Steam Engineering
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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 Unit Temperature (T) 100.0 C 212.0 F Pressure (P) 1.0134 bar 14.698 psi Saturated Liquid ( 958.27 59.823 f) 3 Density kg/m lb/ft3 Saturated Vapor ( 0.59770 0.037313 g) Saturated Liquid 0.0010435 0.016716 (vf) Specific 3 m /kg ft3/lb Volume Saturated Vapor 1.6731 26.800 (vg) Saturated Liquid 418.94 180.11 (uf) Internal Evaporated (ufg) 2087.6 kJ/kg 897.50 Btu/lb Energy Saturated Vapor 2506.5 1077.6 (ug) Saturated Liquid 419.05 180.16 (hf) Enthalpy Evaporated (hfg) 2256.9 kJ/kg 970.3 Btu/lb Saturated Vapor 2676.1 1150.5 (hg) Saturated Liquid 1.3068 0.31213 (sf) kJ/kgBtu/lbEntropy Evaporated (sfg) 6.0483 K 1.4446 R (mayer) Saturated Vapor 7.3550 1.7567 (sg)
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
Temperature Specific Density -
Specific Enthalpy of
Specific
Page 1 of 18
Liquid - Evaporation Steam ρVolume hl - he hs (kg/m3) (m3/kg) (kJ/kg) (kJ/kg) (kJ/kg)
Entropy of Steam - s (kJ/kgK)
pressure (kN/m2)
(oC)
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
Page 2 of 18
1050
182.0
0.186
5.39
772
2006
2778
6.566
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.
Page 3 of 18
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 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 Specific Specific enthalpy Specific enthalpy Absolute Boiling Density volume of liquid water of steam pressure point (steam) (steam) (sensible heat) (total heat) (bar)
(°C)
Latent heat of vaporization
Specific heat
(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
Page 4 of 18
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
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
Page 5 of 18
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
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 Page 6 of 18
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. 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 2.16.1 Steam 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 Page 7 of 18
of
specific
enthalpy
(units
are
in
kJ/kg).
From steam tables, at the inlet pressure of 10 bar a: 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.
Page 8 of 18
Fig. 2.16.1 The T - S diagram for Example 2.16.1
Further investigation of kinetic energy in steam What 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 2.16.2 Consider 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
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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 interest Thermodynamic 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/s h = 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 Page 12 of 18
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.
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 valves If 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 valves In 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 Page 13 of 18
and
its
specific
volume.
Example 2.16.3 Find 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.
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
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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. 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.
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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.
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 Page 16 of 18
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 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
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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 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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