Chemical Engineering February 2014 - 35

0.6
0.5
0.4
Ma2 = 0.04
0.3
0.2
Ma2 = 0.7
0.1
5
10
15
20
25
Molecular weight
and T1, shown respectively in Equations
(17) and (18).
υ
12υ=
α
β
TT
=
12
α
β2
(18)
For an ideal gas, we know that dh
= cpdT and cp = γR/(γ - 1). Integrating
Equation (2) from inlet to outlet
yields Equation (19).
2γR
γ −
1 12 2
2
()
2
(19)
Assuming that T2 > T1 in Equation
(19), T1 and υ1 can be eliminated
by inserting Equations (17) and
(18) into Equation (19), resulting in
Equation (20).
2
γR
−
T2 11
γ 1
γRT
−1
() υ−= −
α
β
2
2
2


γυ
2
2
2 =
βα
αβ
−
−
(21)
By grouping T2 and υ2, Equation
(21) becomes Equation (22).
2M 1
γ Ma−1
=
2
2
βα
αβ
−
−
(22)
Equation (23) simplifies matters by
denoting the lefthand side of Equation
(22) as φ.
ϕ=
2M 1
γ 1 Ma
−
2
2
(23)
With only one Mach number in
Equations (25) and (26), we can now
plot the adiabatic flow equation
using Mak's chart. A typical graphical
representation of Equations (25)
and Equation (26) is shown in Figure
1, for known outlet and inlet
22
2
α=
22
2
α
β
2
2


(20)
Rearranging Equation (20) gives
Equation (21).
2
+
TT υυ−= − 1
(17)
Solving for the roots of Equation
(22), and knowing that both α and
φ are greater than zero, the expression's
valid root is given in Equation
(24).
α=
−+ ++ϕϕ ϕβ
2
41
()
22
(24)
By inserting Equation (15) into
Equation (13), we get Equation
(25).
f =−1
DMa
L 11 1
γα
2
2
γ +1
2γ
lnα







−
1+
1+
−


γ 1
2
γ 1
2
Ma
2
2
αMa
2
2





(25)
The term α is expressed as in Equation
(24), or, alternatively, as shown
as in Equations (26) and (27).
f =−2 1
DMa
L 11
+
γ +1
2γ
lnα





γα1


1+
1+
2
γ −1
2
21 ϕ
1


2
γ 1 Ma1
−
α
2
Ma1
++ +
+





(26)
ϕβ ϕβ ()
()
22 42
41 ϕβ
(27)
Isothermal versus adiabatic
With the adiabatic flow equation
plotted in Mak's chart, we now
can compare isothermal and adiabatic
flow in graphical form. A virtual
fluid with γ = 1.4 and M = 20
is chosen for the comparison. The
fluid passes through a pipe with a
length of L and an internal diameter
of D. The real flow condition is
unknown, instead, we assume isoChemiCal
engineering www.Che.Com february 2014 35
30
35
40
45
Figure 6.
As molecular
weight decreases,
so
does critical
p2/p1
conditions. In Figure 1b, the dashed
line represents the boundary between
the subsonic and supersonic
regions. Ma2 is equal to one along
the dashed line.
From Equations (25) and (26), we
can also see that p2/p1 is affected
by pipe data (fL/D), Mach number
(Ma1 or Ma2), specific heat ratio (γ)
and molecular weight (M). Figure 1
clearly shows how p2/p1 varies with
different fL/D and Mach number.
In the following section, the effect of
molecular weight and specific heat
ratio will be investigated.
Four virtual fluids with the same
specific heat ratio (γ = 1.4) and different
molecular weight (2, 10, 20
and 40) are selected for comparison.
They pass through the same pipe
(diameter and length) at the same
mass flowrate. The outlet pressure
(p2) and Mach number (Ma2) for all
four fluids are identical. The calculated
p2/p1 change with fL/D is
shown in Figure 2. The clustering of
the four curves in Figure 2 indicates
that the effect of molecular weight
on pressure drop is negligible.
Similarly, the p2/p1 values of
four fluids with the same molecular
weight and different γ is plotted
against fL/D. In contrast to Figure
2, Figure 3 shows four curves with
wide separation, indicating that,
unlike molecular weight, specific
heat ratio significantly affects pressure
drop. At the same pipe length,
p2/p1 decreases with increasing specific
heat ratio. Since all four fluids
have the same outlet pressure (p2),
a smaller p2/p1 in Figure 3 corresponds
to larger p1 and larger pressure
drop. At fL/D values of around
one, the calculated p1 for a fluid with
γ of 1.4 is about 6% higher than that
for fluid with γ of 1.1.
Critical p2/p1
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Chemical Engineering February 2014

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