Steps
- 1Define what we are trying to find and label the given values. Enthalpy change is described as ∆H = Hfinal – Hinitial. Tinitial = 40o C, Tfinal = 190o C, Pinitial = .1 MPa, Pfinal = .3 MPa2Find the Steam Tables in the back of your textbook that have values for saturated water. Usually labeled Saturation Temperature and Saturation Pressure. (First page may look like the image below)3Locate the temperature 40o C in the left hand column of the table.4Follow it across the table to the enthalpy columns. Labeled HL, HVap or HV. We just know that our fluid is initially a liquid, so we will use the liquid enthalpy value, HL, as our initial enthalpy value. HL = 167.53 kJ/kg5Now locate the Superheated Steam table.6Identify the values relative to our final pressure (0.3 MPa)
7Locate our final temperature (190o C)
8Recognize that 190 is not listed in the temperature column, therefore we must interpolate. Interpolating values gives us a best guess, when the desired temperature or pressure is between two available values. Interpolating follows the formula, HD (desired enthalpy) = [(H_hi-H_low)/(T_hi-T_low )*(T_final-T_low ) ]+H_low For our example problem, Tfinal = 190o C9Locate the temperature values just above and below 190o C, Thi and Tlow. In this case they are 200o C and 150o C.10Now find the corresponding enthalpy values for 150o C and 200o C, Hhi and Hlow.11Follow the equation above to find the interpolated enthalpy value at 190o C. [(2865.9-2761.2)/(200-150)*(190-150) ]+2761.2 H190 = 2844.96 kJ/kg12Subtract our initial enthalpy value at 40o C ( 167.53 kJ/kg) from our final enthalpy value at 190o C (2844.96 kJ/kg), to find the change in enthalpy needed to heat water from its liquid phase to superheated steam. The answer is given below.
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