kJ/kg.

Answer 1

Part a

Refrigerant 134a

Temperature T = 10 ° C,

Specific volume of fluid vf = 0.0007929 m3/kg

Specific volume of gas vg = 0.049466 m3/kg.

vf < v < vg

0.0007929 < 0.025 < 0.049466

R134a is saturated mixture

P = Psat = 414.89 kPa

Specific volume balance

v = vf + x (vg - vf)

0.025 = 0.0007929 + x (0.049466 - 0.0007929)

x = 0.497

Specific internal energy balance

u = uf + x (ug - uf) = 149.88 kJ/kg

Part b

From the ideal gas equation

PV = R T

T = PV/R

= (100 kPa x 1 m3/kg) / (0.287) = 348.43 K

From the interpolation between T(340 & 350 K) & u (242.82 & 250.02 kJ/kg) to determine u at 348.43 K

u = 248.89 kJ/kg

Part C

P = 0.35 MPa = 350 kPa

From the steam table, saturated temperature

T sat = 138.86 ° C

T (820) > Tsat (138.86)

Water is superheated vapor

At P = 0.3 MPa

And T = 800 ° C

h = 4159.3 kJ/kg

At P = 0.3 MPa

T = 900 ° C

h = 4397.3 kJ/kg

After interpolation

P = 0.3 MPa

T = 820 ° C

h = 4206.9 kJ/kg

At P = 0.4 MPa

T = 800 ° C

h = 4158.9 kJ/kg

At P = 0.4 MPa

T = 900 ° C

h = 4396.9 kJ/kg

After interpolation

P = 0.4 MPa

at T = 820 ° C

h = 4206.5 kJ/kg

After interpolation between 0.3 MPa and 0.4 MPa

P = 0.35 MPa.

T = 820 ° C

h = 4206.7 kJ/kg

Part d

R134a

T = 4°C

From the saturated properties

Internal energy of Saturated liquid

uf = 56.97 kJ/kg.

u(55.1) < uf (56.97)

R134a is compressed liquid .

For R134a, no data is available for compressed liquid

Use saturated data

T = 4°C

v = vf = 0.00078 m3/kg