Question

For H2O, determine the specified property at the indicated state.

(a) T = 140°C, v = 0.5 m3/kg. Find p, in bar.
(b) p = 30 MPa, T = 80°C. Find v, in m3/kg.
(c) p = 10 MPa, T = 600°C. Find v, in m3/kg.
(d) T = 80°C, x = 0.4. Find p, in bar, and v, in m3/kg.

For H2O, determine the specific volume at the indicated state, in m3/kg.

(a) T = 440°C, p = 20 MPa.
(b) T = 160°C, p = 20 MPa.
(c) T = 40°C, p = 2 MPa.

Answer 1

Answer:

3.613 bar

1.0290 m3/kg

0.038378 m3/kg

1.363 m3/kg

0.012166 m3/kg

0.002083 m3/kg

0.001177 m3/kg

Explanation:

part a

Table A-3: vf = 1.0435 x 10-3 m 3/kg , vg = 1.673 m 3 /kg. Since vf < v < vg, the state is in the two phase liquid-vapor region, as shown.

From Table A-3, the pressure is the saturation is pressure at 140 C: p = 3.613 bar.

Answer: p = 3.613 bar

part b

The pressure is higher than the critical  pressure, as shown on the diagram. Hence, the  state is in the compressed liquid region.

From Table A-5: v = 1.0290 m3 /kg.

Answer: v = 1.0290 m3 /kg

part c

Since the temperature is higher than Tsat at 100  bar, the state is super-heated vapor. T sat = 311.1 C

Interpolating in Table A-4, we get

v = 0.038378 m3/kg

Answer: v = 0.038378 m3/kg

part d

vx = vf + x(vg – vf)

v = 1.0291 x 10-3 + (0.4)(3.407 - 1.0291 x 10-3 )

v = 1.363 m3/kg

Answer: v = 1.363 m3/kg

a ) T = 440 C , p = 20 MPa

Tsat(@20 MPa) = 365.75 C

T > Tsat(@20 MPa) hence, super-heated steam

Interpolate the results

v = 0.012166 m3/kg

Answer: v = 0.012166 m3/kg

b) T = 160 C , p= 20 MPa

Tsat(@20 MPa) = 365.75 C

T < Tsat(@20 MPa) hence, compressed liquid region

v = 0.002038 m3/kg

Answer: v = 0.002038 m3/kg

c) T = 40 C , p = 2 MPa

Tsat(@2 MPa) = 212.38 C

T < Tsat(@2 MPa) hence, compressed liquid region

v = 0.001177 m3/kg

Answer: v = 0.001177 m3/kg

Answer 2

The properties of water and steam used to do work can be determined from the steam table, given input of the temperature, pressure, specific volume, as well as other thermodynamic variables

The responses to the required values of pressure and specific volume are as follows;

Part I

(a) Given T = 140°C, v = 0.5 m³/kg, to find p

From the steam tables, we have, at 140°, $v_f$ = 0.00107976, $v_g$ = 0.508519, $p_s$ = 3.61501

Given that the $v_f$ < v < $v_g$, the fluid has two phases and the experienced pressure $p_s$ = 3.61501 bar

(b) Given that at 30 MPa, and 80°C the steam is in the superheated heated region

From the single phase region of the steam tables, we have;

v = 1.01553 × 10⁻³ m³/kg

(c) Given that at p = 10 MPa = 100 bar, and T = 600°C, we have from the single phase region of the  steam table

v = 3.83775 × 10⁻² m³/kg

(d) At T = 80°C, and x = 0.4, we have;

v = 0.00102904 + 0.4 × (3.40527 - 0.00102904) ≈ 1.363

v = 1.363 m³/kg

Part II

(a) At T = 440 °C, p = 20 MPa. from the single phase region of the steam table, we have;

v = 1·22459 × 10⁻² m³/kg

(b) At T = 160°C, p = 20 MPa = 200 bar, the steam is in the superheated region, and we have;

v = 1.08865 × 10⁻³ m³/kg

(c) At T = 40°C and p = 2 MPa = 20 bar, we have

v = 0.00100700 m³/kg = 1.007 × 10⁻³ m³/kg

Learn more about the steam tables here;

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