Environmental Engineering Reference
In-Depth Information
Question #3.5 T pc and p pc
Given that gas gravity to air is 0.8 assume “Miscellaneous gas” and:
a. Find T pc = ___ °F
b. Find p pc = ___ psi
The values of p pr and T pr thus found can be entered in Fig. 3.14 to find the Z factor.
By way of example the value of Z can be found starting with the given values:
p Twf = 2,000 psia, p pc = 650 psia, T Twf = 200 °F (660 °R), T pc = 410 °R then,
1. Find p pr = p Twf / p pc = 2,000/650 = 3.07
2. Find T pr = T Twf / T pc 660/410 = 1.61
3. Enter abscissa (top) at 3.07 ( p pr )
4. Go down to T pr of 1.61, between 1.6 and 1.7 lines
5. Read Z = 0.828
Question #3.6 p pr , T pr , and Z
Given:
T pc = 420 °F, p pc = 662 psi, T Twf = 149 °F (°R = °F + 460), p Twf = 2,913 psi
a. Find p pr = ___
b. Find T pr = ___
c. Find Z = ___
Once Z is established, 1/ B g can be calculated either from the equation:
T
T
p
p
11
BZ
=´ ´
sc
Twf
g
Twf
sc
or by use of the nomogram given in Fig. 3.15 . Note that values for T in the above
equation must be in degrees Rankin (°R). To convert from Fahrenheit to Rankin, use
°R = °F + 460. By way of example 1/ B g can be found graphically given that
p Twf = 140 kg/cm 2 , T Twf = 93 °C, and Z = 0.828 by the following steps:
 
Search WWH ::




Custom Search