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Modern Engineering Thermodynamics

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404 CHAPTER 11: More Thermodynamic Relations<br />

Make your percent error calculations here at 20.0°F intervals<br />

between 40.0 and 700.°F.<br />

c. By 1899, very careful p sat – T sat measurements had been<br />

made, and K. Thiesen curve fit these data to the following<br />

equation:<br />

ðT sat +459:6Þlog 10 p sat = 5:409ðT sat −212Þ<br />

<br />

−ð8:71×10 −10 Þ ð689−T sat Þ 4 −ð477Þ 4 <br />

where p sat is in atmospheres and T sat is in °F. Make your<br />

percent error calculations here at intervals of 20.0°F between<br />

20.0 and 700.°F.<br />

d. By 1915, G. A. Goodenough had developed the following<br />

more complex equation, which he claimed fit steam quite<br />

well over the entire range from 32.0°F to the critical point<br />

(705°F).<br />

log 10 p sat = A − B/T sat − Clog 10 T sat − DT sat + ET 2 sat − F<br />

where F = 0.0002{10 – 10[(T sat – 829.6)/100] 2 +<br />

[(T sat – 829.6)/100] 4 }. Here, p sat is in psia, T sat is in R,<br />

and<br />

A = 10:5688080 log 10 D = 7:6088020 − 10<br />

log 10 B = 3:6881209 log 10 E = 4:1463000 − 10<br />

C = 0:0155<br />

Make your percent error calculations at any convenient<br />

temperature interval between 32.0 and 705°F. Note: The use<br />

of base 10 logarithms in these equations is the way these<br />

equations were originally written, and it has been continued<br />

here for historical accuracy.<br />

87.* Equations (11.31) for mercury are given by<br />

1. ln p sat = 23.6321 − 7042.6208/T sat − 0.1207 (ln T sat )<br />

− 58,060.290/T 2 sat.<br />

2. p = RT/v − (T/v 2 ) exp[10.3338 − 312.0954/T − 2.0795<br />

(ln T)].<br />

3. v f = [12,813.6070 − 2.4531(T sat − 600)−0.000267(T sat − 600) 2 ] −1 .<br />

4. c v ° = 62.168 J/(kg·K) = constant.<br />

where p and p sat are in Pa, T and T sat are in K, and v and v f<br />

are in m 3 /kg, and R = 41.4453 J/(kg· K). Develop a<br />

computer spreadsheet that returns v, u, h, ands values when<br />

p and T are input. Make sure your program checks to see<br />

what region (saturated or superheated) your input data are<br />

in. For T sat =750.K,youshouldgeth fg =291kJ/kgands fg =<br />

0.388 kJ/kg. These values are independent of your reference<br />

state values. Note: More ambitious programs can now be<br />

produced by adding subroutines that return the remaining<br />

properties when any pair of independent properties (T-s, p-h,<br />

etc.) are input.

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