4.6 – Steam Tables¶
4.6.0 – Learning Objectives¶
By the end of this section you should be able to:
- Understand what steam tables tell us.
- Use steam tables to solve thermodynamic problems.
- Interpolate between data points.
4.6.1 – Introduction¶
Water, and in specific, steam is used in many processes. Most commonly, it is used in the transfer of energy. Because of this, scientist and engineers have created extensive data tables on a variety of conditions.
4.6.2 – Table Description¶
Shown below is a saturated steam table. A saturated steam table will include data of both the liquid and vapour phases of water at a given temperature and pressure.
In [2]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df = pd.read_excel('../figures/Module-4/SatTandPofSteam.xlsx', sheet_name='Sheet2', index_col=None, na_values=['NA'])
df.head(20)
Out[2]:
| Temperature (C) | Pressure (MPa) | Volume (l, m3/kg) | Volume (v, m3/kg) | Internal Energy (l, kJ/kg) | Δ Internal Energy of Vapourization (kJ/kg) | Internal Energy (v, kJ/kg) | Enthalpy (l, kJ/kg) | Δ Enthalpy of Vapourization (kJ/kg) | Enthalpy (v, kJ/kg) | Entropy (l, J/g*K) | Δ Entropy of Vapourization (kJ/kg) | Entropy (v, J/g*K) | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 0.01 | 0.000612 | 0.001000 | 205.9900 | 0.000 | 2374.900 | 2374.9 | 0.000 | 2500.900 | 2500.9 | 0.000000 | 9.155500 | 9.1555 |
| 1 | 5.00 | 0.000873 | 0.001000 | 147.0100 | 21.019 | 2360.781 | 2381.8 | 21.020 | 2489.080 | 2510.1 | 0.076254 | 8.948546 | 9.0248 |
| 2 | 10.00 | 0.001228 | 0.001000 | 106.3000 | 42.020 | 2346.580 | 2388.6 | 42.021 | 2477.179 | 2519.2 | 0.151090 | 8.748710 | 8.8998 |
| 3 | 15.00 | 0.001706 | 0.001001 | 77.8750 | 62.980 | 2332.520 | 2395.5 | 62.981 | 2465.319 | 2528.3 | 0.224460 | 8.555840 | 8.7803 |
| 4 | 20.00 | 0.002339 | 0.001002 | 57.7570 | 83.912 | 2318.388 | 2402.3 | 83.914 | 2453.486 | 2537.4 | 0.296480 | 8.369520 | 8.6660 |
| 5 | 25.00 | 0.003170 | 0.001003 | 43.3370 | 104.830 | 2304.270 | 2409.1 | 104.830 | 2441.670 | 2546.5 | 0.367220 | 8.189380 | 8.5566 |
| 6 | 30.00 | 0.004247 | 0.001004 | 32.8780 | 125.730 | 2290.170 | 2415.9 | 125.730 | 2429.770 | 2555.5 | 0.436750 | 8.015250 | 8.4520 |
| 7 | 35.00 | 0.005629 | 0.001006 | 25.2050 | 146.630 | 2276.070 | 2422.7 | 146.630 | 2417.870 | 2564.5 | 0.505130 | 7.846570 | 8.3517 |
| 8 | 40.00 | 0.007385 | 0.001008 | 19.5150 | 167.530 | 2261.870 | 2429.4 | 167.530 | 2405.970 | 2573.5 | 0.572400 | 7.683100 | 8.2555 |
| 9 | 45.00 | 0.009595 | 0.001010 | 15.2520 | 188.430 | 2247.670 | 2436.1 | 188.430 | 2393.970 | 2582.4 | 0.638610 | 7.524690 | 8.1633 |
| 10 | 50.00 | 0.012352 | 0.001012 | 12.0270 | 209.330 | 2233.370 | 2442.7 | 209.340 | 2381.960 | 2591.3 | 0.703810 | 7.370990 | 8.0748 |
| 11 | 55.00 | 0.015762 | 0.001015 | 9.5643 | 230.240 | 2219.060 | 2449.3 | 230.260 | 2369.840 | 2600.1 | 0.768020 | 7.221780 | 7.9898 |
| 12 | 60.00 | 0.019946 | 0.001017 | 7.6672 | 251.160 | 2204.740 | 2455.9 | 251.180 | 2357.620 | 2608.8 | 0.831290 | 7.076810 | 7.9081 |
| 13 | 65.00 | 0.025042 | 0.001020 | 6.1935 | 272.090 | 2190.310 | 2462.4 | 272.120 | 2345.380 | 2617.5 | 0.893650 | 6.935950 | 7.8296 |
| 14 | 70.00 | 0.031201 | 0.001023 | 5.0395 | 293.030 | 2175.870 | 2468.9 | 293.070 | 2333.030 | 2626.1 | 0.955130 | 6.798870 | 7.7540 |
| 15 | 75.00 | 0.038595 | 0.001026 | 4.1289 | 313.990 | 2161.210 | 2475.2 | 314.030 | 2320.570 | 2634.6 | 1.015800 | 6.665400 | 7.6812 |
| 16 | 80.00 | 0.047414 | 0.001029 | 3.4052 | 334.960 | 2146.640 | 2481.6 | 335.010 | 2307.990 | 2643.0 | 1.075600 | 6.535500 | 7.6111 |
| 17 | 85.00 | 0.057867 | 0.001032 | 2.8258 | 355.950 | 2131.850 | 2487.8 | 356.010 | 2295.290 | 2651.3 | 1.134600 | 6.408800 | 7.5434 |
| 18 | 90.00 | 0.070182 | 0.001036 | 2.3591 | 376.970 | 2117.030 | 2494.0 | 377.040 | 2282.460 | 2659.5 | 1.192900 | 6.285200 | 7.4781 |
| 19 | 95.00 | 0.084608 | 0.001040 | 1.9806 | 398.000 | 2102.000 | 2500.0 | 398.090 | 2269.510 | 2667.6 | 1.250400 | 6.164700 | 7.4151 |
Another type of table is a super heated steam table at a given pressure. Here is an example of this type table.
In [3]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df = pd.read_excel('../figures/Module-4/Steam_3MPa.xlsx', sheet_name='Sheet2', index_col=None, na_values=['NA'])
df
Out[3]:
| Temperature (C) | Pressure (MPa) | Volume (m3/kg) | Internal Energy (kJ/kg) | Enthalpy (kJ/kg) | Entropy (J/g*K) | |
|---|---|---|---|---|---|---|
| 0 | 233.85 | 3 | 0.066664 | 2603.2 | 2803.2 | 6.1856 |
| 1 | 250.00 | 3 | 0.070627 | 2644.7 | 2856.5 | 6.2893 |
| 2 | 300.00 | 3 | 0.081179 | 2750.8 | 2994.3 | 6.5412 |
| 3 | 350.00 | 3 | 0.090556 | 2844.4 | 3116.1 | 6.7449 |
| 4 | 400.00 | 3 | 0.099379 | 2933.5 | 3231.7 | 6.9234 |
| 5 | 450.00 | 3 | 0.107890 | 3021.2 | 3344.8 | 7.0856 |
| 6 | 500.00 | 3 | 0.116200 | 3108.6 | 3457.2 | 7.2359 |
| 7 | 550.00 | 3 | 0.124370 | 3196.6 | 3569.7 | 7.3768 |
| 8 | 600.00 | 3 | 0.132450 | 3285.5 | 3682.8 | 7.5103 |
| 9 | 650.00 | 3 | 0.140450 | 3375.6 | 3796.9 | 7.6373 |
| 10 | 700.00 | 3 | 0.148410 | 3467.0 | 3912.2 | 7.7590 |
| 11 | 750.00 | 3 | 0.156320 | 3559.9 | 4028.9 | 7.8758 |
| 12 | 800.00 | 3 | 0.164200 | 3654.3 | 4146.9 | 7.9885 |
| 13 | 850.00 | 3 | 0.172050 | 3750.3 | 4266.5 | 8.0973 |
| 14 | 900.00 | 3 | 0.179880 | 3847.9 | 4387.5 | 8.2028 |
| 15 | 950.00 | 3 | 0.187690 | 3947.0 | 4510.1 | 8.3051 |
| 16 | 1000.00 | 3 | 0.195490 | 4047.7 | 4634.1 | 8.4045 |
4.6.4 – Interpolation¶
Let’s suppose you wanted to find out what the internal energy of steam is at \(725.00 ^{\circ} C\) and \(3.0000 \space \text{MPa}\). Since the value is not given in the table, you must interpolate. The formula for interpolation is:
\[y = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) + y_1\]
In this case the formula for internal energy at \(725.00 ^{\circ} C\) is:
\[U_{3 MPa, 725 C} = \frac{U_{3 MPa, 750 C} - U_{3 MPa, 700 C}}{T_2 - T_1} (T - T_1) + U_{3 MPa, 700 C}\]
and the answer is:
\[U_{3 MPa, 725 C} = \frac{(3559.9 - 3467.0 ) \space kJ/kg}{(750 - 700) \space C} (725 - 700) \space C + 3467.0 \space kJ/kg = 3513.45 \space kJ/kg\]
4.6.5 – Problem Statement¶
Question¶
Using the steam tables below, find the change in internal energy when steam is first cooled isobarically at \(750.00 ^{\circ} C\) and \(5.0000 \space \text{MPa}\) to \(725.00 ^{\circ} C \space\) and \(5.0000 \space \text{MPa}\) and then expanded isothermally at \(725.00 ^{\circ} C\) from \(5.0000 \space \text{MPa}\) to \(1.0000 \space \text{MPa}\).
In [4]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df = pd.read_excel('../figures/Module-4/Steam_5MPa.xlsx', sheet_name='Sheet2', index_col=None, na_values=['NA'])
df
Out[4]:
| Temperature (C) | Pressure (MPa) | Volume (m3/kg) | Internal Energy (kJ/kg) | Enthalpy (kJ/kg) | Entropy (J/g*K) | |
|---|---|---|---|---|---|---|
| 0 | 263.94 | 5 | 0.039446 | 2597.0 | 2794.2 | 5.9737 |
| 1 | 300.00 | 5 | 0.045346 | 2699.0 | 2925.7 | 6.2110 |
| 2 | 350.00 | 5 | 0.051969 | 2809.5 | 3069.3 | 6.4516 |
| 3 | 400.00 | 5 | 0.057837 | 2907.5 | 3196.7 | 6.6483 |
| 4 | 450.00 | 5 | 0.063323 | 3000.6 | 3317.2 | 6.8210 |
| 5 | 500.00 | 5 | 0.068583 | 3091.7 | 3434.7 | 6.9781 |
| 6 | 550.00 | 5 | 0.073694 | 3182.4 | 3550.9 | 7.1237 |
| 7 | 600.00 | 5 | 0.078704 | 3273.3 | 3666.8 | 7.2605 |
| 8 | 650.00 | 5 | 0.083639 | 3365.0 | 3783.2 | 7.3901 |
| 9 | 700.00 | 5 | 0.088518 | 3457.7 | 3900.3 | 7.5136 |
| 10 | 750.00 | 5 | 0.093355 | 3551.6 | 4018.4 | 7.6320 |
| 11 | 800.00 | 5 | 0.098158 | 3646.9 | 4137.7 | 7.7458 |
| 12 | 850.00 | 5 | 0.102930 | 3743.6 | 4258.3 | 7.8556 |
| 13 | 900.00 | 5 | 0.107690 | 3841.8 | 4380.2 | 7.9618 |
| 14 | 950.00 | 5 | 0.112420 | 3941.5 | 4503.6 | 8.0648 |
| 15 | 1000.00 | 5 | 0.117150 | 4042.6 | 4628.3 | 8.1648 |
In [5]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df = pd.read_excel('../figures/Module-4/Steam_1MPa.xlsx', sheet_name='Sheet2', index_col=None, na_values=['NA'])
df
Out[5]:
| Temperature (C) | Pressure (MPa) | Volume (m3/kg) | Internal Energy (kJ/kg) | Enthalpy (kJ/kg) | Entropy (J/g*K) | |
|---|---|---|---|---|---|---|
| 0 | 179.88 | 1 | 0.19436 | 2582.7 | 2777.1 | 6.5850 |
| 1 | 200.00 | 1 | 0.20602 | 2622.2 | 2828.3 | 6.6955 |
| 2 | 250.00 | 1 | 0.23275 | 2710.4 | 2943.1 | 6.9265 |
| 3 | 300.00 | 1 | 0.25799 | 2793.6 | 3051.6 | 7.1246 |
| 4 | 350.00 | 1 | 0.28250 | 2875.7 | 3158.2 | 7.3029 |
| 5 | 400.00 | 1 | 0.30661 | 2957.9 | 3264.5 | 7.4669 |
| 6 | 450.00 | 1 | 0.33045 | 3040.9 | 3371.3 | 7.6200 |
| 7 | 500.00 | 1 | 0.35411 | 3125.0 | 3479.1 | 7.7641 |
| 8 | 550.00 | 1 | 0.37766 | 3210.5 | 3588.1 | 7.9008 |
| 9 | 600.00 | 1 | 0.40111 | 3297.5 | 3698.6 | 8.0310 |
| 10 | 650.00 | 1 | 0.42449 | 3386.0 | 3810.5 | 8.1557 |
| 11 | 700.00 | 1 | 0.44783 | 3476.2 | 3924.1 | 8.2755 |
| 12 | 750.00 | 1 | 0.47112 | 3568.1 | 4039.3 | 8.3909 |
| 13 | 800.00 | 1 | 0.49438 | 3661.7 | 4156.1 | 8.5024 |
| 14 | 850.00 | 1 | 0.51762 | 3757.0 | 4274.6 | 8.6103 |
| 15 | 900.00 | 1 | 0.54083 | 3853.9 | 4394.8 | 8.7150 |
| 16 | 950.00 | 1 | 0.56403 | 3952.5 | 4516.5 | 8.8166 |
| 17 | 1000.00 | 1 | 0.58721 | 4052.7 | 4639.9 | 8.9155 |
Answer¶
The internal energy at \(750.00 ^{\circ} C\) and \(5.0000 \space \text{MPa}\) is
\[U_{5 MPa, 750 C} = 3511.6 \space kJ/kg\]
Since \(725.00 ^{\circ} C\) is not present in the first table, we must interpolate
\[U_{5 MPa, 725 C} = \frac{U_{5 MPa, 750 C} - U_{5 MPa, 700 C}}{T_2 - T_1} (T - T_1) + U_{5 MPa, 700 C}\]
\[U_{5 MPa, 725 C} = \frac{(3511.6 - 3457.7 ) \space kJ/kg}{(750 - 700) \space C} (725 - 700) \space C + 3457.7 \space kJ/kg\]
\[U_{5 MPa, 725 C} = 3,484.65 \space kJ/kg\]
now that we have \(U_{5 MPa, 725 C}\), we must find \(U_{1 MPa, 725 C}\) using interpolation
\[U_{1 MPa, 725 C} = \frac{U_{1 MPa, 750 C} - U_{1 MPa, 700 C}}{T_2 - T_1} (T - T_1) + U_{1 MPa, 700 C}\]
\[U_{1 MPa, 725 C} = \frac{(3568.1 - 3476.2 ) \space kJ/kg}{(750 - 700) \space C} (725 - 700) \space C + 3476.2 \space kJ/kg\]
\[U_{1 MPa, 725 C} = 3,522.15 \space kJ/kg\]
Now that we have all the necessary values, we just need to sum the changes in internal energy
\[\Delta U_{tot} = \Delta U_{1} + \Delta U_{2} = (U_{5 MPa, 725 C} - U_{5 MPa, 750 C}) + (U_{1 MPa, 725 C} - U_{5 MPa, 725 C})\]
\[\Delta U_{tot} = (3,484.65 \space kJ/kg - 3551.6 \space kJ/kg) + (3,522.15 \space kJ/kg - 3,484.65 \space kJ/kg)\]
\[\Delta U_{tot} = -10.6 \space kJ/kg\]
In [ ]: