Try these examples
Example 1.
Calculate the steam quality of a 200 psia system when the orifice exit temperature is 250 F.
Liley, using the method described in his article, calculates the steam quality to be 0.9636. Applying Eqn. 4 with TE = 250 and PS = 200 gives an answer of 0.9649a difference of 0.13 percent.
Example 2
uses thermodynamic data obtained from the Engineering Data Book. It calculates steam quality at a pressure higher than that described in either of the two referenced articles:
Example 2. Calculate the quality of steam at 566.1 psia, when the throttling calorimeter temperature is 300 F.
The calculations use available data from the GPSA Engineering Data Book. At 566.1 psia, the enthalpy of the condensate is 464.5 Btu/lbm while the enthalpy of the saturated steam is 1204.1 Btu/lbm. The enthalpy of expanded steam at 300 F and 14.696 psia is 1,192.8 Btu/lbm. Substituting into Eqn. 2 gives: 1,192.8 = (1-X)(464.5) + 1,204.1 X. Solving for X yields 0.9840. Using Eqn. 4 instead results in an answer of 0.9836a difference of 0.004 percent.
Example 3.
Steam at 500 psia is to have a quality of not less than 0.9775. What should be the expanded steam temperature?
Eqn. 4a provides the answer. The calculated temperature is 289.5 F. This means the steam temperature at the orifice exit must be no less than 290 F.
Example 4.
A saturated steam system has a temperature of 460 F. Expansion in a throttling calorimeter results in a steam temperature of 300 F. Calculate the steam quality.
Start with Eqn. 5 to calculate system pressure when the temperature is 460 F. This gives a calculated pressure of 466.58 psia. This answer differs by 0.06 percent with the value noted in the Steam Tables. The calculated pressure and expanded steam temperature are then entered in Eqn. 4. The steam quality is 0.9844.
Example 5.
Calculate steam quality when saturated steam at 420 psia shows a temperature difference of 165 F between steam temperature and calorimeter temperature.
Use Eqn. 5a to calculate the saturated steam temperature: TS = 449.50 F. Then TE = 449.5 - 165 = 284.50 F. Now substitute PS = 420 and TE = 284.5 into Eqn. 4. Answer: X = 0.9756. Also refer to Figure 1 for comparison.
References:
(1) Ganapathy, V., "Calculate the moisture content of steam," Chemical Engineering, p. 127, August, 1993.
(2) Meyer, C.A., McClintock, R.B., Silvestri, G.J., and Spencer, R.C., Jr., ASME Steam Tables, 5th edition, A.S.M.E., New York, 1983.
(3) Engineering Data Book, Section 24, "Thermodynamic Properties," Revised 10th edition, Gas Processors Association, Tulsa, Oklahoma, 1994.
(4) Liley, Peter E., "A simple equation for steam quality," Chemical Engineering, p. 140, August, 1994.
Frank A. Rusche is a process engineer with Kellogg-Brown and Root in Houston, Texas. He can be reached at Frank.Rusche@Halliburton.com or 281-575-3432.
Figures: Kellogg-Brown and Root
