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:
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