Michigan State University
DEPARTMENT OF CHEMICAL ENGINEERING AND MATERIALS SCIENCE
CHE 201: Material and Energy Balances, Section 001
Fall, 2021
Written Homework Problem Set 4; 105 points
Due Date: Friday, October 15th by 5:00 PM EST on Crowdmark
1. (20 pts.) Fifteen moles/min of hydrogen gas enter a compressor at STP and are compressed to
120 K and 10 atm. Assume ideal gas behavior.
%
v78
a. (15 pts) Calculate the volumetric flow rate of the gas at the entrance and the exit to the
compressor, in m3/h.
b. (5 pts) Calculate the mass flow rate of the gas at the entrance and the exit to the compressor
in kg/day
A)
Entrance :
Approach
15
4
pts
is
# 1:
5mF
mm÷×
Ex
1m
;¥E
Approach
v.
✓
=
=
In
20.2ms
-5
s+an%%
9
#2
ñR¥=l60nm-thx-j.ME
E¥÷÷÷¥÷¥
pressure
S
=
either
for approach (
acceptable) , I pt
answer
336m¥
20.2m£
✗
1
8 pts setup
Exit:
Ñ
Ñ
=
=
ÑPRT
( using
law)
K
F- 150
P
5
=
ideal gas
atm
(5mm ( 0.082056 m¥Em_)( 150k )
-
l
✓
=
37.0m¥
)
5
×
atm
692nd
✗
¥01m? 2.22m¥
-
2 Pts
final
answer
b)
Mass
stays
constant
b/c it
is
conserved:
nÉ=ñeg+=ññ¥-×¥i×-× -715
fe¥gnÑ
g-+3
the
same
Ñientr/exit
3PM
=
43632g
uynfntuersions
dB
×
-114
1000g
=
43,6K£
_day
Ipt
answer
meriting
2. (35 pts) A steel cylinder contains propane gas (C3H8) at 150 psig. The cylinder and gas have a
mass of 260 lbm. Think of this as State 1. The supplier refills the cylinder with ethylene until
the pressure reaches 900 psig, at which time the mass of the cylinder and gas is 295 lbm; call
this State 2. The temperature during the refilling is constant at 25oC. Determine:
oñ
a. the weight of the empty cylinder (in lbm)
b. the volume of the cylinder in ft3
NOTE: the mass of the cylinder, of course, remains the same in the two states, but the
mass of gas is different in each state.
(From
Ethane
Tc
MW
table
305.4 K
=
1)
B.
}
"
Pe = 48.2
atm
"
getting
3Tc
Pa
30.07 91m01
=
statF¥._= motifs
,
,
k=
Pts
2
1=0
l150Psis%?YPagim)(!¥7si
Pr1=
Looking
at
Fig
→
5.4.2
zpts
21--0<-9.2
we
Fig
if
.
5.4.32
between
→
0-2-0.24 )
=
0=23
)
from 0.9-0.94
statfrzZ-zoukfgos.dk
Pr2=l900•s;g!Y¥LK!¥-÷s
From
Zpts
find that
(anywhere
=
( ok
ZP
2- Pts
=
1.29
TPB
7%-122
?⃝
"
÷
,
-
"
.
'
=
0.98
.
.
state
~
i
i
"
-
i
.
.
.
!
;µ
Ppt 1.
29
of
Volume
""
cylinder
is
same
in
"" " "
MT t Nz ELMWE)
.÷;÷-
}
Z-i.npi.IT#- Zzhpz RTz(p,anang ymg g, ng
Zpts
Zpts
zpts
state 2
{
295lb
P, V
state 1 :
state 2 !
=
=
Pzv
=
>
Z , n RT
,
,
Zzhz RTs
both States:
"mnowns:
hi e) Me
Mt
,
,
V
Heyns
Ipt
,
!%¥→%m?*m=!÷I-,
=
"
☒
nz
L295.tt?b#mol
66.2
=
Equating
volumes
:(
R
cancels
(0.92712%-7,5*1*130414)
5Pt§
've
165
for
I.
kf5
Mt
-
out both
sides) :
)( 626.27km
(0.22
29.5 Mt )
-
.
=
psia
¥4k)
=Éa
"
0.006 Mt
1. 38
=
=
0
.
071
0.0053
-
mi
mt=25_8bm_
0.0002Mt
b)
Volume
V=
zptsmg
"
✓=
of
Zznz
cylinder
RTZ
:
ZNp;R
or
I
condition 2
convert
31°C →
'
,
"
or
=
547012
:÷i÷÷¥÷÷÷¥I*
to 57ft >
-
3
pts
-
propanol
3. (25 pts.) A gas composed of 15 mol% rooms
butanol and 85 mol% CO2 is at 485 K. Calculate the
pressure of this gas mixture if the specific volume of the gas mixture is 195 cm3/gmol.
Y prop
0 15
=
F- 485
•
Ñ
Ycoz -0.85
From
Tre
propane:b
I
1959¥
B. 1
Table
coz :
=
K
=
Pc
304.2K ,
=
72.9 atm
Te=536-7k,Pe=95atm
4
pts
Pe
Kay
Using
3.PB
3P
's
Pc'= 72.910.85)
304.210.85)
t
-49.95 ( 0.157=69.IM
536.710.15)
solwtionapproa.ch#1Tr1--T/Td
=
Pr
'
=
he Tc
rule :
+
To'=
identifying
PIPE
485/339.1
=
=
=
1.43
1%9.5 atm
3
339.1K
pvn
To
ZRT
=
get
R
into
0.08206
sit
Imaging
P=zR
e
5. Pr
Pr
Looking
gpts
correct
µM9
apart
{
If
>
¥4m
✗
'
.
=
'
chart
at
=
Z
2.94 Z
5.4.33
2.0
Pr
If
Pr '=
28
Pr '
2.
=
emisatm
atm
Tr= 1.43
Pir / 2- = 4.61
7=0.76
If
=
82.06
=
E)
*
204.1
Pr '= 3.5
'
.
(204.1
=
pop;=p
69
'
Z(•%fmI÷E¥(485
=
P
units
correct
=
2-
=
0
.
Piz
80
Or -775
2ns
=
Pilz
=
2.978
3
t-G.37169.gs?qtm)--1#.mORApproachH-2Fh
Therefore
After
establishing
Pc '=
6915
Tr '=%
=
from
Tc '= 339,1
atm ,
1.43
Kay
that
be
's
K
Pr '= 769,5
rule
:
atm
④ Find
it
idea
'=R÷÷ YE
=
fr ideal
t.95emygmolfa.obamtge-231339.fm)
=
69.5-a
⑤
if ideal
Use
0
=
487
.
5 pts
t-ig-543.TN owing
Spts using
correctly
chart
z
=
0.77
Tr
'
=
1.43
Ñrideak
( ok
z
0.487
0.75-0-80
if
)
P=zR÷=° tµaYa;÷Ims)"•
SPB
OR
B
>
Use
Fig
5.4.3
.
Pr '= 2,4
P=
=
If
2.35Pa
'
-
Spts
,
'
=
t-163-a.tn
?⃝
find Pr -2.5
to
for
using
chart
2.
35169,5
5
pts
)
atm
4. (25 pts) Your team is designing a vessel that has a volume of 5 L and is to be filled with 3 kg
of dry ice. The vessel is sealed and heated to 45°C.
a. (20 pts) If the safety factor of this work is to be 3.0, what should be the minimum vessel pressure
rating? Note: Safety factor is a multiplier of the maximum operating conditions reported. If a table
is reported to hold 100 lb and the SF is 1.5, it must be able to hold 150 lbs.
b. (5 pts) If you cannot find a vessel with that safety rating, what design parameters could you
modify?
a) Dry
→
coz
4491inch
,
ice
MW=
Tc= 304.2K
-
Pc=72.9_atm
#
2.
pts
tr-lus-s.IT?;-E--t.osPr--P/zz.qapm-ZPB
Use
ideal
critical
volume ?
÷÷÷÷÷÷÷÷÷
"
.
Rtc
4
ride at -0.205
Ipt
✓
Using fig
3 pts
5.4.3
.
2-
←
Pr
or
p
=
•
( ok if 0.30-0.35)
31
pI
identifying
2-
0
~
Pr Pe
.
1.6°
=
( ok if
72.9
1.5
-
atm
%116.co#tM2ptsreYIFhsnPy,wzorPrFdP
3P
Therefore
be :
pressure
,
rating should
3.0-(116-64)=3501%12
2. Pts
of
correct use
safety rating
b)
-
Generally
total
we
Lower
much
Pr
and
the
the pressure
rating
leg not as
heating) would have
temperature
similar
5- pts
increased
we
volume of the vessel ,
reduce unideal and
would
their
-
if
for
.
effect
at
least
7-
idea
)
1.75