This preview shows pages 1–2. Sign up to view the full content.
Homework 3 solution
2.24
(a) T=300K, n=PV/RT=0.1e6 Pa*0.1m
3
/8.314/300 =4.01 mols
.
(b) T
2
=T
1
(P
2
/P
1
)
R/Cp
=300(1/10)
2/7
=155.4K
n
2
=P
2
V/RT
2
=0.1E6 Pa*0.1m
3
/8.314/155.4 =7.74 mols
2.27
Energy balance:
;
out
out
in
in
dn
H
dn
H
dnU
−
=
Since n is constant (n=P
i
V/RT
i
=0.1E6*0.002/8.314/300=0.08 mols),
dt
dt
dn
H
H
ndU
in
)
/
)(
(
−
=
(Attn: n and n
in
are different here.)
dt
dt
dn
T
T
C
dT
nC
in
P
V
)
(
)
(
=
−
;
Integrate from initial temperature T
i
to T, dn/dt is constant mass flow rate
t
dt
dn
nC
C
T
T
T
T
V
P
in
i
in
)
(
ln
=
⎥
⎦
⎤
⎢
⎣
⎡
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
=
−
t
dt
dn
nC
C
T
T
T
T
V
P
i
in
in
)
(
exp
)
(
T(t)=35050*exp(0.4062t) K (t in min)
P(t)=nRT(t)/V=332.56*T(t)=0.11640.01663*exp(1.7452t) MPa
Steadystate temperature is 350 K. It will be 345 K within 5 K of steadystate
value:
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
)
(
/
ln
dt
dn
nC
C
T
T
T
T
t
V
P
f
in
i
in
f
=1.319 min=79.1s
3.1
T
1
=15
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/10/2008 for the course CHBE 2110 taught by Professor Gallivan during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Gallivan

Click to edit the document details