Properties of saturated steam at the temperature or pressure you specified are listed in both Metric and Standard units. If you can't find the unit you are using, click the number of that property to convert.

Property Metric Unit Standard UnitTemperature (T) 100.0 C 212.0 F

Pressure (P) 1.0134 bar 14.698 psi

Density

Saturated Liquid (f)

958.27 kg/m3

59.823 lb/ft3

Saturated Vapor (g)

0.59770 0.037313

Specific Volume

Saturated Liquid (vf)

0.0010435 m3/kg

0.016716 ft3/lb

Saturated Vapor (vg)

1.6731 26.800

Internal Energy

Saturated Liquid (uf)

418.94

kJ/kg

180.11

Btu/lbEvaporated (ufg) 2087.6 897.50 Saturated Vapor

(ug)2506.5 1077.6

Enthalpy

Saturated Liquid (hf)

419.05

kJ/kg

180.16

Btu/lbEvaporated (hfg) 2256.9 970.3 Saturated Vapor

(hg)2676.1 1150.5

Entropy

Saturated Liquid (sf)

1.3068

kJ/kg-K(mayer)

0.31213 Btu/lb-REvaporated (sfg) 6.0483 1.4446

Saturated Vapor (sg)

7.3550 1.7567

Properties of Saturated Steam - SI Units

A Saturated  Steam Table with steam properties as specific volume, density, specific enthalpy and specific entropy

The steam table below list the properties of steam at varying pressures and temperatures:

Absolute pressure

Temperature(oC)

Specific Density - ρ -

Specific Enthalpy of Specific Entropy Liquid - Evaporation Steam -

Page 1 of 18

(kN/m2) Volume (kg/m3)hl -

(kJ/kg)- he -

(kJ/kg)hs -

(kJ/kg) of Steam - s -0.8 3.8 160 0.00626 15.8 2493 2509 9.058

2.0 17.5 67.0 0.0149 73.5 2460 2534 8.725

5.0 32.9 28.2 0.0354 137.8 2424 2562 8.396

10.0 45.8 14.7 0.0682 191.8 2393 2585 8.151

20.0 60.1 7.65 0.131 251.5 2358 2610 7.909

28 67.5 5.58 0.179 282.7 2340 2623 7.793

35 72.7 4.53 0.221 304.3 2327 2632 7.717

45 78.7 3.58 0.279 329.6 2312 2642 7.631

55 83.7 2.96 0.338 350.6 2299 2650 7.562

65 88.0 2.53 0.395 368.6 2288 2657 7.506

75 91.8 2.22 0.450 384.5 2279 2663 7.457

85 95.2 1.97 0.507 398.6 2270 2668 7.415

95 98.2 1.78 0.563 411.5 2262 2673 7.377

100 99.6 1.69 0.590 417.5 2258 2675 7.360

101.33 100 1.67 0.598 419.1 2257 2676 7.355

110 102.3 1.55 0.646 428.8 2251 2680 7.328

130 107.1 1.33 0.755 449.2 2238 2687 7.271

150 111.4 1.16 0.863 467.1 2226 2698 7.223

170 115.2 1.03 0.970 483.2 2216 2699 7.181

190 118.6 0.929 1.08 497.8 2206 2704 7.144

220 123.3 0.810 1.23 517.6 2193 2711 7.095

260 128.7 0.693 1.44 540.9 2177 2718 7.039

280 131.2 0.646 1.55 551.4 2170 2722 7.014

320 135.8 0.570 1.75 570.9 2157 2728 6.969

360 139.9 0.510 1.96 588.5 2144 2733 6.930

400 143.1 0.462 2.16 604.7 2133 2738 6.894

440 147.1 0.423 2.36 619.6 2122 2742 6.862

480 150.3 0.389 2.57 633.5 2112 2746 6.833

500 151.8 0.375 2.67 640.1 2107 2748 6.819

550 155.5 0.342 2.92 655.8 2096 2752 6.787

600 158.8 0.315 3.175 670.4 2085 2756 6.758

650 162.0 0.292 3.425 684.1 2075 2759 6.730

700 165.0 0.273 3.66 697.1 2065 2762 6.705

750 167.8 0.255 3.915 709.3 2056 2765 6.682

800 170.4 0.240 4.16 720.9 2047 2768 6.660

850 172.9 0.229 4.41 732.0 2038 2770 6.639

900 175.4 0.215 4.65 742.6 2030 2772 6.619

950 177.7 0.204 4.90 752.8 2021 2774 6.601

1000 179.9 0.194 5.15 762.6 2014 2776 6.583

1050 182.0 0.186 5.39 772 2006 2778 6.566

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1150 186.0 0.170 5.89 790 1991 2781 6.534

1250 189.8 0.157 6.38 807 1977 2784 6.505

1300 191.6 0.151 6.62 815 1971 2785 6.491

1500 198.3 0.132 7.59 845 1945 2790 6.441

1600 201.4 0.124 8.03 859 1933 2792 6.418

1800 207.1 0.110 9.07 885 1910 2795 6.375

2000 212.4 0.0995 10.01 909 1889 2797 6.337

2100 214.9 0.0945 10.54 920 1878 2798 6.319

2300 219.6 0.0868 11.52 942 1858 2800 6.285

2400 221.8 0.0832 12.02 952 1849 2800 6.269

2600 226.0 0.0769 13.01 972 1830 2801 6.239

2700 228.1 0.0740 13.52 981 1821 2802 6.224

2900 232.0 0.0689 14.52 1000 1803 2802 6.197

3000 233.8 0.0666 15.00 1008 1794 2802 6.184

3200 237.4 0.0624 16.02 1025 1779 2802 6.158

3400 240.9 0.0587 17.04 1042 1760 2802 6.134

3600 244.2 0.0554 18.06 1058 1744 2802 6.112

3800 247.3 0.0524 19.08 1073 1728 2801 6.090

4000 250.3 0.0497 21.00 1087 1713 2800 6.069

Absolute Pressure = Gauge Pressure + Atmospheric pressure.  Specific enthalpy or Sensible Heat is the quantity of heat in 1 kg of water according to the selected

temperature.

Example - Boiling Water at 100°C and 0 bar

At atmospheric pressure - 0 bar gauge or absolute 101.33 kN/m2 - water boils at 100°C. 419 kJ of energy is required to heat 1 kg of water from 0°C to the saturation temperature 100°C. 

Therefore, at 0 bar gauge (absolute 101.33 kN/m2) and 100°C - the specific enthalpy of water is 419 kJ/kg.

Another 2,257 kJ of energy is required to evaporate the 1 kg of water at 100°C to steam at 100°C. Therefore, at 0 bar gauge (absolute 101.33 kN/m2) - the specific enthalpy of evaporation is 2,257 kJ/kg. 

The total specific enthalpy of the steam at atmospheric pressure and 100oC can be summarized as: 

 hs = 419 + 2,257 = 2,676 kJ/kg 

Example - Boiling Water at 170°C and 7 bar

Steam at atmospheric pressure is of limited practical use. It cannot be conveyed by its own pressure along a steam pipe to the points of consumption.

At 7 bar gauge (absolute 800 kN/m2) - the saturation temperature of water is 170°C. More heat energy is required to raise the temperature to the saturation point at 7 bar gauge than needed for water at atmospheric pressure. From the table a value of 720.9 kJ is needed to raise 1 kg of water from 0°C to the saturation temperature

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170°C.

The heat energy (enthalpy of evaporation) needed at 7 bar gauge to evaporate the water to steam is actually less than the heat energy required at atmospheric pressure. The specific enthalpy of evaporation decrease with steam pressure increase. The evaporation heat is 2,047 kJ/kg according the table. 

Note! Because the specific volume of steam decreases with increasing pressure, the amount of heat energy transferred in the same volume actually increases with steam pressure. In other words the same pipe may transfer more energy with high pressure steam than with low pressure steam.

Properties of Saturated Steam - Pressure in Bar

The Saturated Steam Table with properties as boiling point, specific volume, density, specific enthalpy, specific heat and latent heat of vaporization

Absolutepressure

Boiling point

Specific volume (steam)

Density (steam)

Specific enthalpyof liquid water (sensible heat)

Specific enthalpy of steam

(total heat)

Latent heat of vaporization

Specific heat

(bar) (°C) (m3/kg) (kg/m3) (kJ/kg) (kcal/kg) (kJ/kg) (kcal/kg) (kJ/kg) (kcal/kg) (kJ/kg)

0.02 17.51 67.006 0.015 73.45 17.54 2533.64 605.15 2460.19 587.61 1.8644

0.03 24.10 45.667 0.022 101.00 24.12 2545.64 608.02 2444.65 583.89 1.8694

0.04 28.98 34.802 0.029 121.41 29.00 2554.51 610.13 2433.10 581.14 1.8736

0.05 32.90 28.194 0.035 137.77 32.91 2561.59 611.83 2423.82 578.92 1.8774

0.06 36.18 23.741 0.042 151.50 36.19 2567.51 613.24 2416.01 577.05 1.8808

0.07 39.02 20.531 0.049 163.38 39.02 2572.62 614.46 2409.24 575.44 1.8840

0.08 41.53 18.105 0.055 173.87 41.53 2577.11 615.53 2403.25 574.01 1.8871

0.09 43.79 16.204 0.062 183.28 43.78 2581.14 616.49 2397.85 572.72 1.8899

0.1 45.83 14.675 0.068 191.84 45.82 2584.78 617.36 2392.94 571.54 1.8927

0.2 60.09 7.650 0.131 251.46 60.06 2609.86 623.35 2358.40 563.30 1.9156

0.3 69.13 5.229 0.191 289.31 69.10 2625.43 627.07 2336.13 557.97 1.9343

0.4 75.89 3.993 0.250 317.65 75.87 2636.88 629.81 2319.23 553.94 1.9506

0.5 81.35 3.240 0.309 340.57 81.34 2645.99 631.98 2305.42 550.64 1.9654

0.6 85.95 2.732 0.366 359.93 85.97 2653.57 633.79 2293.64 547.83 1.9790

0.7 89.96 2.365 0.423 376.77 89.99 2660.07 635.35 2283.30 545.36 1.9919

0.8 93.51 2.087 0.479 391.73 93.56 2665.77 636.71 2274.05 543.15 2.0040

0.9 96.71 1.869 0.535 405.21 96.78 2670.85 637.92 2265.65 541.14 2.0156

1 99.63 1.694 0.590 417.51 99.72 2675.43 639.02 2257.92 539.30 2.0267

1.1 102.32 1.549 0.645 428.84 102.43 2679.61 640.01 2250.76 537.59 2.0373

1.2 104.81 1.428 0.700 439.36 104.94 2683.44 640.93 2244.08 535.99 2.0476

1.3 107.13 1.325 0.755 449.19 107.29 2686.98 641.77 2237.79 534.49 2.0576

1.4 109.32 1.236 0.809 458.42 109.49 2690.28 642.56 2231.86 533.07 2.0673

1.5 111.37 1.159 0.863 467.13 111.57 2693.36 643.30 2226.23 531.73 2.0768

1.5 111.37 1.159 0.863 467.13 111.57 2693.36 643.30 2226.23 531.73 2.0768

1.6 113.32 1.091 0.916 475.38 113.54 2696.25 643.99 2220.87 530.45 2.0860

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1.7 115.17 1.031 0.970 483.22 115.42 2698.97 644.64 2215.75 529.22 2.0950

1.8 116.93 0.977 1.023 490.70 117.20 2701.54 645.25 2210.84 528.05 2.1037

1.9 118.62 0.929 1.076 497.85 118.91 2703.98 645.83 2206.13 526.92 2.1124

2 120.23 0.885 1.129 504.71 120.55 2706.29 646.39 2201.59 525.84 2.1208

2.2 123.27 0.810 1.235 517.63 123.63 2710.60 647.42 2192.98 523.78 2.1372

2.4 126.09 0.746 1.340 529.64 126.50 2714.55 648.36 2184.91 521.86 2.1531

2.6 128.73 0.693 1.444 540.88 129.19 2718.17 649.22 2177.30 520.04 2.1685

2.8 131.20 0.646 1.548 551.45 131.71 2721.54 650.03 2170.08 518.32 2.1835

3 133.54 0.606 1.651 561.44 134.10 2724.66 650.77 2163.22 516.68 2.1981

3.5 138.87 0.524 1.908 584.28 139.55 2731.63 652.44 2147.35 512.89 2.2331

4 143.63 0.462 2.163 604.68 144.43 2737.63 653.87 2132.95 509.45 2.2664

4.5 147.92 0.414 2.417 623.17 148.84 2742.88 655.13 2119.71 506.29 2.2983

5 151.85 0.375 2.669 640.12 152.89 2747.54 656.24 2107.42 503.35 2.3289

5.5 155.47 0.342 2.920 655.81 156.64 2751.70 657.23 2095.90 500.60 2.3585

6 158.84 0.315 3.170 670.43 160.13 2755.46 658.13 2085.03 498.00 2.3873

6.5 161.99 0.292 3.419 684.14 163.40 2758.87 658.94 2074.73 495.54 2.4152

7 164.96 0.273 3.667 697.07 166.49 2761.98 659.69 2064.92 493.20 2.4424

7.5 167.76 0.255 3.915 709.30 169.41 2764.84 660.37 2055.53 490.96 2.4690

8 170.42 0.240 4.162 720.94 172.19 2767.46 661.00 2046.53 488.80 2.4951

8.5 172.94 0.227 4.409 732.03 174.84 2769.89 661.58 2037.86 486.73 2.5206

9 175.36 0.215 4.655 742.64 177.38 2772.13 662.11 2029.49 484.74 2.5456

9.5 177.67 0.204 4.901 752.82 179.81 2774.22 662.61 2021.40 482.80 2.5702

10 179.88 0.194 5.147 762.60 182.14 2776.16 663.07 2013.56 480.93 2.5944

11 184.06 0.177 5.638 781.11 186.57 2779.66 663.91 1998.55 477.35 2.6418

12 187.96 0.163 6.127 798.42 190.70 2782.73 664.64 1984.31 473.94 2.6878

13 191.60 0.151 6.617 814.68 194.58 2785.42 665.29 1970.73 470.70 2.7327

14 195.04 0.141 7.106 830.05 198.26 2787.79 665.85 1957.73 467.60 2.7767

15 198.28 0.132 7.596 844.64 201.74 2789.88 666.35 1945.24 464.61 2.8197

16 201.37 0.124 8.085 858.54 205.06 2791.73 666.79 1933.19 461.74 2.8620

17 204.30 0.117 8.575 871.82 208.23 2793.37 667.18 1921.55 458.95 2.9036

18 207.11 0.110 9.065 884.55 211.27 2794.81 667.53 1910.27 456.26 2.9445

19 209.79 0.105 9.556 896.78 214.19 2796.09 667.83 1899.31 453.64 2.9849

20 212.37 0.100 10.047 908.56 217.01 2797.21 668.10 1888.65 451.10 3.0248

21 214.85 0.095 10.539 919.93 219.72 2798.18 668.33 1878.25 448.61 3.0643

22 217.24 0.091 11.032 930.92 222.35 2799.03 668.54 1868.11 446.19 3.1034

23 219.55 0.087 11.525 941.57 224.89 2799.77 668.71 1858.20 443.82 3.1421

24 221.78 0.083 12.020 951.90 227.36 2800.39 668.86 1848.49 441.50 3.1805

25 223.94 0.080 12.515 961.93 229.75 2800.91 668.99 1838.98 439.23 3.2187

26 226.03 0.077 13.012 971.69 232.08 2801.35 669.09 1829.66 437.01 3.2567

27 228.06 0.074 13.509 981.19 234.35 2801.69 669.17 1820.50 434.82 3.2944

28 230.04 0.071 14.008 990.46 236.57 2801.96 669.24 1811.50 432.67 3.3320

29 231.96 0.069 14.508 999.50 238.73 2802.15 669.28 1802.65 430.56 3.3695

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30 233.84 0.067 15.009 1008.33 240.84 2802.27 669.31 1793.94 428.48 3.4069

Example - Boiling Water at 100°C, 0 bar Atmospheric Pressure

At atmospheric pressure (0 bar g, absolute 1 bar ), water boils at 100°C, and 417.51 kJ of energy are required to heat 1 kg of water from 0°C to its evaporating temperature of 100°C. 

Therefore the specific enthalpy of water at 0 bar g (absolute 1 bar ) and 100°C is 417.51 kJ/kg, as shown in the table.

Another 2 257.92 kJ of energy are required to evaporate 1 kg of water at 100°C into 1 kg of steam at 100°C. Therefore at 0 bar g (absolute 1 bar) the specific enthalpy of evaporation is 2 257.19 kJ/kg, as shown in the table. 

Total specific enthalpy for steam: 

 hs = 417.51 + 2 257.92 = 2 675.43 kJ/kg 

Example - Boiling Water at 170°C, 7 bar Atmospheric Pressure

Steam at atmospheric pressure is of a limited practical use because it cannot be conveyed under its own pressure along a steam pipe to the point of use.

At 7 bar g (absolute 8 bar), the saturation temperature of water is 170.42°C. More heat energy is required to raise its temperature to saturation point at 7 bar g than would be needed if the water were at atmospheric pressure. The table gives a value of 720.94 kJ to raise 1 kg of water from 0°C to its saturation temperature of 170°C.

The heat energy (enthalpy of evaporation) needed by the water at 7 bar g to change it into steam is actually less than the heat energy required at atmospheric pressure. This is because the specific enthalpy of evaporation decreases as the steam pressure increases. The evaporation heat is 2046.53 kJ/kg according the table. 

Note! Because the specific volume also decreases with increasing pressure, the amount of heat energy transferred in the same volume actually increases with steam pressure.

Practical use of entropy

It can be seen from Module 2.15 that entropy can be calculated. This would be laborious in practice, consequently steam tables usually carry entropy values, based on such calculations. Specific entropy is designated the letter ‘s’ and usually appears in columns signifying specific values for saturated liquid, evaporation, and saturated steam, sf, sfg and sg respectively. These values may equally be found in charts, and both Temperature - Entropy (T - S) and Enthalpy - Entropy (H - S) charts are to be found, as mentioned in Module 2.15. Each chart has particular use in specific circumstances.

Page 6 of 18

The T - S chart is often used to determine the properties of steam during its expansion through a nozzle or an orifice. The seat of a control valve would be a typical example.

To understand how a T - S chart is applied, it is worth sketching such a chart and plotting the steam properties at the start condition, reading these from the steam tables.

Example 2.16.1Steam is expanded from 10 bar a and a dryness fraction of 0.9 to 6 bar a through a nozzle, and no heat is removed or supplied during this expansion process. Calculate the final condition of the steam at the nozzle outlet? Specific entropy values quoted are in units of kJ/kg °C.

At 10 bar a, steam tables state that for dry saturated steam:

As no heat is added or removed during the expansion, the process is described as being adiabatic and isentropic, that is, the entropy does not change. It must still be 6.1413 kJ/kg K at the very moment it passes the throat of the nozzle.

At the outlet condition of 6 bar a, steam tables state that:

Specific entropy of saturated water (sf) = 1.9316

Specific entropy of evaporation of dry saturated steam (sfg) = 4.8285

But, in this example, since the total entropy is fixed at 6.1413 kJ/kg K:

By knowing that this process is isentropic, it has been possible to calculate the dryness fraction at the outlet condition. It is now possible to consider the outlet condition in terms of specific enthalpy (units are in kJ/kg).

From steam tables, at the inlet pressure of 10 bar a:

Page 7 of 18

Specific enthalpy of saturated water (hf) = 762.9

Specific enthalpy of evaporation of dry saturated steam (hfg) = 2014.83

As the dryness fraction is 0.9 at the inlet condition:

From steam tables, at the outlet condition of 6 bar a:

Specific enthalpy of saturated water (hf) = 670.74

Specific enthalpy of evaporation of dry saturated steam (hfg) = 2085.98

But as the dryness fraction is 0.8718 at the outlet condition:

It can be seen that the specific enthalpy of the steam has dropped in passing through the nozzle from 2576.25 to 2489.30 kJ/kg, that is, a heat drop of 86.95 kJ/kg.

This seems to contradict the adiabatic principle, which stipulates that no energy is removed from the process. But, as seen in Module 2.15, the explanation is that the steam at 6 bar a has just passed through the nozzle throat at high velocity, consequently it has gained kinetic energy. As energy cannot be created or destroyed, the gain in kinetic energy in the steam is at the expense of its own heat drop.

The above entropy values in Example 2.16.1 can be plotted on a T - S diagram, see Figure 2.16.1.

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Fig. 2.16.1 The T - S diagram for Example 2.16.1

Further investigation of kinetic energy in steamWhat is the significance of being able to calculate the kinetic energy of steam? By knowing this value, it is possible to predict the steam velocity and therefore the mass flow of steam through control valves and nozzles.

Kinetic energy is proportional to mass and the square of the velocity.

It can be further shown that, when incorporating Joule’s mechanical equivalent of heat, kinetic energy can be written as Equation 2.16.1:

Equation 2.16.1

Where:E = Kinetic energy (kJ)m = Mass of the fluid (kg)u = Velocity of the fluid (m/s)g = Acceleration to due gravity (9.80665 m/s²)J = Joule’s mechanical equivalent of heat (101.972 m kg/kJ)

By transposing Equation 2.16.1 it is possible to find velocity as shown by Equation 2.16.2:

Equation 2.16.2

Page 9 of 18

For each kilogram of steam, and by using Equation 2.16.2

As the gain in kinetic energy equals the heat drop, the equation can be written as shown by Equation 2.16.3:

Equation 2.16.3

Where:h = Heat drop in kJ

By calculating the adiabatic heat drop from the initial to the final condition, the velocity of steam can be calculated at various points along its path; especially at the throat or point of minimum pass area between the plug and seat in a control valve.

This could be used to calculate the orifice area required to pass a given amount of steam through a control valve. The pass area will be greatest when the valve is fully open. Likewise, given the valve orifice area, the maximum flowrate through the valve can be determined at the stipulated pressure drop. See Examples 2.16.2 and 2.16.3 for more details.

Example 2.16.2Consider the steam conditions in Example 2.16.1 with steam passing through a control valve with an orifice area of 1 cm². Calculate the maximum flow of steam under these conditions.

The downstream steam is at 6 bar a, with a dryness fraction of 0.8718.

Specific volume of dry saturated steam at 6 bar a (sg) equals 0.3156 m³/kg.

Specific volume of saturated steam at 6 bar a and a dryness fraction of 0.8718 equals 0.3156 m³/kg x 0.8718 which equates to 0.2751 m³/kg.

The heat drop in Example 2.16.1 was 86.95 kJ/kg, consequently the velocity can be calculated using Equation 2.16.3:

Equation 2.16.3

Page 10 of 18

The mass flow is calculated using Equation 2.16.4:

Equation 2.16.4

An orifice area of 1 cm² equals 0.0001 m²

Point of interestThermodynamic textbooks will usually quote Equation 2.16.3 in a slightly different way as shown in Equation 2.16.5:

Where: u = Velocity of the fluid in m/sh = Heat drop in J/kg

Considering the conditions in Example 2.16.3:

This velocity is exactly the same as that calculated from Equation 2.16.3, and the user is free to practise either equation according to preference.

The above calculations in Example 2.16.2 could be carried out for a whole series of reduced pressures, and, if done, would reveal that the flow of saturated steam through a fixed opening increases quite quickly at first as the downstream pressure is lowered.

Page 11 of 18

The increases in flow become progressively smaller with equal increments of pressure drops and, with saturated steam, these increases actually become zero when the downstream pressure is 58% of the absolute upstream pressure. (If the steam is initially superheated, CPD will occur at just below 55% of the absolute upstream pressure).

This is known as the ‘critical flow’ condition and the pressure drop at this point is referred to as critical pressure drop (CPD). After this point has been reached, any further reduction of downstream pressure will not give any further increase in mass flow through the opening.

In fact if, for saturated steam, the curves of steam velocity (u) and sonic velocity (s) were drawn for a convergent nozzle (Figure 2.16.2), it would be found that the curves intersect at the critical pressure. P1 is the upstream pressure, and P is the pressure at the throat.

Fig. 2.16.2 Steam and acoustic velocities through a nozzle

The explanation of this, first put forward by Professor Osborne Reynolds (1842 - 1912) of Owens College, Manchester, UK, is as follows:

Consider steam flowing through a tube or nozzle with a velocity u, and let s be the speed of sound (sonic velocity) in the steam at any given point, s being a function of the pressure and density of the steam. Then the velocity with which a disturbance such as, for example, a sudden change of pressure P, will be transmitted back through the flowing steam will be s - u.

Referring to Figure 2.16.2, let the final pressure P at the nozzle outlet be 0.8 of its inlet pressure P1. Here, as the sonic velocity s is greater than the steam velocity u, s - u is clearly positive. Any change in the pressure P would produce a change in the rate of mass flow.

When the pressure P has been reduced to the critical value of 0.58 P1, s - u becomes zero, and any further reduction of pressure after the throat has no effect on the pressure at the throat or the rate of mass flow.

When the pressure drop across the valve seat is greater than critical pressure drop, the critical velocity at the throat can be calculated from the heat drop in the steam from the upstream condition to the critical pressure drop condition, using Equation 2.16.5.

Page 12 of 18

Control valves

The relationship between velocity and mass flow through a restriction such as the orifice in a control valve is sometimes misunderstood.

Pressure drop greater than critical pressure drop It is worth reiterating that, if the pressure drop across the valve is equal to or greater than critical pressure drop, the mass flow through the throat of the restriction is a maximum and the steam will travel at the speed of sound (sonic velocity) in the throat. In other words, the critical velocity is equal to the local sonic velocity, as described above.

For any control valve operating under critical pressure drop conditions, at any reduction in throat area caused by the valve moving closer to its seat, this constant velocity will mean that the mass flow is simultaneously reduced in direct proportion to the size of the valve orifice.

Pressure drop less than critical pressure drop For a control valve operating such that the downstream pressure is greater than the critical pressure (critical pressure drop is not reached), the velocity through the valve opening will depend on the application.

Pressure reducing valvesIf the valve is a pressure reducing valve, (its function is to achieve a constant downstream pressure for varying mass flowrates) then, the heat drop remains constant whatever the steam load. This means that the velocity through the valve opening remains constant whatever the steam load and valve opening. Constant upstream steam conditions are assumed.

It can be seen from Equation 2.16.4 that, under these conditions, if velocity and specific volume are constant, the mass flowrate through the orifice is directly proportional to the orifice area.

Equation 2.16.4

Temperature control valvesIn the case of a control valve supplying steam to a heat exchanger, the valve is required to reduce the mass flow as the heat load falls. The downstream steam pressure will then fall with the heat load, consequently the pressure drop and heat drop across the valve will increase. Thus, the velocity through the valve must increase as the valve closes.

In this case, Equation 2.16.4 shows that, as the valve closes, a reduction in mass flow is not directly proportional to the valve orifice, but is also modified by the steam velocity and its specific volume.

Example 2.16.3Find the critical velocity of the steam at the throat of the control valve for Example 2.16.2, where the initial condition of the steam is 10 bar a and 90% dry, and assuming the downstream pressure is lowered to 3 bar a.

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But as the dryness fraction is 0.8701 at the throat condition:

The velocity of the steam through the throat of the valve can be calculated using Equation 2.16.5:

Equation 2.16.5

The critical velocity occurs at the speed of sound, consequently 430 m/s is the sonic velocity for the Example 2.16.3.

Noise in control valves If the pressure in the outlet of the valve body is lower than the critical pressure, the heat drop at a point immediately after the throat will be greater than at the throat. As velocity is directly related to heat drop, the steam velocity will increase after the steam passes the throat of the restriction, and supersonic velocities can occur in this region.

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In a control valve, steam, after exiting the throat, is suddenly confronted with a huge increase in space in the valve outlet, and the steam expands suddenly. The kinetic energy gained by the steam in passing through the throat is converted back into heat; the velocity falls to a value similar to that on the upstream side of the valve, and the pressure stabilises in the valve outlet and connecting pipework.

For the reasons mentioned above, valves operating at and greater than critical pressure drop will incur sonic and supersonic velocities, which will tend to produce noise. As noise is a form of vibration, high levels of noise will not only cause environmental problems, but may actually cause the valve to fail. This can sometimes have an important bearing when selecting valves that are expected to operate under critical flow conditions.

It can be seen from previous text that the velocity of steam through control valve orifices will depend on the application of the valve and the pressure drop across it at any one time.

Reducing noise in control valves ,There are some practical ways to deal with the effects of noise in control valves.

Perhaps the simplest way to overcome this problem is to reduce the working pressure across the valve. For instance, where there is a need to reduce pressure, by reducing pressure with two valves instead of one, both valves can share the total heat drop, and the potential for noise in the pressure reducing station can be reduced considerably.

Another way to reduce the potential for noise is by increasing the size of the valve body (but retaining the correct orifice size) to help ensure that the supersonic velocity will have dissipated by the time the flow impinges upon the valve body wall.

In cases where the potential for noise is extreme, valves fitted with a noise attenuator trim may need to be used.

Steam velocities in control valve orifices will reach, typically, 500 m/s. Water droplets in the steam will travel at some slightly lower speed through a valve orifice, but, being incompressible, these droplets will tend to erode the valve and its seat as they squeeze between the two.

It is always sensible to ensure that steam valves are protected from wet steam by fitting separators or by providing adequate line drainage upstream of them.

Summing up of Modules 2.15 and 2.16

The T - S diagram, shown in Figure 2.16.1, and reproduced below in Figure 2.16.3, shows clearly that the steam becomes wetter during an isentropic expansion (0.9 at 10 bar a to 0.8718 at 6 bar a) in Example 2.16.1.

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Fig. 2.16.3 A T-S diagram showing wetter steam from an isentropic expansion

At first, this seems strange to those who are used to steam getting drier or becoming superheated during an expansion, as happens when steam passes through, for example, a pressure reducing valve.

The point is that, during an adiabatic expansion, the steam is accelerating up to high speed in passing through a restriction, and gaining kinetic energy. To provide this energy, a little of the steam condenses (if saturated steam), (if superheated, drops in temperature and may condense) providing heat for conversion into kinetic energy.

If the steam is flowing through a control valve, or a pressure reducing valve, then somewhere downstream of the valve’s seat, the steam is slowed down to something near its initial velocity. The kinetic energy is destroyed, and must reappear as heat energy that dries out or superheats the steam depending on the conditions.

The T - S diagram is not at all convenient for showing this effect, but the Mollier diagram (the H - S diagram) can do so quite clearly.

The Mollier diagram can depict both an isenthalpic expansion as experienced by a control valve, (see Figure 2.15.6) by moving horizontally across the graph to a lower pressure; and an isentropic expansion as experienced by steam passing through a nozzle, (see Figure 2.15.7) by moving horizontally down to a lower pressure. In the former, the steam is usually either dried or superheated, in the latter, the steam gets wetter.

This perhaps begs the question, ‘How does the steam know if it is to behave in an isenthalpic or isentropic fashion?’ Clearly, as the steam accelerates and rushes through the narrowest part of the restriction (the throat of a nozzle, or the adjustable gap between the valve and seat in a control valve) it must behave the same in either case.

The difference is that the steam issuing from a nozzle will next meet a turbine wheel and gladly give up its kinetic energy to turn the turbine. In fact, a nozzle could be thought of as a device to convert heat energy into kinetic energy for this very purpose.

In a control valve, instead of doing such work, the steam simply slows down in the valve Page 16 of 18

outlet passages and its connecting pipework, when the kinetic energy appears as heat energy, and unwittingly goes on its way to give up this heat at a lower pressure.

It can be seen that both the T - S diagram and H - S diagram have their uses, but neither would have been possible had the concept of entropy not been utilised.

Wet Steam Quality and the Dryness Fraction

An introduction and definition of vapor or steam quality and dryness fraction. Includes formulas for calculating wet steams enthalpy and specific volume

To produce 100% dry steam in an boiler, and keep the steam dry throughout the piping system, is in general not possible. Droplets of water will escape from the boiler surface. Because of turbulence and splashing when bubbles of steam break through the water surface the steam space will contain a mixture of water droplets and steam. 

In addition heat loss in the pipes will condensate steam to droplets of water.

Steam - produced in a boiler where the heat is supplied to the water and where the steam are in contact with the water surface of the boiler - will contain approximately 5% water by mass.

Dryness fraction of Wet Steam

If the water content of the steam is 5% by mass, then the steam is said to be 95% dry and has a dryness fraction of 0.95.

Dryness fraction can be expressed as:

ζ = ws / (ww + ws)       (1)

where

ζ = dryness fraction 

ww = mass of water (kg, lb)

ws = mass of steam (kg, lb)

Enthalpy of Wet Steam

The actual enthalpy of evaporation of wet steam is the product of the dryness fraction - ζ - and the specific enthalpy - hs - from the steam tables. Wet steam have lower usable heat energy than dry saturated steam.

ht = hs ζ + hw          (2)

where

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ht = enthalpy of wet steam (kJ/kg, Btu/lb)

hs  = enthalpy of steam (kJ/kg, Btu/lb)

hw = enthalpy of saturated water or condensate (kJ/kg, Btu/lb)

Specific Volume of Wet Steam

The droplets of water in wet steam will occupy negligible space in the steam and the specific volume of wet steam will be reduced according the dryness fraction.

v = vs ζ        (3)

where

v = specific volume of wet steam (m3/kg, ft3/lb)

vs = specific volume of the dry steam (m3/kg, ft3/lb)

Example - Enthalpy and Specific Volume of Wet Steam

Steam at pressure 5 bar gauge has a dryness fraction of 0.95.

Total enthalpy can be expressed as:

ht = (2,085 * 0.95) + 670.4 kJ/kg = 2,651 kJ/kg

Specific volume can be expressed as:

v = 0.315 * 0.95 = 0.299 m3/kg

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