The substance is steam (H2O). NOTE: The purpose of this problem is to illustrate that there are conditions where water vapor is
far from ideal-gas behavior. For this reason, in this course, we will always use tables or NIST to determine properties of H2O for all conditions to avoid misuse of the ideal-gas EOS.
a. For the four conditions below, calculate the reduced pressure and the reduced temperature.
HINT: See Appendix E. Based on these values, decide if ideal-gas behavior is reasonable, i.e., that Pv = RiT within +/- 5%.
i. P = 15 MPa T = 1200 K
ii. P = 15 MPa T = 800 K
iii. P = 10 MPa T = 600 K
iv. P = 0.10 MPa T = 600 K
b. Use the generalized compressibility chart (Fig. 2.18 p. 92) to estimate a value of the compressibility factor Z for these conditions. How do these values compare with your thinking in part 1?
c. Use the NIST webbook to determine the compressibility factor Z for steam at the conditions above. How "ideal" is steam at these various conditions? How do these Z-values compare with your estimates from part 2?
a. For the four conditions below, calculate the reduced pressure and the reduced temperature.
HINT: See Appendix E. Based on these values, decide if ideal-gas behavior is reasonable, i.e., that Pv = RiT within +/- 5%.
i. P = 15 MPa T = 1200 K
ii. P = 15 MPa T = 800 K
iii. P = 10 MPa T = 600 K
iv. P = 0.10 MPa T = 600 K
b. Use the generalized compressibility chart (Fig. 2.18 p. 92) to estimate a value of the compressibility factor Z for these conditions. How do these values compare with your thinking in part 1?
c. Use the NIST webbook to determine the compressibility factor Z for steam at the conditions above. How "ideal" is steam at these various conditions? How do these Z-values compare with your estimates from part 2?
1 answer:
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Answer:
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
Now, we will find (enthalpy at the outlet for the isentropic process):
Now, the isentropic efficiency of the turbine can be given as follows: