o
(t
N*
*
.J
3
O)
c)
O
(,.)
N)
o)
I
--r.
o
O
O
O
I
O
O
O
O
O
I
.-L
SLAYthe
{
c:
P.E.
L
o
-{
C
wrwv.Slar,thePE.com
CI
U)
o_
0)
:<
o)
o
J
MECHANICAL EI{ GINEERII\G
N)
THBRMAL AND FLUID SYSTEMS
@
PRACTICE EXAM (SOLUTIONS)
I
O
.-t
N)
i\)
(o
-U
,.r,r,, rr'. S
J
ar,iire['
1.,
co*l
t,-,.;p1 r:11i;i .r:l
lill? .tli
right-t reservctj
(t
rT
N
*
:F
J
)=
o)
O
O
iri:: f.iechiiriica: ,'i it*r:r-:rii ar:J Iir.iiLi Sr,sicurs
-
l,r'ilrticc i:l:::i:t Sciurrons
G)
N)
O)
I
_L
O
O
O
O
I
MECHANICAL ENGINE ERING
THERMAL AND FLUID SYSTEMS
o
O
MORNING SESSION ANSWER KEY
O
O
O
Question
I
..1
Question
001
B
421
B
{
.-L
aa2
D
422
D
003
E
023
IJ
C
C-
004
C
024
B
o'
00s
D
025
c
{
006
B
026
A
o
({,
oo7
D
o27
I
008
s
028
B
009
A
429
B
010
c
030
(,
011
A
031
A
012
c
432
B
013
A
033
c
014
D
034
B
c
o_
o)
5
o)
-t
o
:t
p
N)
O
.--l'
@
-t
N)
lu
(o
"tl
lr"u,u'.
S
iar'! he
I) ! :.
ct:nt
015
A
035
a
016
B
036
D
o17
B
o37
D
018
B
038
B
019
D
039
D
020
B
040
ti.rp:'rigiti
?0 i
7,{li
ritlris rcsi:rvrd'
(t
C'
N{-
)t
i'i-i fuitciriinit:iil
.-J
3
O)
O
O
(r)
l\)
-
1i:erril;.ri rnci l--lL:iii Sr,-sig:1',5 .'.ilrl',:rii:.:
i:r..irl Soiiiiit:r-ls
',..,.rr,, Siat
t!:ti'l:
e
ut-rr
SOLUTION 001. THE CORRECT ANSWER IS: (B)
Statements (A) through (D) are all true, but three of them are unrelated to strain hardening.
Statement
(B) is the definition of strain hardening.
o)
I
-l
O
O
O
O
I
O
O
'O
O
O
I
THE CORRECT ANSWER IS: (B)
-l'
J.
C-
c
o'
c-{
o
a
oo)
5
SOLUTION 002. THE CORRECT ANSWER IS: (D)
Statement
CI
C)
)
I
(D) is true. The equation of state Pv,:ZRT provides accurate
temperafures and pressures than
results for a vvider range
of
the ideal-gas equation, Pv:RT, does. Hence, the compressibility
factor is a correction factor to account for deviation from ideal-gas behavior at a given temperature and
N)
pressure.
@
.-r
N)
THE CORRECT ANSWER IS: (D)
O
.-t
l\)
(o-U
ivii",r, S ia.,.iirci)ll coirr
f-'op:',lgiit
r.-
-lii!? '\li
ri3iri.< r'cstlveij
u
g
Nt
I)l: \lclltrlrii-:al - [itcir:i:ri .lir.i [-ilirri Si::ttrr.r - I'r.retlrr ::\i]:l Sr]ii.ilifii-],\
{-
l-rr'.
n
.
S
lai'lh*i)!l.cr-.r':i
.-J
3
O
O
(,
o)
I\)
o)
t
_\
O
O
o
O
I
O
O
O
O
O
t
..-r
{
-t
(-
c
o'
c-{
o
U'
SOLUTION. 003. THE CORRECT ANSWER IS: (B)
The problem statement indicates that the salvage value matches the book value, which is the difference
between original purchase price and accumulated depreciation. For the double-declining method, the
book value at the end of
year can be calculated directly (that is, without calculating the
accumulated depreciation) as:
I
BV,:CI
rn'here C is the
t
t1j
-;J
initial cost, and r is the asset lifetime in years. For our case,
I
r lr5
. t -+l
r)J
BV,r:9235,0001 I
=$27,469
The salvage value is a lump sum payment occurring in the future, thus it is denoted rvith the symbol
F
.
F*,** .B=$27,469,which no'uv needs to be converted to a present value. This
is done by multiplyingFby the "single payment present worth" factor: (etf ,i%,n), which is
The salvage value is then
tabulated in the literature and can be looked up directly. Alternatively, you can use the equation
(
f tf
,
i %,
n):{t+r
)-'
:(
1
+ 0.06 )-'5
:
0.417 3 . Therefore:
P***u :$
o_
o)
5
theT'-1l-r
27,469 x0'4173
:
$ 1 1,462
THE CORRECT ANSWER IS: (B)
o)
c)
:t
I
N)
O
@
N)
l\)
CO
T
rri,...r Sl.it
lhrl'[-
ci-.nr
L'i.ri;r'rigl:i
i .i(iiI ;\!I tigit;s iescn'c,.1
C'
(t
N*
*
J
3
o)
O
O
(r)
t\)
o)
I
--l
O
O
O
O
O
O
O
O
O
I
I
l)l', ltdci:hlnici:i
'
i
ilrlir,l! r;iri l:iuiii
:11
:icir..!
- !)ri:iiti:t i:r,i:i:t -to!,";iti:trs
11'sr.'11,
ilil','1!igPll clri:i
SOLUTION 004. THE CORRECT ANSWER IS: (C)
The provided drarving is an example of an illustration technique known as multi-view orthographic
projection, which is quite common in engineering drau'ing. In this technique, the viervs are positioned
relative to each other according to either of tr.vo schemes: "first-angle" or "third-angle" projection. The
truncated cone symbol shown in the provided drarving indicates the scheme being used is the thirdangle projection. The third angle projection scheme is more popular in the United States. A simple way
to visualize this technique is to place the object in the bottom of a bowl. Sliding the object towards the
right up to the bowl's edge reveals the right side view.
By inspection, you can confirm that the only view that could be obtained by using third
angle
projection from both of the provided views is C.
{
-1,
--a
C-
c
THE CORRECT ANSWER IS: (C)
o'
c-{
o
a
oo)
5
o)
-t
o
J
I
t\)
O
--.l
Oo
.J.
N)
in
(o
-U
rv*'u'. Sial, tl:,:lll
i. cc,;.:r
i,."ril::
riltrt
't lt i? :\ii iigirls itset'r'i.:i
5
N
{{-
J
J
=
O)
O
O
u)
N)
o)
I
-L
O
O
O
O
I
O
O
O
O
O
I
.-\
ir.
(-
c
o'
c-l
o
a
o0)
I'l:. L'ltcl:l:r:i-':ii * 'l-h,-"r'r:iirl iir:i.i l'r:riri Sr,.i.irrr:- !ri::ci:cr: ir\:lt:: Si)iltttclih
SOLUTION. 00s. THE CORRECT ANSWER IS: (D)
This is a linear programming problem. Linear programming (LP) is a quantitative business analysis
technique typically taught in industrial engineering and MBA courses. In LP we attempt to optimize a
function subject to a set of constraints.
First, identify whar is it that must be optimized (in this particuiar problem, optimize = maximize). This
must be defined in the problem statement. Clearly, for this problern we must maximize profit.
The next step involves identifying the variables of the problem. This is most easily determined by
reading the intenogative part of the whole problem (usually the last sentence) "The number af parts A
and B that must be made daily" Thus, if we let
I
l\)
O
.-L
@
x:
the number of parts A made in a day, and
y:
gt.
number of parts B made in a day, then the profit function is.
P(x, y):59:r+30y
Otn task is to find the(-r.y)pair that
associated
rnakes
Ptr,y)as high as possible within
the constraints
wi& the problem. This brings us to the next step: identifoing the constraints.
The first constraint is the capacity constraint. The machine cannot make more than 100 parts per day:
5
o)
-t
c)
r'"ri.rr 5i:r., iil,-'iti: Lrrtrr
x*y<
100
Also, there is another capacity constraint; the number of parts has to be greater than or equal to 70:
x+ y>70
The last constraint is the cost. The plant can only spend up to $1200 a day in this operation:
2Qx+ l0;,s1200
and of course, the implied constraints x
>0 and y> 0 (one cannot make
a "negative" nurnber
of parls).
.-\
f\)
i$
(o
T
The next step is to graph the feasible region (this is what we call the collection
of (x, y) that satis$ all
the constraints). The implied constraints tell us that we need only to consider the upper right quadrant
in the (x, y) plane. First, find ail the axis intercepts:
ri.,,.'ri'. 5 1:r, llrc PI'i.t:lrnl
t
l'-rp_r:
i5i:t :.' li;
i? Ail nghts resi-'n'cii
cr
ct
N
*
)F
*.
I ;; ivl(:Cl1i:i,.ii:i:j - l-jtCin:l!
:illr.i ij.ii:iJ
Jf
J-!
=
o)
- iir:iirtiii,i::rtrn Stiiut:o1s
x+Ysloo
j.B
x+y270
O
O
O
O
,20 x* fi y<1200
r,,,u,r.i,.Slrt-r.liie
i)l:.eorl
*:*--,=-____=
Intercept
loo-- loo
70
7A
t)fiht:,^_rn
1200120:60 ,4
t20Alt0:
I20
For the first constraint'
plot 100 on both the x and
yaxeq and connect the
points with a straight
Next decide which side
line.
of the line is the side that
contains sorutions ro the
inequarity. pick a point
which is not on the line'
ln this case the origin is
the simprest such point
with which to work.
5
8
8
I
Does the
P
in all of the points in
the upper right quadrant
which are on the same side
of the line as the origin:
I
.
120
c
C
S
Constraints
O
G)xr)
O
=
o
S.,
---_______j_j..;li:i
100
rB0
I
o60
a
c}q
CI
5n
grro''4ari,h
5
3
r
ne lower vqp4urry
capacity constrarnt
constraint has y o-J ., i-+^-_
hasxandyintercepts which
are both equal
.P
12A
N)
O
_r
to
70:
.
100
Oo
-.t
f\\
rv
80
60:
a^
i$
(O
40'
T
>.
0:
In this case' the origin
is not u JorutiiX ,o lfl,,
is nor on the correct side,
"flrouilr. riH.
other side is the correct
the
side' The region of ftasible
sorutions has been cut down
to the shaded area
above' Next' graph the
cost constraint. It has
an;r intercept of 60
ffii"
and a 7 intercept
t,,'R,r.;. S j a.i.ihc{}l:1.
of
120:
C<tr-r.:
tlcpi rillrt
,,,,
l{i l7 ;\ti riqhls icjcirrtrj
u
C'
N
*x-
11i',, 1..
$l;;r,:1liq:!)l;
ir!I
J
3
o)
140
O
O
(,^)
120
100
l\)
O)
80
I
_t
O
O
O
O
O
O
O
O
AN
4
20
I
U
100
120
c)
We have used all the constraints and determined the feasible region as the shaded area shown above.
.-l
{
-L
(:
L
o
c-{
o
a
is a feature
[t
ollinear programrning that the maximum and minimum of the function being optimized are
always in a corner of the feasibie region. So, we have to identi$ the comers. By inspection we see that
(0,70) and (0,100) are two corners. The next comer is the intersection of20x+10y:1200and
x* y:79 which can be quickly
intersection of 20x+ lA y:!200
found to be (50,20). Similarly, the final corner is (20,80), which is the
and
r+Jl:100
.
oo)
5
o)
-t
o
I
N)
At this point, it is wise to see if
correspond to any
some
of the answsr choices can be ruled out
because they do not
of the feasible comers. We see that choices A and B do not represent any of the
feasible comers, thus they can be ruled out now and not considered any further!
Likewise, it pays to check if any of the feasible corners is not in the answer choices. We see that (0,70)
and (0,100) are not any of the answer choices, so there is no need to consider those in our anaiysis.
O
-r,
@
Now, we evaluate the objective function in all the feasible corners:
t\)
-l
P (50,20)= 5o(sO)+
i$
(o
-U
"
P
30(20)=5 3.1s9
(zo,so;:59(zo)+:O(s0):$3,400
The winner is (20,80); that is 20 of part A, and 80
of
part B.
Although there is no need to do this following step (because of the multiple choice format) the
objective function atthe points rve have ruled out is P(0,70):SO(O)+:O(70)=52,190
P (0,100)= s0(0) +30 ( 100
ri',',rr.Si:r . rl:ci)l' rorti
)=5 3,000
*O
.
{. ,]11i
1;3;,t
i
:..t l
l.
;f ii ilgi:::; i*sOi\',rti
(t
I
N)F
J
)F
3
O
O
(^)
o)
l:'1.:
i\'lctilanrcal ' 'l-l:crr':rl.ri ;nd irliii.i S\ 1la;'ils *- Priltittc I;r:rn 5rr!utt<ur:,
"vrvr','
S
ia'. tirr:l)I
r.cixt
THE CORRECT ANSWER IS: (D)
N)
o)
I
--l
O
O
O
O
O
O
O
O
O
I
I
Alternate solutio,x: Under certain circumstances, it might be faster to just plug in the answer choices
into the objective function and choose the one that provides the. optimal value rvhile still respecting the
constraints.
Once you have defined the objective function and the constraints in mathematical form, set up a table
{
-L
(-l'
with the results fiom plugging in the answer choices:
c
Answer Variable Constraint I
Choice r y x*y<100?
o
A
c{
B
30 70
70 30
50 2Q
20 80
Constraint 2
x*y>70
Constraint
2A
3
Objective Function
x+ fi y< 1200
50
x+34
y
yes
yes
No point doing this
yes
yes
No point doing this
yes
yes
yes
$3,100
yes
yes
yes
$3,400
o
g)
c
o)
D
o)
Note that, in the process of filling in the table we can rule out choices A and B, because they do not
C)
satisfy the third constraint (choices A and B yield costs of $1,300 and $1,700, respectively).
I
This "plug-n'-chug" approach certainly is quicker, but it still requires careful reading of the problem
N)
statement for the correct mathematical formulation of the objective function and the constraints.
o-
5
:t
O
.-r
@
-L
N)
$
(o
-U
ltir,rr. Si;t\ ihcPll.corn
lio1,..ri3irt
'
2.1
t7,tll rr5l;ls,.r*,,.'.i
CT
(t
N
**
J
=
-J
O)
O
O
(-^)
N)
O)
SOLUTION 006. THE CORRECT ANSWER IS: (B)
The question is about a temperature diference. The conversion of temperature values between the
scales is:
o
Tr=32+iT,
I
_L
O
O
O
O
I
O
O
O
(3
t
..-t.
{.--I
(1)
whereas for temperature changes, the relationship is:
q
A
fn: TLf ,
(2t
In our case, the lon'est temperature ditference is 25Yo of the normal
L T r: 9.25 r30"F : 7. 5oF . Therefore A ?''c : { 5 / 9) x 7. 5.F : 4. I 6"C
temperature difference;
THE CORRECT ANSWER IS: {B)
C-
c
o'
c-{
o
a
oo)
5
o)
-l
Further clarification:
If you are only familiar with the use of the temperature values conversion formula (equation l), then
this altemate approach can clariff alittle bit: If the ambient temperature is, say, 70"F then the normal
c)
chamber temperature is 40'F (because
_o)
difference was cut to 7.5oF, hence the chamber temperature peaked at (70-7.5): 62.5oF. Since 70oF
N)
2l.l'C,and62.5"F:16.9'C,thedif&renceintheCelsiusvaluesis2l.ll-16.9:4.2
O
it is 30"F lower). During the process
upset the temperature
:
--.l
@
70"F
-r.
N)
70'F
I
t$
(o
-r-2l.l.c
-u
A:30 | A:16.7
---21
1"C
i
I
62.5"F
I
i
I
I
I
40'F
',i r..'."r.. Siar,
l.itePi;.rtt;rt
f
4.4'C
lii
t'rrilr,r i'riri .i l
?i-i
i7 'Ai!
Ligl:i:. r';:crl.ijd
q
C'
N)F
*
._i
Fl: ir'!crl.:anrca! -
.i
rvu,r.. Siar
ircri:;al ::;;t.; ljii:iii Sls:c;-r:s-'lli'rii-trl:a1 l:rnri, Sr:ir.:trt:r's
3
SOLUTION 007. THE CORRECT ANSWER IS: (D)
O
O
The absolute pressure is the atmospheric value plus the measured (gauge) pressure.
o)
G)
N)
o)
I
_1,
O
O
O
O
I
O
O
O
O
O
t
Pu*: P*,
l.ire
l'll tc:lr
+Pnur*"
When a gauge pressure is negative (below atmospheric, it is typical to report the magnitude and to note
that
it
's
avacuum, as is done in the problem statement).
P
*,:70.7 kPa-25 kPa:45.7
kPa
Now use the appropriate unit conversion factor to convert to the desired units:
THE CORRECT ANSWER lS: (D)
.-L
!----r
c-
C
o'
c-l
o
a
oo)
5
o)
o
I
N)
O
.--l
@
..J
N)
in
(o
-U
r','rr rr.. 5
l;r, !ir,:
i) i i
etrnl
ii
('opr i r,-'ii: i' -l,l i 7 .{il I i5'i:t: !('jr! \ rri
(:r
oN)t
)F
J
-)
=
O)
O
O
(,
N)
irl:
ME:eiranlrriij
-
-l'hcr-nr::i
;i;tLj i:iLiiii 5.,,:Ierris
-.-
l)rilr:!lcc i] \it;it tt:i:.!tiil::
',i rr
",
S
l
lii l itt
l) l r crir ir
SOLUTION 008. THE CORRECT ANSWER IS: (B)
The fluid velocity in the thin gap of a cylindrical viscometer varies linearly. The fluid layer adjacent to
the inner cylinder will have a velocity equal to the tangentialvelocity of the cylinder, rr,rR, where ru is
o)
I
the rotational speed of the inner cytinder (in radls) and R is the radius of the inner cylinder. The fluid
O
O
O
layer adjacent to the outer cylinder will be stationary. If the gap betuaen the cylinders is ft, then the
--r
o
I
O
O
O
O
O
I
{
-l
velocity gradient across the gap is (o R)lh
When a torque
is applied to the inner cylinder, the fluid in the gap experiences a shear stress
r
"
which is related to the velocity gradient:
,:u@P
Since the torque is:
I=stress Xarea X moment arm
-1,
C-
c
o'
I
.
T:tX2r
RLxR
of,
c-{
2rR3oLu
, :---h-
o
(t)
o-
(})
5
o)
c)
J
p
Additionally, the power P, transmitted by a shaft rotating rvith an angular velocify
torque
ln is
a
P:oT
Combining these two expressions, we get:
p-
2ro'RtLpt
O
(t)
h
where, from the problem statement we have
t\)
i\)
(o
T
by applying
given by.
N)
@
co
arso
,.,:80
rrr
:
1200 rpm .l t min /OO sl-12 x t
cp.l!!ls/gl=
I
a7880cPl
So we can now plug in all the numbers into equation
z't
P_
*;.,1'11' $
j
11;{lixi)l:. eorn
rzs
z,-'t'(o s i.
revl:
125 .7 s-
|
r.ozx r0-3rbf s/ft^2
(l):
l+*l)'{or l#l*zi"){
r
oz,
ro-'F)
0.0079 in
l:
(.',rl:rii:ilr r
-)ii
!l ..\l: irtil'r rJ\rr'\i.-i
ry
o
N
{-
*
.-i
J=
O)
O
O
(r)
t\)
o)
I
jll: lle ci,-:iltrr":ai -
l'he
r'rlli
i:i-rii I:ir-iid Sr,-s1rrr.l:;
- l)r:trirlr
rr -rrr.t Slilr
f--r;,irn SoJlrtrons
llici)l: rrl:n
P:4etri"ln#ol=ornono
THE CORRECT ANSWER IS: (B)
---r
O
O
O
O
I
O
O
O
O
O
I
.-L
--L
SOLUTION 009. THE CORRECT ANSWER IS: (A)
(-
c
The Pitot tube inserted in a flow field "traps" some of the fluid,
o'
so the inside of the tube represents a stagnation point. [n the
c-{
analysis
of
compressible
flow, the stagnation pressure is
o
a
typically denoted with Po. If we denote with P the free-stream,
o)
static pressure, then we can calculate the pressure ratio:
o-
5
P
Po
o)
C)
=t
Pitot tube
lr7+r4.7!
1.25+14.7
With this value you can interpolate in the isentropic compressible flow factors table, or use the
I
isentropic compressible flow graphs (both graphs and tables are available for download in the "free
N)
resources" section of www.SlalrthePE.com), to obtain the value of the Mach number
O
-\
oo
Ma:0.58
THE CORRECT ANSWER IS: (A)
-L
!Y
N)
(o
T
ri',i.
ri
S
I
;rr
lhc['[:.ctrirr
!i
Lop.\'fisiit i::
ltill :\ll rigfrts rcse;i,*.1
U
C'
N*
*
J
3
O)
O
O
(-l)
O)
t
.-:
SOLUTION 010. THE CORRECT ANSWER IS: (B)
The equations for normal shock waves are rvritten for a frame of reference on the shock wave. When
vierved by an observer on the shock wave, we have the air
- upstream of a stationary
shock wave
-
approaching with a velocity of600 m/s:
O
O
O
O
O
O
O
O
O
shock
wave
.i
will figure out the velocity dorvnstream of
r-i observer
So first, we
;1
the shock \\ave, l/r, for a frame of reference fixed to the
.x--,
I
shock wave.
tl
v,
"
<--,_:*_
I
vr400mls
' ', '/')'/:'//t/l//l////zt/zl; /7...Ii/
',
)','z'
Since we know the temperature upstream of the shock
\ /ave, we can calculate the speed of sound upstream of
'
the shock wave.
{
(-!
.-L
c
o
-{
C.
CI
{n
oo)
J--QO +ztZ)K :343 m/s
R'J,:
a" 1il l.4 Qg7 kgK
therefore, the Mach number upstream of the shock wave is: M ,: V ,l c ,: I .75 . With this value you can
,,:
i
k
use the normal shock wave factors
for compressible flow table or graph (both graphs and tables
available for download in the "free resburces" section of r+rvw.SlaythePE.com) to obtain the Mach
number downstream of the shock wave: M z:A.628. At this point, we can calculate
Vr:Mrc,:Mri
o
=)
c)
:t
p
where
Tr:iTrlT.,l.I,
Y
J-x
-":a.628 tlt l.+xzn t(gK
438 K:263.5 m /s
@
Bu! this is the velocity
-r.
T\)
shock wave. Using the notation customarily used in kinematics,
T
?".: 1.495x293 K:438 K
and the velocity downstream of the rvave is:
N)
(o
tr/, as:
kR^rf ,
and the ratio i TrlT,l:1.495 from the table. Thus
O
t'\)
are
seen by an observer on the wave, that is, the relative velocity
r,ve
with respect to the
write this as: t 21r.*.:*263.5 imls
where the'notation "/s.w." in the subscript means "with respect to the shock wave", and
vector
in
the
x
(horizontal, left-to-right
/ rtr.n.:/ r,nuu*-y'
s.rv./srund
V
zr
is positive) direction. From
?
is the.unit
kinematics, we know that
therefore
s**t:/
r,,,,
*/
...u./s,*nd
:?
zat.s +
e00 )
i mls :
336. 5
i m /s
THE CORRECT ANSWER IS: (B)
lr
* ri'
!!3r.:11;1!)i:..*:..trt
!:i
i',,p..Ir:
:1
, lili-: .iil itll;{.. lus\'i\cd
o
ct
N{,F
.J
3
o)
fri: \it,:it.:ri;t:ll .' i'ircir,'rll l*r! I il::r.l S..::!rr:..:; !ii:riiicr
1 is in stagnant (no velocity), atmospheric
pressure air (no gauge pressure). Point 2 is just
1
\
I
/\
+l*
inside the hose, at a given vacuum pressure and r,r'ith
v__2
an unknown, but non-zero velocity.
This problem is solved with the Bernoulli equation. For steady, inviscid flow.
tr'
,'
z
Setting
V
t:0 , pt:0
applications) we obtain this expression
-;.J.
Noruv, insert
pz:-0.3lbf1in2
o
:
o
^:l'.titl
tr/ I
U'
o_
o)
5
and use
I
i
c-{
p,
- -+ gh,
*p,
nr"
-, _trt r' +f
i* - shr:
,
and neglecting the change in potential enerry (as is customary with gases in most
.-l
c
ton-r
Point
G)
N)
O)
{
c-
tlrtl)I'.
SOLUTION 011. THE CORRECT ANSWER IS: (A)
O
O
-l
O
O
O
O
I
O
O
O
O
O
I
r.,'ri rr'.Sii:t
i::.tarn Soluircns
t:
I
I
for V r:\ii-2 pJ pl
.
p:0.07651bm/ftr for ai
rzinl' lrz.ztu*.tlt'l
^ | ^ "rbf l
-'l-";tj
l'ftll rhr l:rern/s
I
_ _-_tb
0.076s:
"'"'"'ft3
The problem statement asks for the velocity in m/s, so we have to convert our result from ft/s
vz:ter*lr*Frr+
o)
o
::'-
I
t\)
o
oo
.-l
N)
i\)
(o
!
r,,'ri
ri S i:n.ihcf,!:.ctr;n
1i
l\-'1r_1 E11i,1
, l{'if
,\ii iillris
rtliCr\ r'ti
(:r
u
N
:+
J
)
!'1
*
o)
O
O
(^)
f\)
o)
!
-1,
O
O
O
O
O
O
O
O
O
I
\r\\,.
Si
ii\il!
:e
ili.l r:oi!l
SOLUTION 012. THE CORRECT ANSWER lS: (C)
The equivalent length for the valve is the length of straight pipe that produces the same friction
pressure drop as the valve. From the problem statement, the pressure loss across the 1,000 ft of pipe is
70 psig
-
25 psig
:
45 psi. This drop of 45 psi is the sum of trvo effects: I ) friction within the pipe, and
2) the presence of the valve.
The flow velocity
tr/
is obtained with the relationship betr,r,een volumetric flow rate Q, and pipe cross-
sectional area- A.
r
o
A
-\
\--.]!
--l
(C
The total head loss due to friction and the valve can be calculated with the Darcy-Weisbach equation:
o
.{
hto"",rnrol:-f
'+#
C
CI
6
which can also be wrisen in terrns of pressure drop, rather *r]r, heud tors,
o0)
5
o)
-t
o
f
LP,,u.rror:t
++,
where L,o,o!: L o** Luq.*t"" includes the equivalent length of the valve. tnserting this in the expression
above, and solving
fot L*.,ot,
we obtain.
p
N)
O
--L
@
.-r
!Y
N)
(o
,,.**:\ryf-1,,":
-1000ft
0.018
*
222ft
10.1q .62.41ry
F
ft'
THE CORRECT ANSWER IS: (C)
-g
\r!\'lv Sl3\'i liei:'L:, itl:i
ii:
(ic;>-irigi:i ,' :(i
i; ;\ii rtgi::s :ci*i vt-'ri
5
cr
N)F
f ii :"4ctharirci:i ,.'l'irr.rli,,l ;irril lrlriiii Sr,stcrl:;-. i,raciirc
I:_r.ii;l lioiuiir;rrs
*
J
3
rr rr rr tl.r. ti.,.i)1.
. -.,.
SOLUTION 013. THE CORRECT ANSWER rS: (A)
Assuming for now that at all times the sphere is
a spatially isothermal body (its temperature is at the
O)
O
O
(^)
N)
same value throughout the sphere, but this value
changes with time), we can approach this problem by
.-J
using the lunlped capacitance method. For a lumped
capacitance, the temperature history
object is given by:
o)
I
O
O
O
O
I
O
O
O
O
O
t
-J,
{
.-L
c-{
o
U'
the
rQ)-r- _^-.^l | *,\.1
r:T_:e.o[1;*],1
where
r- is the temperature
of the bathing fluid,
?",
is the initial spatially uniform temperature of the
body,ft is the heat transfer coefficient,A"is the surface area
and p,
v,
andc are the body,s density,
volume and specific heat respectively. We can re-arrange
the expression above and solve for
7,:
h:1n[r(t)-r-1.1-on")
L T,-f_ Jt A,t l
cCo'
r(r)of
Using the standard formulas for a sphere we see
p :558 lbm /ftr and
c
thatV/A,:rl3:0.t66in,
also for solid copper
: 0.092 Btu/( lbm. "F ) . Therefore:
o-
o)
5
o
=)
-t
C)
N)
with this value of the convective coefficient we can now check if the
assumption of a spatially
isothermal body is valid. For this we check the
Biot number Bi:hL,lk, where L,:vlA,is
--r
characteristic length (for this problem
-)
I
O
oo
-.J
N)
l\)
(o-
!
a
VlA":yl3-0.166in) and& is the thermal conductivity of the
body' For copper we can use ko22}Btu/(hft"F). Hence, *0.000g
Bi
. In general, if the Biot number
is less than 0' 1, the error associated with using the
lumped capacitance method is negligibly small.
THE CORRECT ANSWER IS: (A)
rvr,,
rl..S1*r.t!ri:l;l : c$lir
Ii
{'oqi:ri3}:t ,r,:2iii? Ali ri:hts rr:sen,rii
F
N
>F
)F
.-i
3
O
O
(,
o)
N)
irl Vlrtliiuircili - I'irr:ii'r:ll ::::ii I ii,!,.i 5\.iir;ls
-- L':iii:tt.{ irtit:ir :,i,'ilir,)t:s
SOLUTION 014. THE CORRECT ANSWER IS: (D)
Consider the outer face of the wall. The
cl
'rad
heat that arrives to that surface from
o)
I
the left
-J.
O
(f
O
O
O
O
O
O
O
equal the heat leaving the
-
due to conduction
-
T:?
I
m&
d|
Q...,^^
must
rre*.
surface
h.onu:3.5
Btt(#
h "F)
s'con\
torvards the right:
-'' tr="tot
I
I
T :77"F
The heat leaving towards the right is
the sum of the heat leaving due to convection and the heat leaving due to radiation:
Qc-d:Qronrt{rra
--t
{
-l'
(-
C
where
o'
values:
g
c(D--t
o=1.714x10-eBtu/(hrftt"Ro)
q**
a,
*i:
: 3.5 *1=
ai
ft'h"F
h
"*,
Al.r
r-
r -\ + e .4 o
I
r:- ril
is the Stefan-Boltzmann
uz - t ti"
F
+
0. 8 A
{t
.7 l 4.1
0-,
)
constant. Substituting
{en + +oo)o -(tt
--9!ghrft'"R- "
alt
+ +e0
known
)u
j "R
o
oo)
ey:oroBIu
5
o)
A
also,
c)
Zcnd ,
-T:-nrrnt-
I
N)
O
Now, combine equations
(l)
and (2), and solve for 7',
@
T
=-7,
L-
(2)
:
r,:r,+ft,[.*H,tJ
-1,
N)
i\)
(o
T
(t)
ft'h
r -zt2ol
, :6e8'F
+;## (.tt
ffiJ
u'"1*Ll
u. / -:--
ft'h'F
\
!
THE CORRECT ANSWER IS: (D)
i i;i:.. t'ir":i:i r.') :ilil. Ali ri1:lris icstin'cri
u
cr
N)F
{-
J
3
o)
O
O
(-^)
l]!r l!i:ei:l:nreiii
i
ircit:r:rl :lr;il l;lLri,i ili:rl.:i:...
. ilii:iitii.'.: irratn Slrlutltlt-i:
'.r',1 r,,
SOLUTION OI5. THE CORRECT ANSWER IS. (A)
Start with a schematic representation:
f\)
alr
insulation
O)
I
I
.-1,
O
O
O
O
I
O
O
O
O
O
I
insula
r:r
-\i
c-l
CI
a
o-
o)
5
(I)
=
C)
J
i
tl
LN2
iq
i
T-.,
ti
/t
T*,r:77K
The problem statement does not provide information regarding the wall thickness of the container, only
that it is "thin". This rneans that any temperature gradients within the container wall can be neglected;
that is, we wilt assume there is negligible resistance to heat translbr through the container wall.
Furthermore, we will assume the temperature of the container wall is the same as that of the liquid
nitrogen (LN2), so there is negligible resistance to heat transfer from the LN2 to the container.
We take the approach of a themral circuit in which there are two resistances in series: conduction
through the insulation, and convection from the outer face o[the insulation to the outside air:
R"on"
T-', R-na
'o-{\\A--c-\\\A--s T*'^
t\)
c)
.I
*q
@
i\)
'(o
r:r
:i
I
N)
ra
{m,r
T*.,
!
-1,
(-
o'
-:300K
,6J
tron
honv =20 W(m2 K)
\
i
--l.
c
Siiir thc[)li r:or:t
The thermal resistances are
p _ r it_t\
,rl
-u
r\cond-
4r k\r,
and
n^^-
:--1
h4n
ri
^
So, the rate of heat transfer from the outside air to the LN2 is:
t-rr,rv.: ltr lhe
l)l:.il'rr
il;
i,'og;r,r'ui::
t' liil?
:1il rights resetreij
o
5
N{-
.i
)t
-)
SOLUTION 017. THE CORRECT ANSWER IS: (B)
O
O
(,)
In
O)
t\)
o)
t
-t
O
O
O
O
t
O
O
o
O
O
I
1D transient heat conduction problems, the simplicig,of the lumped capacitance method makes it
attractive because there are no spatial temperature vanations, and the temperature of the solid. only
varies with time (This is what r,ms done in problem 013). However, before employing this approach,
one must check the maenitude of the Biot number- Bi-'*'hich is defined as:
hL
U,:-{
rvhere
I
is the characteristic length defined as the ratio of the solid's volume to surface area. This
definition reduces to the half-thickness I for a plane wall of thickness
using the lumped capacitance method is small
-1,
if Bt<0.1
2
L . The error associated with
.
With the given data \4€ can check ttre value of Br:
--L
l,soo#h(ry*)"|*kl
C.
c
Bi:
o
:2.45
Btu
/ a^.
-"
hft"F
{
L
CI
a
ao)
5
o)
-t
()
I
N)
O
-t
Oo
-1,
!Y
N)
(o
-U
which indicates that the lumped capacitance method would not be appropriate.
The mattrematics of transient heat conduction in which there are spatial gradients are quite
complicated. Nevertheless, for some simple geometries and initial conditiom, a graphical
representation
of the solution to the partial differential equations can be practical. These are called
Heisler charts and they're available in the back of any heat transfer textbook. We also have some in the
"Free Resources" section ofwww.SlaythePE.com.
In
these graphs, they-axis
is used for
(dimensionless) temperature and the x-axis represents (dimensionless) time. Each curve on the graphs
corresponds to a specific value of the Biot number. However, because of the logarithmic scales,
it
is
sometimes difficult to use the charts accurately.
A much
better alternative is to use the equation on which the charts are based. The equation for the
time-dependence of the centerline temperature is:
r'l
{g!l
To0):T",+(T,-7.,)C,explr;lu j , *.t"t",.y[
t,
=
(l)
Lr ]
u'here
fo(r) is the mid-plane
irnmersed in the bath,
ivl.' ri'
Sl:xiitc Pl-:.r*i;:
I-.
temperature,
I, is the initial temperature of the solid before
is the temperature of the bath fluid.
-11
a
being
is the thermal diffusivity of the solid"
'i i:r.'',;'..1,,
t'
2i)i1 . Ali fr:!ils i*si:i"rdJ
(t
(t
N*
)t
.J
3
o)
O
O
(r)
N
o)
I
O
O
O
O
O
O
O
o
O
I
-t
i)f: [ice ha;rc:li -- iitrirrai
an<i ]rltriri Sr,slciris
wu,rru.SlaythePE.com.Fromthegraphs-with
iav
tirrill:. eii::r
Bi:2.05-weobtainCr*l.l8andq-i=l.tS
ln this particular problem we know Ze(l)and need to calculate r . Therefore, we re-write equation
(l)
as.
,:\,nf.rffifo]
and plug in the knorvn
(2)
values:
io.er in jt
,:4+,"f,,r"ffi]:,u,
tn xt.t8 L
_I
{
zLv-
o.oz
-l
o'
rvi.,'* I
Sli,-:irotts
andcr and 6iare parameters that depend on the Biot number. These t\^/o parameters are usually
tabulated or graphed. These graphs are avaitable in the "Free Resources" section of
I
c-s
c
- irlactric l'.r;i;r:
THE coRRECT ANSwER IS: (B)
c-{
o
a
o-
0)
5
o)
-t
o
I
N)
O
.-r
@
..-r
N)
i$
(o
T
rvu'rv.
Siavthe.l)l:.lrnr
:-1
Coprlrgl.rl
..,'2ili? .'\i:
r'rgitis lescrlcd
ct
cr
N
**
\\,\.'.\ Si.:\ iirri)l' ct,l.:
--:
3
SOLUTION 018. THE CORRECT ANSWER IS: iB)
O
O
Draw a schematic representation to help visualize the thermal resistances
O)
G)
t\)
o)
I
tl
.-1,
O
O
o
O
O
O
O
O
O
I
r .(
A,
r1l
r.1\
^ ,/:,Tr,.l
TsJ
\l
I
. ti
I
R.ttll
r*
4onu.,nri.t.
R"onu,o*.,d.
q
!
(-t
c
o'
The heat transfer rate from the inside air to the inside face of the wall (by convection), must equal the
conduction heat transfer rate from the inside face to the outside face (this is an energtr' balance on th€
{
inside face ofthe wall)
o
u,
In equation form:
c
r*.r-r".r_Tr.r-Tr.z
o-
Ir \tt-l ir\
o)
5
\h^AJ
o
=)
r-T -T
h.-K'"l'"'2
'-ln
L T*,t-7".t
-
C)
I
/^- Btu \
. l"hftuF156-20 .^ Btu
o'":l
I lft llo-N:r'6hftroF
l(l /-
N)
O
oo
N)
\kAl
|
r
THE CORRECT ANSWER IS.
(B)
i\)
-co
T
1.1r..,
1r. g j31,
iiig lll l. ct:rir
1:
liirp.,'i.i:ii1
r' I(i I?.'r\II ri3lii:. r'*sr';'trrii
u
u
N
J
{{-
iil:
fr,Jcr:irtnici.ii
'- i-l:clil::i li;r.l
i:1r.itij
jitittr:s.- l)fiic1;cr i:r.:;t: Sitluiicr:s
if ir.1., -{!;i1
ilit?li rrtl:r
3
SOLUTION 019. THE CORRECT ANSWER IS: (D)
O
O
(,)
Draw a control volume around the tank and ',wite the mass conservation equation for the Iiquid:
O)
N)
o)
I
That is, the rate at which
-I
O
O
O
O
I
O
O
O
O
O
I
{-1,
-l
(-
accumulates inside the control volume
is equal to the rate at which it enters
I
minus the rate at which
:
h(4'
__f>
gr: I m3/s
mass of
it
leaves. The
jet fuel inside the tank, M, is
equal to the liquid volume
I
V,# times the fluid
in the tank
density p
a(fto'-nl
4o; :ee:pez = T:Q:Qz
o'
-l
where lz is the height of the liquid inside the tank. The above can be simplified and used to solve for the
oo)
5
exiting flow rate:
N)
O
where
dlldt
rsthe rate at which the
1
l-"(tom)'.0.5 mrn
l60mrnl 4
C)
I
Qz:Qr-+ #
tt,
o,=rd]
,
s
o)
-t
J
.
Therefore. the mass balance can be uryitten as:
c
o
@
mass
liquid level rises. Therefore,
:2g.7Jr'
mtn
The problem statement asks for the flow rate in gpm, so the correct answer is 5,476 gpm.
THE CORRECT ANSWER IS: (D)
-l
@
!Y
N)
(o
-U
'.,,
t
rr..Sl:tvtlre Pi,...ctwt
:i
(io1;r, r rrii t
r:
2
0I?
i\l i ri-ehts r.rserr''t:ij
C'
(r
N
*
.,1\\\\ \iit\
**
J
SOLUTION O2O. THE CORRECT ANSWER IS. (B)
O
O
(,
Drarv a control volume around the humidifier and label all the 'water streams:
o)
t\)
iit(:
i)i
riz, ; liquid water
O)
I
l-
l
O
O
O
O
I
O
O
O
O
O
I
-t
fll
c
I
--l_*
n?.
I
I
I
l**__l
I
rnri
drain
-l
:-l
.J.
(-
c
where iz rand m uare the moisture contents
mass balance is:
o
c-{
o
@
2*:Z*.
in
ou'
therefore, iz r*mr=m
o*tnr,
thus:
o-
m
o)
5
o)
-t
c)
=r
I
of the entering and leaving air streams, respectively. The
(r)
r:mo+mr-fir,
We can obtain&rusing the psychrometric chart. At 95"F, and 20o/orelative humidity, the hurnidity
ratio is approximately cur:43 5 grains of moisture per pound of dry air. Since
I lbm:7,000
grains,
each pound rnass of incoming air brings 48.5/7,000=0.0069 pounds mass ofwater.
;,,:{
" \ o ooor$Sft|;,"'
lbm-dry alr I
N)
O
,;,,={ooootffi)k
- \
, , "*'
-r.
oo
N)
i\)
where !'o,.=50,000CFMfrorn
.U
psychrometric chart.
the problem statement
and uu;,*,I4.2ft3llbm-dryairfrom the
CO
f+r
s0,000+rnln
,r,,:{o.oool lbm-water
lbm-dry air
W.e can perfonn
rr..'.,,.
li.r. ;i,.:!)[
ct',:i
fti
'-
14?
' '.- lbm-dry
:/4
1^, .lbm
mln
air
a similar calculation for m, , but we only have the relative humidity- To fully
3(:
(.,"r1'riglrt rr' !{)iI
:\ll
Ltglti. t'eseI'r'eij
(t
u
N
)t-
{-
.i
3
o)
O
O
(-^)
N)
O)
Ili'. \'lcciriirii.-:Lil
-
'l-hci-r:r:,:i
.i:..lJ
i:iirir: S..:r'irr:rs .. llraciic* l:.ra:l Soii:trtr;s
rv*'rr'.Sl:ti thcPf: coin
determine state 4, we have to recognize that the evaporative cooling pfocess occurring in the humidifier
takes place at constant enthalpy (i.e.,
it is adiabatic). So, in a psychrometric chart, follow a constant
enthalpy Iine from state 3 up to the 80% relative humidity line to locate state 4:
I
O
O
O
O
I
O
O
O
O
-1,
I
-L
-1,
(\:
o
{
c
(D
a
oo)
5
o
=)
?ii
c)
itfl
-9
f\)
o
@
75
65
ss
Y &ui, 6 Tt&rpcRATu ti
From this we find the humidity ratio at 4,
co
o
tr.
9$
9*
..
is approximately 88 grains of moisture per pound of dry
air. Since I lbm:7,000 grains, each pound mass of exiting air carries 88/7,000*0.01257 pounds mass
of water:
N)
i\)
(o
,i,,:{0.01255
!
I
50,000-9
mtn
Ibm-water
*,:\0.012s7
lbrn-dry air
7a
:qa.zlbY
^i
tr-
a
rJ.O;-----:--
mtn
.
lDm-ory arr
Norv insert these values for
\\'ti\\ :,ia\ lltLl'l:
r-r)tTt
&,
and m
o
in the mass balance. equation (1)
t?
i:.,i'.',1,-;"".i :ii
r
i
ar r
rif i,,i,.:i*':o..r
(t
-N
J
**
h,:
o
=)
46.2lbrn lmin + m z-
24j
lbm / min
O
O
Also, from the problem statement, 30% of the water sprayed into the humidifier goes out through the
f\)
o)
I
drain. In other rvords
G)
O
O
O
O
O
O
O
O
O
I
"
m,:9.3 ir
, . Combining this with the mass balance,
tu
-t
I
-2l.9lbm/min+0.3
dz
'
,i,,:42*T&J':31.3!!*
- - min
0.1
Since the problem statement requires gallons per minute, we must convert this mass
flow rate into
volumetric flow rate:
",
lbm
J I.J" ----:l.!
'l-n-aw
_L
:-l
mm
:3.75gpm
624+l##rl
-.-1,
(_
c
THE CORRECT ANSWER IS: (B)
o'
c{
o
a
o-
0)
5
0)
o
I
N)
O
.-J
@
.J
T\)
i'j
(o
T
r.,
r*r'.
S
hi.tl;e{nlr. c.rl:r
:$
t),.lpr
rigiri
r,',
:iil? il)
righis ;i:strie'j
a
C'
5
N
*
J
5
l':,i*ti1.,l1: 1:,::i,1\ i:,*ll: :l_::. _* iii
{-
i
r:r I :\i, )n :;()! i ii |i,!,s
i
\\ \\'.\ Siit\ tilcl,l: coi:l
SOLUTION 021. THE CORRECT
ANSWER IS: (B)
o)
O
O
(^)
The
P-v
diagram for the idear Dieser
cycre is shor.vn berow.
t\)
o)
I
-Ji
O
O
O
O
I
O
O
O
O
O
{=58
bar
2
_I
I
I
I
I
I
I
I
I
I
I
I
I
,
--J'
!
P,=14.7 psia
--l
c_
c
I
-1
I
o
-{
c
o
a
o-
CI
o)
-l
o
J
I
v2
The problem staternent asks
for the compression ratio, r, which
is the ratio of the maximum volume
over the minimum volume.
That is, r:V ,lV
"
consider process l-2' which
is the isentropic compression
of an ideal gas u,ith constant specific
heats.
We can use the ideal-gas
isentropic relation: V
,/V r:(p2l pr)r,o
Now, plug in the numbers (keeping
careful track of the units):
N)
, i ,, ou,.f
i',',
/r:f
__l .lgarll
v2 I r4lp" -/
O
r+.s prif
-.t
oo
--J
N)
i$
(o
T
Vr=Vo V
V.
THE CORRECT ANSWER
rs. (B)
rt'rt'rr:.
S
j:rvliii:l
;1..
cr,;l
:t;
:rS
o
o
N
\r.\\i! \i.',\
{{-
.J
3
o')
O
O
(,
f\)
i.:i)i :.ciri:
SOLUTION 022. THE CORRECT A4NSWER IS: (D)
To determine the percent excess air, rve first need the theoretical
- or stoichiometric -
air used. This is
obtained from the stoichiometric reaction for octane:
O)
crH,,
I
.-r
O
O
O
O
I
O
O
O
O
O
I
li
*
12.5(O,+3.76N,)
*
8CO, + 9H,O + 47N.
On a molar basis, the theoretical air-fuel ratio is 12.5 mol of air per mo[ of fuel, while for the actual
process as given
in the problem statement
percentage of theoretical air being used
it is 16.32 mol of air per mol of fuel, Therefore,
the
is 16.32112.5: l3lo/o. [n other words, the percent excess air
is
3r%
THE CORRECT ANSWER IS: (D)
-1,
:l
--L
(-
c
o
c-{
o
(t,
Gg,
5
o)
-t
o
)
9D
N)
O
-l'
oo
-l'
!Y
N)
(.o
-U
=
ri r:',r' Si:lvii:e Pi:.cQit
-li)
{.,.;prrri*lrf i.) 2{i l 7 ,1li r i::iri: tt:eivcii
u
(t
N)F
'.,,'\i
{-
\\' SiavihePi: eoni
.-J
3
O
O
(r)
O)
SOLUTION 023 THE CORRECT ANSWER IS: (D)
A sketch of the system
as described in the problem statement would look something like this:
N)
o)
3
J
O
O
O
O
I
O
O
O
O
O
I
--t
{
c-
T
c-{
22oC
Pz= 105 kPa
Vz= 0'1 mg/h
(Gas) ffir= fr,
To determine the time for the mass in the tank to be depleted down to 50 kg, we will need to calculate
the mass florv rate leaving the tank,m,. We are given thevolumetric flow rate downstream of the
heaters and regulato rc, V r. We can use
l/, to determine fin",themass flow
c
o'
f
fuz= Pz'V
rate at that location:
(1)
z
At the conditions in location 2, the CO, behaves as an ideal gas, therefore.
o
a
(f-
pz
Pz:p.or4
o)
5
where R.o, is the particular gas constant:
p
o)
c)
J
I
-Runiu*r "*':-M[:@
S-31447kJl(knol'K) :0.
lg9,-4;T
"'--(kg'K)
so the density at state 2 is:
N)
Pz:
O
-L
@
N)
__
So, from equation
lr>
(o
:3.05€
170 kPa
o rseG%-Qz+ztz)r
m
l:
o.r$:o.lOsF
v"
ilr:3.05€
vJ
"'""" h
h
-U
m3
A
mass balance around the control volume indicated
with the dashed lines shows thatmr:m.,. At
rate of 0.305 kglh it takes roughly 1475 hours to consume 450 kg.
THE CORRECT ANSWER IS: (D)
ir..,r
..r'.
5
i:r!ti..cl'
i
:.
cii:c
:ll
:..
il1;;irl:lrt
ir,':
]ii 1t
r\11 rig!:rs teset vrr'!
a
ct
(:r
N
)F
x-
J
3
())
O
O
(,
N)
o)
t
-l
O
O
O
I
O
O
O
O
O
I
-1,
\l
-L
(_
c
o'
-{
L
o
u)
o-
CI
5
o)
C)
I
N)
O
-l
oo
-t
N)
i$
(o
!
rvr.i rr'.Siiir..ii:e iti.i.
ii.'nl
::,
il,:;J;'r'riliri
:a.'
:ii i I ;'ril : ri:ilis
iaji-i! r'cri
o
o
N
)F
:F
ftii
i\,1te
haiiiu:l - i i*i::rl j aarj i:lui'l
S., sir' j:':s
- !)r'ltltri:i
i::.:i;l::-r
.rji;iutiitns
:lt]\1i-re i)lr r:i)rr
\!,11.1.
._J
3
SOLUTION 024. THE CORRECT ANSWER IS: (B)
O
O
Label the relevant locations as follows:
o)
G)
N)
Vaoor to turbine
o)
I
I
i
-.1
O
O
O
O
O
O
O
O
;..-.
I
I
Separator
P= 70 Psia
o
I
Liquid to re-injection well
{
-1,
:-r
a
c
From Production Well:
Saturated liquid water
o'
-l
t-
o
@
4500F
The energy balance around the separator is (rate at which enerry enters equals rate at rvhich
a-
it
leaves):
(t)
rhrltr= furh3+ m4h4
o)
5
and the mass balance is:
mr:mr*mo
o)
o
:t
The problem statement asks for the ratio
I
N)
O
-r,
@
f\)
f\lin2.
(a rl
Ql
With this in mind, re-write equation
in r)- h r+( m ol
n r\.
h
o:
h
(l)
as:
(3)
"
and equation (2) as:
(molnr\:l-(nrl*r)
(4)
By combining equations (3) and (4) we obtain the following equation for m/mr'.
i\)
(o
{h"-h":
(mrlth"):,ffi
T
u,here ht:hsgtolo,o:1180.8
Btu/lbm
and ho=hrrotop,in:272.'l
Btu/lbm . To obtain
lz, we must
recognize that the throfiling process is isenthalpic; that is, hr:h, (a throttling device is simply a
partially closed valve that results in a pressure drop. Pcrform an energy balarrce around suclt
neglecting heat transfer, and kinetic energy changes. Since there is no work, you get kr=
ri
i,,
r..
Sliti
;-i:c
i'l r c.rtrr
,t,l
{.'c;1r-r,rrgiii,i
I lil
i
I
h
3l i
a device,
r).
Lt
g.ltis t esc*i"r.i
u
o
N
)F
J
*
iti: I,1tti-rl::ri,:;i
l-trtrr-:rl! iii:r.i i:il:tr-i Sr
i:fil'i:
i)irlr.i;r:a i1.'ii:lt Soiiltir;r::;
* r',ri' .iiin'il.ltf:l: t'i;i:t
:f
The enthalpy of the incoming geotherrnal water is the enthalpy of saturated liquid lvater at 450"F.
O
O
(,
Therefore,
o)
h,: 43g.rBtr,r/lbm,
t\\
and
(mrlmrj=
o)
t
.-r.
O
O
O
o
iqza.z-ztz.ti
i1180.8-272.7i
THE CORRECT ANSWER IS: (B)
I
O
O
O
O
O
I
_L
{.J,
c-
c
o'
c{
o
CI
o_
o)
5
CI
-t
o
I
N)
O
.-J
oo
-I
N)
irJ
(o
-U
i', ..r..v S i a'rlhcl)l r co:-r:
-r.i
=0. i73 :17.3Yo
5
u
N
)F
{-
.J
3
o)
O
O
(.r)
N)
i'i-l ivircir;irrii::li
*
iirelr;iiii lil'ij lit;iJ -\\
-
si(:'r.j-i!
i]:-::ul;cr i:.xa:n Sojtrlrt;r':s
SOLUTION 02s. THE CORRECT ANSWER IS: (C)
ln the basic vapor compression refrigeration cycle the expansion device discharge'is the same as the
evaporator inlet. Similarly, the compressor inlet is the evaporator outlet:
o)
I
-L
O
O
o
O
I
O
O
O
O
O
I
expansion
valve
compressor
!
tl
'-:'
,
..t
!
-!
(-
Therefore, from the problem statement at the evaporator inlet
o'
T.:S"F andx,:l.0. The enthalpy change across the
c-{
"rbfrigeration effect" and it is related to the load Qo,ur6 follows:
c
o
a
Q"uuP:
oo)
5
o)
-i
c)
I
N)
So, the mass flolv rate of refrigerant
ll*,,
li'I'"u'V'-
{:5"F
andx,:0.3, while at the outlet
evaporator,
(.hr-h,), is known as the
h
')
can be calculated as:
.
Q",uo
nl*u--WQ
Using the providedP-h diagram we obtain hf2lSBtu/lbm andhrx$lJBtullbm. Also, a "ton of
refrigeration" is simply 12,000 Btu/h. Thus,
O
I rz.ooo Btu/h
I
*
;^ -"onl
,"ter,(ols-2rc ier;#-
@
'
.-1,
!Y
15l
lb/h
l\)
(o
T
.i,lir iit,"l*l'ii ;.;;,
{-oi:"ri!'.i:i C
:t")
i
; Ali rights tsse t""'i-r,.i
o
(t
N
J
*)t
i)l: lrJti'irliticlii jhllrr:.:i
:.1::.i
i:ii:;t.i
\i.tc;ii:. -. l)i:iatrr:a
I:s;iir.r
ltiilt;oiis
_)
SOLUTION 026. THE CORRECT ANSWER IS: (A)
O
O
A heat engine can operate between two thermal reservoirs and produce power. The most efficient heat
o)
G)
O)
I
-l
O
O
O
O
O
O
O
O
O
I
engine is a Carnot engine- When a Carnot engine operates between a reservoir at a high temperature
T
,, and another reservoir at a lorv temperature ?", , its thermal efficiency is given by:
?l**:
l-nT,
We can substitute the numbers, using absolute ternperatures
Tn**: | -
I
!:t - ffi:o'071 :7'lYo
THE CORRECT ANSWER IS: (A)
{
-.I
(-l'
c
o'
-{
c
(D
(t)
oo)
5
o)
=
o
I
N)
O
-l
oo
N)
l\)
(o
-u
=
t?rr:.,'r'lgl:". :,.'
11.)
!
?
,,\ii ilg!tis t*sarr.-a.j
CT
6
N)e
*
-i
3
1:'!: lr,!cuf,.rriic"r;
-
!-iicrrrr-r!
:t;li i :ri:J
SJ\
-
-ii.i:ilri
i'r'::rtli,'c 1...":irtr
!iiiiti:ltis
o)
SOLT"ITION 027. THE CORRECT ANSWER IS. (B)
O
O
(-^)
A heat pump takes heat from a low temperature reservoir at a
t\)
o)
I
O
O
O
O
O
O
O
o
O
I
it at a rate 0 ,, b a high temperature
reservoir. It consumes power at a rate f *,nu.oto do this. The
rate
Q Land delivers
.::'l
T
T
r,and
,,
sarne rate the heated space looses heat to the surroundings, so
.
For
a
a
TH
Qn
The heat pump must deliver the heat to the heated space at the
y:55,ffi0 Btu/h
-a-
.
I
Q
:
.
Heated
Space
-L
absolute temperatures of the reservoirs arc
.
Heat Pump
heat pump, the coefficient of
performance is:
!
-L
(-L
c
o
COP*,n,.n:;:!t
/
hcal.
(l
)
pump
o
Therefore, if we knew the value of the COP we could solve the
c-{
problem. The problem statement specifically asks for the "theoretical minimum power" consumed by
o
a
the heat pump. The Carnot heat pump is the ideal model and is the one that consumes the theoretical
o)
minimum pow€r. For a Camot heat pump, not only is equation (1) valid, but there is an additional
o-
expression for COP, applicable only to the Carnot heat
o)
-t
coP crr.r
C)
tcat
J
I
l\)
O
:-
-j
Q.",.
I
I
T,
iJ"
*-'::;
' IE
lL
-;-g
TH
1
I
{
ti
-L
Q'
rL
Plug the numbers in equation (3) to get:
coP cu*,
@
heat
pmp
N)
i\)
(o
pmp
pump:
Now insert this value of COP in equation
T
W/
tt
heat pump -
it'r,..
rr Silr l[rcl]ir.citil
55,000
.'| --25+460
78+460
:10.15
( I ):
Btu/h
10.15
=5,4,'*Fful:",n0
t-:
(i:;:r,r ig:lii i;: :i! i7 .'\i1 righls lci*1\:L'r':
c'
C'
N
**
.J
O)
O
O
G)
N)
o)
I
-L
O
O
O
O
I
O
O
O
O
I
r1'\!ri iiiii". l:lai)i; iitni
SOLUTIOIY 028. THE CORRECT ANSWER IS: {B}
With a relatively large process diagram such as the
one for this problem,
it is very likely that a lot of
unnecessary information is being presented. Don't
it
let this overwhelm you. Read the question carefully
for hints telling you what piece of equipment should
\l -x)m
be your focus. Here they want to know how much
Examining the energy balance on the FWH is
a good idea. Letmbe the
mass flow
483'F
-1,
through the high-pressure turbine,
-t
(_
fraction that gets sent
o'
(t -x)rA gets sent to the low pressure turbine and
eventirally (t-r)ftalso enters the FWH through
{
and
to the FWH.
.x
the inlet labeled'2" inthe figure.
o-
Therefore, the energr balance for the FWH is:
o)
5
I
N)
O
-.J
;
solving for x
sat.
liq.
Rate at which
energy exits
xmhr+(t -
x) mno
t y"';'ih3 t tt'ltr, - it' th4
, lvri\,',q -*,f1t
x(h,- h rJ:(t - x)(no- nr)
.
oo
*=.--J\!)-\h,- hr)+lhr-hr)
N)
i\)
(o
,ti,t1
-.r)riz
I
which
enters
x mhr+(l-x)izh, :
of,
1
t4
the
Rate at
energy
o)
o
ir
Therefore,
c
o
a
11oo "F
xtn
of the turbine discharge goes into the FWH.
therefore
c
580 psia
Since pump 002 handles a saturated liquid, tve can
-U
h
r:
h
writeh4-hz:c *ou*lTo-Trl:l?6Btu/lbm
. Also,
:463 gtu /lbm , so our expression for x becomes:
,{580 psia )
*:
176
Btui lbm
(1)
So our task now is to determinehr, the enthalpy at the high-pressure turbine discharge. Since the
problem statement indicates it is an ideal cycle, rve may assume the turbines operate isentropically.
'.i.,.,
'\ 5la|iiieill:.a{li:l
{-io1;r'rrgi:t r.'
lti il
,'1ii rr*hls tese :vr.:ri
(t
ct
N
**
.-J
3
O
O
(t
N
o)
itl'. tr.lcchiir:ii:i:i
-
l-licr:r:lii ;:i:ei l--iL:iri S'..ter:rs -' !,racirtc l:.r:inl Soir.iiiil;-rs
rv*,r,r.
S
I
a\ lhr,:ill : c,,r:;r
We norv use a Mollier diagram to quickly represent the process in the high pressure turbine:
I IUU
o)
I
-!
O
O
o
O
O
O
O
O
O
I
I
{-t
--a
C-
c
o
:l
C
o
@
o_
o)
5
o)
-t
C)
-t
I
N)
O
-l
oo
-1,
N)
i$
(o
-U
Locate the turbine inlet as the point where pressure is 2200 psia and ternperature is 1100"F. Now draw
a vertical line (isentropic process) downto the knorvn discharge pressure of 580 psia. The enthatpy is
read from the vertical axis ashtxl,350Btu/lbm. Now, plug this value into equation (1) to get
x:0.
rr
I7
il,lr'.5 la1,liri"lLl'.. cit:tr
t9
Cop-vrigi:t t''
2illl
.'1ll riglrts f*sgr"rj*d
5
u
N
)t
>(-
J
3
o)
O
O
o)
N)
o)
I
_I
O
O
o
O
I
O
O
O
O
O
I
l'i: ivllcirllii:iil * iirti::::ri;tii,i ii:ir:c Srii,:::ts . i)i',-lt.lrt:c !:rii*i Stiiiitirt:,:
ir
l *, :!it., iirr:i)lr
ior.li
SOLUTION 029. THE CORRECT ANSWER IS: (B)
For the process described, the initial temperature isTou.r:65"F, and the lorvest allolrable .final
temperature is lrr.,:20.,.'r,,int-F5"F. The dew point temperature
is most easily obtained rvith
a
psychrometric chart. Some charts list the humidity ratio in grains of moisture per pound of dry air, but
others list it inpounds of moisture per pound of dry
air. A "grain" is l/7000h of a pound.
With fn.r=65"F and rrr,:iJgrains/lbm:0.00786lbm/lbm we can use
a
psychrometric chart to
obtain the dew point temperature:
--t
{
--J
c-
c
o'
-{
c
(D
({,
o.
o)
5
o)
c)
5
1S
I
.{c
45
5B
qq
s0
$5 ?0 75 8* 85 *0 *5 l0S 1CI5 11S
ORY BI-}I.8 TE}TPTRATURI
N)
O
-I
-'F
Therefore ?n*,u-*,n,:51"F so the lowest allowable air discharge temperature is Too.r:51
*5:56"F . It
oo
follows from this that the maximum allornable dry-bulb temperature drop for the air is:
t\)
i\)
(o
-IJ
A
?"**:(65-56)"F:9"F
THE CORRECT ANSWER IS: (B)
i..'rl
rt 5ial'.i:efrl:.
c*:r':
4{}
(-i;1r'iili:i ,'
ltr,
l7 Ail n;!:t: rrsei!..r!i
o
ct
N
{-
l'l-- irJrchit;;rclri *. i'i.:i:r-;.:tirl .i:rii i:ii.:rI
>(-
J
3
o)
O
O
(J)
N)
o)
I
-I
o
O
O
o
\\:.lanis , il,l.i,lrer i:r.iitri Si:lritiiuts
SOLUTTON 030. THE CORRECT ANSWER rS (C)
At any point of the semi-spherical shell, the stress is a function
of the container pressure,p. It can be
shown that at any point in a spherical shelt of thickness
l, the state of stress is represented by two
normal stresses, each of magnitude 6: prl(2t) where
r
lo
is the internal radius.
.--.".-....:
The wall thickness can then be obtained as:
' -.o |
<-'r '. I
I
''tl
I
O
O
c)
O
O
I
-J,
t:,Pr,
| o
l----->
I
i
12o)
io
Yllu'z t :0.02m.1]!qq nrm
r' 1'2.lsoMpa
l:zo *,n
-".',,,,|-l * l:
!
.-L
(-
The equation we have used is valid as long as the
containe/s walls are "thin", which is generally taken
to mean that the wall thickness is less than ten percent
of the tank diame ter, d, , or tld,<01 . In this
o'
case, since
c
d,:4m
we have
tld.:0.005
so the "thin-wall', approximation is appropriate.
-{
l--
o
@
o_
o)
5
o)
-t
o
f
_o)
N)
c)
-l
oo
SOLUTTON 031. THE CORRECT ANSWER rS (A)
In a bolted joinl the bolt and the clamped parts act as resistances
(to deformation) in series. The
unthreaded part ofthe bolt (stiffness ft, the
threaded part ofthe bolt (stiffness &, and the plates form
),
)
a composite system. The stiffness courd be
obtained with:
=-l-:a*I kzI *
-I
!Y
N)
(o
tl
ft**pori
"
k
I
kropyrtur" frbouo*
prur.
since the clamped plates are "rigid", they can be considered
to have infinitely large stiffiiess, so we
obtain fr"u*po,i," :(ft ,. k r) | k k
{ r+
r)
..'.,ri-,r.. S
I
:r,,,tl:titi--. errilr
i.,",rr.lti;l,i .,' .tir
i
.r;; i,5i,,r,i...,o,r
o
r.F
N
*
*
iil:. ir.lcclii:niciti
.-J
=)
o
O
O
(.^)
N)
O)
I
-r.
O
O
O
O
O
O
O
O
O
I
I
soLUTroN
* i-!:cit:::ri.ii:i: i.itiJ;i:ri ilrrits -.itriii:itci, !'.ritili Sclltirr:t;032. THE CORRECT ANSWER tS {B)
The "codes and standard"' questions in the test are more like a reading comprehension question and
don't require familiarity rvith the code involved in the question. Careful reading of the excerpt shor.vs
that only when the input is less than 5 million Btu/h is it acceptable to have only one shutoffvalve, as
long as this valve has an overtravel(proof of closure) interlock function, per CF-180(bXl).
SOLUTION 033. THE CORRECT ANSWER IS: (C)
A sketch ofthe system can
be useful:
Pump
.-J
{
-l
(-
c
o'
c-{
o
a
o_
o)
5
o)
-t
C)
f
I
l\)
O
.I
@
t\)
i\)
We need to calculate the highest value of
-U
so that cavitation does not occur. Cavitation at the pump
inlet is avoided when the net positive suction head available (NPSHA) is greater than the net positive
suction head required (NPSHR). The highest value possible of L,Z (lefs denote that as
LZ^n ) makes
NPSHA: NPSHR.
From the provided pump curve at 1100 GPM, we obtain NPSHR:7.7 ft. Therefore, NPSHA:7.7ft
-l
CO
AZ
when A
Z:L,Z*.
NPSHA is the actual total fluid enerry at the pump inlet and, for the configuration
inthesketch'wecancalculateitusingtheconditionsatpoint1as.follows
NPSHA:he,t-
LZ^ni-h,o,
(l)
In the abbve expression, hr,ris the pressure head (in feet) corresponding to the absolute pressure in the
tank, and h,,ooit the vapor pressure corresponding to the water temperature- Equation (1) neglects the
effects of friction and minor losses, as per the problem statement.
qr,,.1
-.',
\
i..'. iirti,[..u.t'rrt
47
o
o
N
J
*
*
-J
O)
O
O
(J)
l]l:
i\r,lcrh.:rric;il
-
-i-licir..rll
;rir,.l ijiL:ir.i
:1...i!a:,lr\,.!rr.iiclttt l-.rani Siiiiiiions
hr'to:#:
I
m
eoo9:zz\
Insert the appropriate unit conversion factors:
<tbf
,4....-
4
I
'
P,o,
-
jt
-, tbrn.ft/rt
I tz in 'lf )L-L--a-----l
".,
", .J -----;'l----;._l
in'| ft
/t-o
r6
l l-
oo.albT.:z.z
--'-
ftt
!
-iI
c-
Now, ua solve for
c)
LZ*in
p'g
equation
:
I
{
17.8ft
st
ooop:zz$
(1): A Z^n:lto]-ft,.qp-NpsHA
A.Z^*:59.4
5
CI
l
, _Pr_
tl n l-
c-{
o
CI
lbf
Similarly, the pressure in the tank rvritten as a head term is:
c
o
o
a
i:,'r'ill: ,:cllr
_ _lbf
/').t
--J.
-.J.
,\11,,
Fronr a steam table, the saturation pressure corresponding to 180"F is approximately 7.5 psia and
the
density is 60.6 lb/ff. Therefore, the vapor pressure head is:
N)
O)
O
O
O
O
I
O
O
O
O
O
\\tlr"
ft-
17.8ft-7.7 ft:33.9ft
THE CORRECT ANSWER rS: (C)
5
I
N)
O
--J
@
_I
N)
l\)
(o
-U
r1'1rr.1.
i
l;i.,,i116Pt:.. i:0:rr
i'ottr,rreirt
t llill r\ii
ri3iris icscrr..rii
(t
g
N
**
J
)=
o)
O
O
SOLUTION 040. THE CORRECT ANSWER IS: {A)
It is a well-knorvn result from Thermodynamics that to
G)
minimize compression work during two-stage
o)
compression
NJ
-L
O
O
O
O
I
O
O
O
O
O
I
Pr
of an ideal gas, the pressure ratio across
each stage of tl're compressor must be the same. That is:
P'-Pz
Pt
where
"1"
P'
denotes the inlet
denotes the discharge
to the first
stage, "2"
of the second stage, and '.x" the
intermediate pressure (1" stage discharge and 2nd stage inlet). When this condition is satisfied, the
--\
\l
!
(-
c
o'
cornpression work at each stage becomes identical. The details of the derivation of this result can be
found in any thermo textbook.
Solving for the intermediate stage pressure.
'l
P,:'il
c-{
tffi
p,:ri
(D
a
o-
15
Pr Pzl
psiax{58+ l5i psia:33 psia:18 psig
(})
5
o)
()
:'
I
l\)
O
-l'
@
---!'
N)
i\)
(o
-U
ir..ru..\Iu',
il:ri'i
i.rl..r
L 13;.:..ir1i:i
I
I
I
I
lu
lo
IN
I
lfl)F
r.. ,,r'ri. .5
[)i: \'1tc!r;irrical -'l-lr.:it::tr] anr: ltir-iiil S\,',iietn\ '. itiaetiic ir.:-ilt-r Srili:tit>;is
!ii,"'ihe
lll:.erint
lG.
tt
t1
lo)
lo
lo
|
lN)
(^)
I
-r.
O
lo)
lr
MECHANTCAL ENGINEERING
THERI4AL AND FLUID SYSTEMS
AF"TERNOON SESSION ANSWER KE}'
O
O
?
(3
Question
Question
O
O
O
O
I
241
C
021
c
202
B
222
A
203
A
223
c
2U
D
224
D
205
c
225
A
206
B
226
c
207
B
227
B
o'
208
B
228
D
-{
c
209
D
229
A
214
D
230
B
211
D
231
C
212
A
232
B
213
D
033
B
CI
214
C
234
C
C)
215
B
235
A
-9
216
D
236
D
I
217
c
237
D
218
D
238
D
@
219
B
239
c
-\
N)
220
B
240
C
l
:
(-
c
o
a
oo)
5
-t
f\)
i\)
CO
T
rr
ri
n
Sl-i, iircl'[:..ccrrr
.ti'
C'opr'r1gil1
r-:)
lii;l
,'\ii ri';.iiis rescrvc.i
o
(t
N
*:F
*
I'jj. \.Jtrir:tu;i;ii
-'
i-irei:::::! :r;i.: iiir.:iii S\'.rii:it''-: ilr'itetitc
i::.iit: Si)i:,iillt:::
:f
SOLUTION 201. THE CORRECT ANSWER IS: (C)
O
o(^)
The conversion efficiency of, power plants in the United States is often expressed in terms of "heat
rate", which is the amount of heat supplied (in Btu) to generate 1 kWh of electricity- The smaller the
o)
I
heat rate" the greater the efficiency. Considering that 1 kwh
O
O
O
O
O
O
o
O
O
I
associated ,*,ith the conversion of shaft power to electric power, the relation befu'een the heat rate and
O)
--r
the thermal efficiency can be expressed as
I
>
-r'
\t
ltltz.t4
4rh*ul=
Btul
l-l kwh
I
Heat Rate(in Btu/kWh)
For the power plant of this problenr, a heat rate of 9,000 Btu/kWh is equivalent to
37 -9
W ni:f
gu.
t*ti*,
**/
.r**
turui*.
*:
250 MW
:
250 X I 03 kW
c
Encompassing the power plant in a global control volume allows us to define some
o
g*
is the heat adtled-to the platrt from a high ternperature source,
c-{
(e.g. the combustion of a hydrocarbon fuel) md
o_
that rnust be rejected to the atmosphere. This
o
(tl
o)
5
9*o
W
"rct
llrn*rl:-;Va
o
0*" b what
w
"rct
=
Power Plant
:65e.4xlo3kw
o :?50+gk\tr
0.379
O
J,
Now we can use an enerry balance to calculate Qou, :
oo
e
-r
!'9
N)
>
-
we need to
w,*
V- :- l?ft*ul
-o)
.(o
T
is the waste heat
calculate. We can calculate heat input from the thermal efficiency.
o)
=
l\)
percent thermal
efficiency. Noq for the power plant in the problem, lr,e have the net po\ryer produced:
-l'
C.
:3,412.14 Btu and disregarding the losses
Q
*,t:
r,:W,o*Q
40e.4x r o'
*
=
0"*
A"*:409.4X103kW
tw .l i{L?laaul *
1,3e7 xl
06
Btu/ h
THE CORRECT ANSWER IS: (C)
rr rr
*
Siln.iitcl)i . r.rtti
1-)
t
:.;ft'' ti:.li-;i.
.(
li.i
i
?'.il i f tg.ilis
1r:;rjj
1
ali
(t
(t
N{-
J
*
Pi:. \'iccl--anic:ii
i,,t r., .-i.rr'i:ci)i: c"i,l
-. i!rci;rrll .ir:ii Iiiiiri S\'-\icllis - llr-l:lltii: l::iil;11 !iliuiJ'.',n:;
3
SOLUTION 202. THE CORRECT ANSWER IS (B)
o
O
In the analysis of combustion of liquid and solid fuels, each combustible element is
O)
(J)
N)
O)
I
.-r,
O
O
O
O
I
O
O
O
O
O
separately. We start with the stoichiometric reactions
examined
for each of the combustible components. For
carborL we have the stoichiometric reaction:
C+O,--'gg'
l2kg*32kg --44 kg
c
where we see that
Therefore,
i2 kg of carbon
02
need 32 kg
I kg of C requires (32112)kgof
co'
of oxygen to produce 44 kg of carbon dioxide.
oxygen. Since
I
kg of our fuel contains 0.82 kg of carbon,
I
the oxygen requirement for the carbon in our fuel is 0.82x(32112)kg=2 186 kg
--J.
-
The next combustible component in our fuel is hydrogen, for which the stoichiometric reaction is:
H2 + 0.5O" -_+ cot
(zte) (totg) (tste)
-1,
(-
C
I kg of our fuel contains
o'
So
-{
c(D
hydrogen, the oxygen requirement for the hydrogen in our fuel is 0.02
I
kg of hydrogen requires (1612\:9 kg of oxygen. Since
x( 1612)kg:Q.16 kg
a
The rest of the components of the fuel are non-combustible.
o)
Adding the oxygen requirements we can obtain the stoichiometric oxygen requirement as
o.
5
o)
-a
:
air requirement is therefore (2.3461A.23): 10.2 kg air per kg of fuel.
J
o=f=uit,
AF--,-,
= 10.2,kg
'- ' so,ch
kg Of fuel
I
Sinue wc are usirig 509'o excess air,
-L
oo
A F*,*,
:
I .5
x
AFod.n
:
I
5.3
Jg1!Lof fuel
',.s
.-J-
N)
N)
2- 186+0.16
2.346 kg of oxygen per kilogram of air. S ince air is 23Yo oxygen (on a mass basis), the stoichiometric
C)
f\)
O
0-02 kg of
THE CORRECT ANSWER IS: (B)
(o
T
i$i
rr
.
S
!,it',ltti'i:.
c.t;,-
53
(.o5,rrrilrt,rl)
:til
7.
:\ii rlgi:ts ttscir'cii
o
q
N)t
*
J
3
O)
O
O
(.^)
irl;
i\.1,:ti-};i:t:it;;:
I
O
O
O
O
I
O
O
O
o
O
I
fu *-rr, h,
C-
irr.:\', it.'ii;:.- i)ti,;l:ii l:x;:t:r.'il!l:lrr:t:s
rS
(A)
tm
h
^,*
r:
it *-rro hr+
m *r* h o
Q-1'2,Aa
-JU L.
sat. vap
or
I n.- n.\
/i?*_,r.:Ii?.,^l
\ n^-n,
I
-l'
hr-h,,
'!
Ar,co2 mixture
4
can be obtained from the table provided.
The enthalpy change for the gas mixture can be calculated with the constant-specific heat assumption
gases:
hr- ho: c o..o(I, -
f.) . the mass flor.v rate for the gas mixture is the sum of the mass
flow rates of the components, ril-,*:(0.5 + i.0)kg/s: i.5 kgi s .
From the table provided,ft,:12.65+0.3x(nz.fi-12.65):78.5kJ/kg
o'
can insert known values into equation ( I ):
c-l
o
a
oo)
o)
o
,
andhr:232.I7kJlkg. We
ru*-,rr:1.59{61
{2}
We see from equation (2), i.la only need the specific heat of the gas mixture to arrive at a solution. For
a mixture of ideal gases, the mixture specific heat is the weighted average of the components' specific
heats using the mass fraction as weighing factors:
cp,.i*:rcorcp.co.+ xn,
J
I
I r ll
LI \ Insl/
ttftl.[fi]{
'o'*:['- ]lo
N)
O
oo
-25'C
Heat Exchanger
c
5
-30 "c
x=3Aa/o
(l)
I
where the enthalpy change for the refrigeranl
of ideal
-r,
I
Perform an energy analysis on the heat exchanger:
-L
-t
{
^1,:.:
soLUTroN 203. THE CORRECT ANSWER
tv
O)
-. iit,::l:t:rl
o
cp, n
vr I
r.r
sz6-,| :0 i4#K
Thus, from equation (2):
t\)
i$
(o
.
.
frR_r3.r: l.)-ke
-5
T
t,r
0.74;+;(sOK)
- ' tgt< r---,
rrrr':t-Sir:t 1lr*f l:
i:rrr-::
I
, l:O.rO!g
(ztz.tt-t8
s)g
.
.Ke |
vt
THE CORRECT ANSWER IS (A)
I
I
'
u
u
N
*{-
.i
3
o)
O
O
(J)
i\ \\ '!'. S ii\ li:r-'l)l r .r.rn:
SOLT]TION 204. THE CORRECT ANSWER IS (D)
Perform an energv balance on the tower:
N)
m^,hr* m,hr* mohr: mur,h,+
{
not too different than that of the cooled *ater, u'e set
hu:hr-
@
Therefore:
th ui,(h,
- h t):
m
:
h:-
tm
r- tu) h,
rlr. , perform a mass balance for
,
u,,
t0
,:
where rrr, and (o)are the humidity ratios (pounds
-{
c
entering and leaving the tower. We can solve for rilo :
o
a
0)
mo: m^,(to.rtherefore,
^,lh-
CI
o
J
t\)
,*
0)
ur,
in,
to
---t
h
r):
it"'
fit, hr+ m,,,(a,
-
at r) h,
- fu,h,
=6th](L-hs\
From the steam table:
From the psychrometric chart:
N)
@,:
' g1FaEl-:
T
a,:
-
(o
(2)
We have enough information to look up or calculate all terms in the expression above.
--J,
!Y
air respectively
r)+ mr- m,
O
@
the
fizr-mo:7.r-fuuo(rr-a.,,) . Now, insertthis in equation (l):
fu
I
in
of water per pound of air) for
o'
5
waterout
water:
Iit 3+ ti't 4+ m
o-
n,t"ru-rp
(l)
.
flow rate of make-up \t'ater,
-r,
L
c
it,h,
Now, assuming the make-up water is at a temperature
Since we dont know the
-J,
i.i€)
whichl-fRate at whichl
["n"rgy enters l-[energy exits I
[Rate at
o)
I
-I'
O
O
O
O
I
O
O
O
O
O
I
B6"F
ll^
I i-.9rr,out lggyo p.1.
tbm-arr
t9
g.oosz
l9l-*ttt
:9 ozzs l?I-*1"'
t#llDm-arr
tDm-alr
The temperature of the cooled water
hr=hlgtsr:62.8
h.=50.3i19.
lbm-arr
hr:hr(T
,
lbm-alr
I,
-_ ^ Btu
h,:26,,"'u
'
lbm-alr
is found from the definition
lb;
r)
of the tower effrciency:
Tr-Tt
l? ro*r",
r,, rr rv.
Sl:ti tlrcltl:..cQ:n
T_T,
rv.h.. I
Ci:ir'-.'r;glit
il :iiI? ;\ii
Lrgili>
trsr:r\rii
6
cr
N
*
J
)F
3
O
O
(^)
l:il \.itiiralrii:ri -
i'!'rcrtn.:i ;lrir.: iriulii St'sitl:-r:. Itr;.ti-il,:c I:ui:r: )ir.il'.riirlr;l;
T
o)
I
.-r,
O
O
O
O
O
O
O
O
O
I
I
Is:
Therefore,
h
r: lt 1{fr)=
/t:l,6A0gpm
38.3 Btu
, We
?'*,0 ., )
/lbm
95
-
0.68 x( 95
-
59.
2):
70.7"F
.
multiply this by the density to get the mass flow rate:
nx,: v, p,oo,.,
:
r,600
4L
x ez
|
!!g
xl
-:
rful
I |,^
"
I
iT*l
|
:
7es,6i0
T
Now insert all numerical values into equation (2):
7 g5,670,6001bm lvafer
--l
*(
ez,s
- rs.:
-"aE
(_
c
c-{
f r-
Now we only need the mass florv rate of rvater rz, . We are provided with the volumetric flow- rate,
-t
\-f
o'
tT ro*o, (
,..r iill iit,:J)i rr':rt
where ?"*0.,:59-2oF from the psychrometric chart. So,
lv
O)
r:T r-
".
|(so
:-zo) -(o ozts-o oo87)x 3s
THE CORRECT ANSWER IS (D)
o
a
o_
o)
5
o
=)
5
-t
o
I
N)
c)
-J,
@
..J
N)
i\)
(o
-U
t1'11
1y !;j.1i
il:1itl:.i:iil:
)
lh*
lL.-.''-+--+=
't'5*ot'' :826,450+
Btu
3j-=
lbm-atr
-:OrO-+JV-=
h
u
cr
N,t
*
J
3
O)
O
O
(,^)
[]'l: lJte
iruiici:i . i'5ci'r:':r:1
solr-rrroN 20s.
l,:rrc!
ijluri.i Sls:i'lt"ts'-
i'iittlii:c i.lltln Sitiuillr:i:;
r,.
u rr Sl.tr litci)ir t tr.r
THE CORRECT ANSWER lS (C)
Label the free surlace in the basin as "1" and the spray nozzle location as"Z"
t\)
o)
l
Cooling
-1,
O
O
O
O
I
O
O
O
O
O
I
_\
{
Pump
-t
C-
c
o'
Now apply the extended Bernoulli equation:
P' ' V" *52-Fft1';.'-n*h^;*'
P' -V" -- -''
-1
pg' 2g ' ' 'hpu*p: ** rf
c-{
o
a
o-
and solve for the head added by ttre purnp:
o)
5
o)
c)
.P
r
*r:#
*
Q
r-
z r)
+
hfridrion
*
(1)
ftminor
where we have neglected the velocity of the basin water surface (in steady state operation
also used the fact that
pr: pz:A
I/,:$
V 2 , as
psig. We know the velocity at the spray nozzles
well as the
N)
hydrostatic head, (z r- zr) ; ro our immediate task is to find the friction loss and the minor losses.
-J
The friction loss is obtained from the Darcy-Weisbach equation:
O
@
.
hki.,i*=J"L
1\)
i\)
(o
-U
), and
V-
DZg
and the minor losses for valves and fittings is obtained rvith the method of loss coefftcients:
minor'fiuings,rtdvcr,"r.=2
So, vi,e need the velocity head in the pipe,
itt:l!
K
V2
Zg
'
Vzl2g to calculate both the friction and the minor
a rlfr-r,ri in;'i
t'
:I.j
!
;
A
I
losses.
i rlghis ! t$f i|.li'i
(t
o
N
*)t
ri\\\\ :,ix!i.llf
._i
-I
=
o)
O
O
(-^)
iql'
L''_\AI : Q' :
Q'
2g 2g 2gA'2gixort4i'
t\)
o)
I
I
rn' I lr *inll'
i-^^
nut
/uu--xli48oss"tl.l
_L
O
O
O
O
O
O
O
O
O
I
*
|1
V1
2g
:0.939 ft
Therefore,
&rii.ti.
:f
*#=oot "ffix0'939ft
: 44'6ft
and
\-l
-t
&mino.: fiuings- *urr*,
c-
c
lJ
s
I
-.J
l)l : r't!:)
","
:I
K
V-
t.
= 20 x0. 939 ft
:
I 8. 8
ft
o'
The head loss across the condenser can be considered another component of the total minor losses- This
-l
C
CI
$
loss is given as a pressure drop, so expressed in head terms:
o-
,^
/rmirnr: condensr-LPp g-
o)
5
o)
C)
ro4xl:zrrb't
in' I lbf s' l*l14$l
| | lft' :23.t
LJ. r l.ft
|
We can now use equation (1) to calculatethe head added by the pump:
V,
h
_o)
N)
O
o o:1fr
+
(=
r-
=,)+
y'rr,,.,* + &.,n*
. :-fti4Oft/sir* ,:r- :, \)+ &".r.n + &*,*,
n*o
-..I
ft
@
(
2x32.2+
-t
N)
i\)
(o
!
h
r, r=6.242 + 3a + 44.6 + I 8.8 +23. 1 -- 123 ft
To calculate the required power, first dse the water horse power (WHP) equation:
*"0:/,**o[ft ]9[gpt ](sc) 3956
123x700x l 0 :zr..ho
--'' --r
3956
'.
The WHP is the power that the pump adds to the vater. The power that the motor must deliver to the
punrp is the brake horse potuer (BHP), also known as the shaft horse power. The BHP is slightly larger
i:i
i ;.r;:.. ii:.:it1 i'' l:lll,\il
:;;l:ir rtscirccj
o
u
N
.i
**
O
=)
O
O
(^)
i)l: l,,ittirii:ric:li --l-!rcrilrli
arir..i
!:i*irj 5r's'ictiir * i't':liriitr i^'ranl Srliuirclr:;
BHP is obtained from the WHP and the pump efficiency as follbws.
BHP:ryffi:H:ztznp
N)
O
O
O
O
O
O
O
O
O
I
*'r., !l alii;t l)!.:.ior'l
ineversibilities- The
than the WHp because some energy musr be used to overcome friction and other
o)
I
--l
i.,
THE CORRECT ANSWER IS (C)
t
-r-J.
C-
c
o
-l
C
o
a
o_
o)
5
o)
-t
o
f
I
N)
O
-r,
@
--l
!Y
l\)
(o
T
r., r.r rr'.
Sllr,ihciti:
ctrtrr
1q
l-op" r'igir:
.,1)
2il
i? .\ii r rsiits ttscr'-,'ciJ
ct
ct
N)F
)F
.-J
3
O)
head when the florv rate is 2,000 gpm. So, (2000 gpm,76
ft) is the left most data point in the red curve.
O
O
(,
paraltel rvith an identical pump, they
o)
I
gpm, 70 ft) is the second data point ftom the left in the red curve. The operating point is found as the
Similarly, the single pump adds 70 ft of head when the florv rate is about 2,750 gpm. When paired in
will add 70 ft of head when the flow rate is 5,500 gpm. So, (5,500
--t
O
O
O
O
O
O
O
O
O
I
intersection ofthe red and sreen curves.
I
--\
{
soLUTroN 2r0. THE CORRECT ANSWER
(_
A
--t
c
o'
c-{
(D
rnass balance shows
interest. Since
by:
1s1:
I
lbrn/h
p AV
rrl
we can solve for tho diameter:
i
f
4fr' :,i
D:^ipntl
o)
I
N)
-
o"o
o.t]l_*&rl:o
Ilft's+ollgxnx5.sL
o8o2ft:
a e623in
The inner diameter of schedule 40,y4" nominal diameter pipe is A.824 in, this would be too small and
the velocity limit of 5.5 fps would be exceeded. The inner diameter of schedule 40,
!Y
diameter pipe
is
-U
exceeded. The
I-in nominal pipe should
1.049
THE ANSWER rS (D)
it i' u' Sialiircl'l-. c,.-tn
l"
nominal
in, this would large enough and the velocity limit of 5.5 fps would not be
N
(o
2000lbmlh
3OO0lbsrh
m:oLD?v
,4
o)
-t
c)
---r
Oo
1000lbm/h
10,OOO
o_
o)
O
m:4,00}lbmlhfor the branch of
flow velocity and and mass flow rate are related
At
5
rS (D)
be selected.
o
(t
N*
*
.-J
3
O
O
O)
ljir. Nlr'rilinic:ii
soLUTroN
21
1.
o)
I
T
o
a
the compressor. To determine this, first determine the discharge
temperafure if the compression process rvere isentropic:
/ _ \&-'
-
Tr,:T,l+l
\r'l
r,,:(30 +2r ) r x{
#)+i :
-h"-h'
conp
'-t
hr- h
'I
t
which for ideal gases simplifies to:
T r"-T
Il*^p:Tz
,
and solve for
Z,
:
Tr:T ,+(Tr,-T r)l q o
"o
:303 K+(414.7 -3a3)K/0.85:434-4K: l6l
_o)
f\)
e*r:fix c o(f r-T ,)
N)
-U
"c
Now, use the enerry balance, equation (1):
.-J
l\)
(o
K
Now use the definition of compressor isentropic effrciency:
5
O
-\
@
414.7
1l
o_
o)
o)
c)
ill:: ct'rnt
(1)
r)
o
c-{
ii-.r
The only thing missing is ?"r, the discharge temperature from
_{
c
:*
THE CORRECT ANSWER IS (D)
e.*:tu(.hr- h r): n c o(T r-
--l
{.-L
(-
i
fuhr:Qo^+ith,
N)
o
S
From an energy balance on the.intercooler:
C^)
O
O
O
I
O
O
O
O
O
I
.,vri r,..
'- i'ilcir':rll alrti lrirriii S\'-it;r-,s * itt:,tc1icr: ij.ti;l:;'r -{illuitr:t,s
:0.3
Bx 1 3-*1 t6t-30)K:39.4 kw
s
KgK
THE CORRECT ANSWER IS (D)
soLUTroN 212. THE CORRECT ANSWER
t
ll'r.i,. S I it.,
tllr
[t! :. rori'r
IS (A)
65
Cl1ri;-r,r
ir:ht
ri
i{,1
!7.
.,1!i
iigiii:
fr1-scfltii
g
o
N
*rf
J
a
o)
O
O
(-^)
f\)
o)
I
Itl', \lecl:arrical
-
l-!.:,.:nrr:,ri :-ii;ri
i: 'rllir iitci'l:.rrirl
First, consider the stoichiometric reaction for ethane:
c, H6+4
(o 2+3.76N'
)
-. ZCO'+ 3H2 O +(3.76) a N,
The coefficient a can be determined from an O balance:
2a:4+3 =
-J.
O
O
O
O
l
O
O
O
O
O
I
.r rr
i:irrid Sr,sttitts-- i:'l:ittrii: i:i-iitrr S,.iliiiirins
a.:3.5
Now for the reaction with excess air, n'e will assume combustion is complete (because there is excess
air), hence, the combustion products contain CO:, HzO, Nz like the stoichiometric case, but now the
products include some excess O:. We multiply the stoichiometric coefficient for air
for the
20olo excess
by
1.2 to account
air:
C2H6+ 1.2x3 .5(Oz+3.76N2)-2COz+3
H"O+ bOr+ 1.2x3.5x3.76
N2
and perform an O balance to determine the coefficient 6:
-L
{
1.2x3.5x2:4+3+2b
-J.
(-
c
Therefore, the mole fraction of water
O
a
S
\z
-d
+
1ll
\1
?
1Y
so the vapor pressure
b=0.7
.
in the products is:
N",o
-/u.o:17-:-:0'14
rcr,r z-:-0.7
o
c-{
vapr
=
3
*1.2x3.5x3.76
of water vapor ls:
pH,o=IH'o P*o6:0'
l4xl4''l psia:2'06
psia
The dew point for the products is the ternperature at which the gases have to be cooled in order to
produce condensation of the rvater vapor. This occurs at the saturation temperature coffesponding to
the partial pressure of water vapor in the mixture. F-rom the steam tables, the saturation ternperature at
this pressure
is l27F
THE CORRECT ANSWER rS (A)
@
N)
le
'(o
-1J
SOLTITTON
',r.,',..r, 5
213. THE
llit'ilit:()l:.ct.t;t
l
CORRECT ANSWER IS (D)
i,,-.r1.r:.ri5l:t
:: liiI ; liii tt;rts;ts*r';tr"l.
(t
(t
N)t-
.i
{-
3
o)
O
O
(,
N)
O)
I
-L
O
O
O
O
I
O
O
O
O
O
I
fil-.
\itrhil;Tluii * i'irr:r-rr:i
::r':ii l:ii.riJ
'.vir,.,t'.
Slslttr:'. !'iiir!i{r,: i::.r;tt St:li:iiclis
S
li:r' ii.:citl:'- cr.tiii
ln this problem, we know the relative composition of the products, but lve don't have the amounts of
fuel or air used during the combustion process. However" we can obtain these amounts from mass
balances.
We can write the combustion reaction by considering 100 kmol of dry products, for
convenience (for example, 100 kmol of dry products contain 10.02 kmol of COr according to the given
composition of products).
xC* H,*+a
{Or+l.leNz)-
CO+5.62Or+ 83.48
10.02 COz+0.88
Nr+, H"O
where the underlined terms represent the components of the products on a dry basis and they are in the
proportions provided by the problem statement. The reaction eqution indicates that, x kmol of fuel
burns
with ax4.76 kmol of dry air to produce
100
kmol of dry products and 6 kmol of water.
The unknown coefficients x, d,and 6 are determined from mass balances:
--.l'
!
-1,
(-
c
o'
c-l
o
a
o_
o)
3.76a:83.48
C:
8x:10.02+0.88
C:
I$x:2b -
1.36 C*
mass
-1,
@
b:12.24
O
"
+ 83.48
Nr+
12.24
HzA
nxui'
AF:
I
O
x:1.36
The air-fuel ratio is the mass of air used per nit mass of fuel bumed:
but we don't have masses in the reaction equation,
N)
-
H,*+ 22.2(O 2+ !76 N" ) -- 1 0.02 COr+ 0.88 CO + 5 .62
o
J
a:22.2
The reaction equation then is:
5
o)
=
Nz:
;ll*","r.
However, these are related by the molar
(or molecular wergh$ as follows: m^,:Nur,M^r, ofld m*r:N ,*rM r*, where N is number of
moles and M is molar mass' The molar mass of air is M u,,:29 kg/kmol ' and the molar mass of octane
is Mru.,:(8kmol)(12kglkmot)+(tt kmol)(1 kg/kmol):1I+kg/kmot . The mass of air is then:
N
m
-l
N)
N)
(o
similarly, the mass of fuel is:
-U
trx ru",:
^,:
Therefore AF
: rt
o,,
I
m,*
:
(
u,,
N
M
^,,:
(22.2 x 4. 76 kmol ) (29 kg/ kmol ) = 3064. 5 kg
*, Mnd
: ( I .36 kmol
3064.5 / I 55.04 ) =
19 .7
)
(1I4
kg/ kmol
)
:
I
55.04 kg
6 kgair I kg fuel
THE CORRECT ANSWER rS (D)
soLUTroN 214. THE CORRECT ANSWER
u.r:. rr'. 5
i
l'[ lhr' [:t [ :. ro*r
IS (C)
!-)
I
i'r-!p\it t!:!;*
!, 1l.it]
o
o
N
)F
J
*
fiij
fu!t'tha*i,,:iii
. iiur:l,.rl i.rni i"luiii
i',..,i'r'..S!sl
-\i'slenis; "- l)i:,:i.itai: i:..;.i:rr 5rri,;i.iot':';
ti:*i)| . co:ir
:J
o)
O
O
1,500 lbm/h
sat. steam
200 psia
C'
N)
o)
I
-r.
O
O
O
O
O
O
O
O
I
o
I
.-t
.-r,
sat. steam
20
20 psia
Heat Exchanger
Throftle
Valve
An enerry balance on the heat exchanger yields:
Q:m(hr-hr)
We have fit
,
(1)
andft, could be obtained from the steam table. We are only missing &, . For this, an
enerry balance on the throttle valve
yieldshr:h, (i.e-, the throttle
process occurs at constant
enihalpy).
(-
c
o'
From the steam tables,hr:11r:hg.gzutpsia:l,199Btu/lband hr:ltrgzoyo*:1,l56Btu/lb. Insert these
c-{
values in equation
(t
g: 1,500?x(
CI
o_
(I
):
t, 199
-
t,
I
s6)#:64,soo
CI
5
+
THE CORRECT ANSWERTS (C)
o)
o
p
l\)
O
-!
Oo
t\)
-L
in
(o
T
SOLLTTION 215. THE CORRECT ANSWER IS (B)
n',,.,,r Sl:tr
iltcl'[
.ctrrti
i:f
C,rr:',:r:;.i l{ii-:' \ll I j,j;:i\ rcsr'i\i.j
(t
(t
N)F
*
.J
3
Pl: il'lr;i:lniciii
-
l'!ic'r'rrrel iir':il I:ii:ir.i Svslcrr:s
o)
-"i)raifiei: i,rriri litriiitl<:t:r
D =15 in
r, Slrlt iirci'i c,rlt:
500F
4O%r.h.
25ft,ts
N)
o)
I
--a
O
O
O
O
I
O
O
O
O
O
I
tr
Q=4kW
147 psia
O
O
(.()
"r
Heater
An energy balance on the heater yields:
Q:*z(hr-h,)
Which we could use to calculate &r. Once $€ have h.we can use the fact that the heating occurs at
constant moisture content, o)y:(o1to locate state2 on a psychrometric chart. Then rve would
readT,
---l
{
-t
C-
c
from the chart.
The enthalpy at the discharge is obtained from the enerry balance:
o'
-l
c
o
a
where h-15.3Btu/lbm is obtained from the psychrometric chart,
oO:+
CD
5
o)
(1)
hr:hr+9
'm
rcwxlr,+r,Hl:
r3,648Btu/h
,
and
-l
C)
J
I
N)
O
:
,,
12.9
therefore, inserting these values in equation
rs
I
bl
s
(l):
;-m-"
@
Btu
-1,
N)
l'\ {- ::--r
h,= --'"
lbm
i\)
(o
T
z
+ :t :'=*=:?::,1:
ft'/lbm
Since rrl r :0.003 lbmo*,/lbmdrv
read
air
13,648+
,
and (o2:@t
Btu
:16.9;:
tDm
, we can locate point 2 in the psychromeffic chart
and
Trxi$.J
THE CORRECT ANSWER IS (B)
soLUTroN 216. THE CORRECT ANSWER
*
rrrl'. 5i
e'"
ihcili:.
ccit
rS (D)
,','..,r,,rrt
L'rtjl
r\ii rili:i.'. rCiCr\ri'.l
cr
N
{-
',
:t
r\\'. Siii\.i!-i.i)J'
a0rir
.-J
a
o
sat. vapor
l212"F
O
(,
N)
o)
I
-l
O
O
O
O
I
O
O
O
O
O
I
6B"F
500F
rel. hum. =
P=14.7
1240 CFM
70o/o
osia
Assuming that
-l^
\.1
I
,
all of the stearn mixes with the air strearn, then
&.*,"r,it* &o"u.: &*.t
c-
o
tn"nto ,n"r*
o
$
o)
\z
<
o)
c)
600/o
a r\iater mass balance
.oul
of,
c{
o-
rel. hum. =
volume in dashed lines shows that:
-l
c
l-r
Heating
coils
?
mr,"*:ftui,
@*tr.t
&s*. :&ui. i{) *rrt.r - 0
which can
k
itt.l
i
written in terms of tlre volumekic flow rate of air:
ir
nlo* : aff1 (0o4.,- {d1n1o
I
Now, use the psychrometric chart to obtain the properties:
I
uui,E l3
ftrllbm
p0.0053 lbm water/(lbm dry air )
N)
a) inr".
-J
ru*o*o0.0087 lbrn water/( lbm dry air )
o
@
.--r
IY
N)
ftrr"*:
(o
'*'#l#l
r.
ft,
t'ib* d*
T
10
oo87-o
ui,
THE CORRECT ANSWER IS (D)
SOLUTION 217. THE CORRECT ANSWER rS (C)
\\\\\\ )i:1, {|ti'j
r',-':r
7i:
Ie.sry
oos3ig+31+:
n
lbm drv alr
on the control
(t
o
N
{-
J
)F
3
O
O
(.*)
O)
N)
o)
I
l'l: llccir:rnic:ii -
-i
irerr:iai :InC
irirrii
S\.\ti:;us
-
rr.ri.ri
!tt'eciite l;ratn Sr;ii.iliril,s
..t
i
aviire
l) f:.
conr
When mixing trvo streams of moist air, the
resultant properties
averages (using the
factors)
will be weighted
flow rates as weighing
MIXING BOX
of the properties of the streams
--L
O
O
O
O
I
O
O
O
O
being mixed. For instance, the temperature
of the mixed stream is given by:
{ nr, \ / it. \
T-,:l
lT.+l-lT.
- m'\
\mimzl- ' \mimzl- '
o
I
t
rel. hum. = 90%
3OO CFM
I
Horvever, this only applies to those properties that exhibit a linear scale on the psychrometric chart: dry
..1
{.-t
(-:
o
and
rlat bulb temperatures, enthalpy, volume, and humidity ratio. Relative humidity is a
exception to this rule. We can use the chart to locate the trvo streams, and obtain the humidity ratios.
Then the mixture temperature and humidity ratio can be calculated as the weighted averages- Once
of the mixed air are determined, we use the chart again to read the
-{
c
those trvo properties
o_
o)
From a psychrometric chart, we obtain:
o
@
resulting
relative humidity.
5
o)
notable
u r:27 .Sgrains waterl(lbm dry air)
and
co
,:140
grains rvater/(lbm dry air)
.
So, now we can calculate the resulting temperature and resulting humidity ratio as r.veighted averages-
o
Although strictly speaking the weighing factors should be calculated with mass flows, little error is
=
incurred
-t
_o,
t\)
O
oo
--r
N)
l\)
(o
T
if
instead they are calculated
with volurnetric flow. This is because over the temperature
ranges associated with grpical air conditioning processes, the variations in density are
Spically small-
Hence:
goo
,oo
l.r.o*/
r",
rt ''--[6FIr,1+cFM,
*i .tt, \r.' - i\ t2001"'
=/ crv, i,/' ''[cnr','r,+cnrr4
"'
' tzoofl*o"r:ur
'
I
/'
Similarly,
1as;-€ragl-:55
5--gc'n!--|rgi
'
' '" tum dry air --'-6lbm dry air
-.,,=l gl27
I tzool"''- lbm dry air \ tzoo /
srains
With this f *,, and o*i* , we located the "mix" state in the psychometric chart, and find:
soLuTroN 218. THE
it rr'ti Slar tltcl-'l: ci,:n
Q
^r*o\ZYo
.
CORRECT ANSWER rS (D)
7\
i ,'r:i r i,'hf " ", l7 ,\1' ri.lirt: :fs(t'\
r'!l
ct
6
N)F
*
.J
O)
O
O
(,
t\l
o)
I
.-r,
O
O
O
O
t
O
o
O
O
O
I
Iti. \4ecirirt;i.l:i iir.'lirr.:i -;::d i'iuri.
The mass flow rate
ir
ii:
Sr iLe
il:.:st'gg I r:i:rr
I
!i
\,,
\r
S i ir!,r
iii,.' !) i:.,. crr
:--
i
is related to the average velocity in the pipe as follows:
yn: p
where
l:,l t','t'
is the cross sectional area,
ar,rd
ll/
(1)
V is the fluid's average velocit-v in the pipe. We don't have the
velocity, but we can obtain it from the definition of the Reynolds number:
*":
(t
D
/'
-' I, - pD
lt =
l:'.
Re
Inserting this last expression in equation (l ), we get:
I :- it:4DRert
4
l\pDl
nt:o{LD'IlRt,"
-14Plugging in the given data:
I t-l*
a:4xr,"-l
-llzinl
4
-!
\l
--1,
r,500.000x
6 s27x1s-'l$:
-13
ft'l
zztYftrll''99,0'l:t
lh
lbfs'll
22.B40tbmth
I
C.
q
o'
{
c
o
SOLUTION 219. THE CORRECT ANSWER
(B}
For this steady state device, lhe enerry balance is:
8O kPa
{n
fnate at
oo)
\<
uitrichl-[nate at whichl
lenerry enters
I
100c
nr/s '
o.o*fof{
2O0
[enerry exits j
:
ArR
*(r,.;L.trnfij
o)
-=
c)
TS
The exit velociy of the diffuser can be considered negligibly small
I
compared to the inlet velocity, per the problem statement; thus, the kinetic energy at the exit can be
O
neglected. So, the enerry balance is:
l\)
tr2
-I
@
ih-hti:+ ='
J
coiTz-r,i:+
N)
in
(o
which \\,e can use to solve for the inlet-to-"ir,"rnp"rature change:
/\r2
T
lzoo-[l- +ooooL
'"""'s2
t-""s1
,r-rr:E -I2-lOoC:\'j:20K
2xrooo# 2ooo#
=
Tz:30oC
THE CORRECT ANSWER IS (B)
SOLUTION 220. THE CORRECT ANSWER IS (B)
ir.rlrr,.
S
!avilre
Pl,:.
r:on:
7:
(-irriri
r!i'rt
t.'.
llil1 - Ail tiglt:s ! c;itlr \ tri
(t
ct
N
r(-
*
.-i
3
O)
P!..-.
L!cclieiiri::ri
-
-i'iie
["n..gy
enters
]-[enetgy
I v'\ |
nln,+
"*its
r'.'\
250
]
t'
(l)
STEAM
:
10lbm/s
200 psia
:
:
0.2ftz.<-->
900 fVs
--)
h-l37L4Btu/lbm
and vr:2.6883ft3/lbm ,
*:ry
:134.4 !
:- Vr: rolb*s xz.6s83S
0.2ft2
tDm
S
Now we can returx to the energy balance, equation (l), and use it to calculate the enthalpy at the exit:
v.'-v- ,'
hz:hr-T
c-{
h,:13714+
:
lbm
9oo2- :34.47
5
+"
s-
,Btu
'rb* I
I
;41
&,:1355.6*L
'
lbm
Now enter the superheated steam table rvith the value of hz:1355.6Btullbm for p,=200psia
-t
c)
obtain Tz:662"F
I
THE CORRECT ANSWER IS (B)
J
psia
7000F
Therefore:
o'
o)
Sll., ti:ci'li c,;t:i
rs
From the steam tables, with 250 psia and 700oF, $€ find
c
o_
o)
i.',,
We rvill need the velocity at the inlet before moving on.
--r
o
a
ri'ri
St;i,".:iictr:.
,l:nllh,+?J
-I
{
(-
- llrrictit'e i:]rli*:
[Rut" ut whichl-[Rut. at whichl
t\)
o)
I
.--I-
Svstc:i'!t
For this steady state device, the energy balance
O
O
q)
O
O
O
O
I
O
O
O
O
O
I
lr:it] ;;ri.l lrluiri
and
.
N)
O
-J.
@
.J.
!Y
l\)
CO
T
soLUTroN 22r. THE CORRECT ANSWER IS (C)
i,.-,1:11'. g11i,1[1gpf;.1:11111
i'r,frr;"' t i: zitiz ait
,i3i.,,n ies!"j'r'.!
a
The heat transfer rate to the process, is obtained from an energy balance on the process heater
O
O
(^)
Rate at
f
l"n"rgy
t\.)
o)
I
-r,
O
O
O
O
8
I
O
{
whichl-[nate at whichl
enters J-[energy e*its .|
h4+
tnsh:: Q wn"rr+
m
4
Q
*nn"= lfi oh r* m,h
r-
fu1 h7
m, h,
*n"u: fu, h i in h 5- 0h 4+ m r) h
In this equation, u'e have everything butnr, (notice thatmu:rn., rvhich is given in the
Q
statement). To obtain rn, ,
(t )
?
5
u€ try an energy balance on the turbine,
problem
will show up in that
because dz,
equatton:
futht:W
-I
:.-l
brir+ m5h5+
it6h6
tu;h:: W,*+ turhr+(nr-
(:
mr)hu
-L
.
L
o
-l
C
o
$
13
(ztts t-zor
0)
->
$
gI I,ooo+
s
kg
o)l;
: 10.6kg
We can now go back to equation (l ) and plug in known values:
10.6x 2739.3-(1.5+10.6)e0-091:
-'--l
-'-t-'
9*,*",,:(1.5x3411.4+
S
=P(Eess
6
f
-o)
H
-
s9(34r t.4-2073.q
o-
5
mr(hr- hu)-W *^
h:- ho
,,.i_-----]-]-
QP'n'u:26'4o9kw
THE CORRECT ANSwER IS (C)
@
.-L
N)
i$
(o
E
SOLUTTON 222. THE CORRECT ANSWER IS (A)
rr
r.',
ri .5 iitr.,liitili:. itli:r
?
i
{,-gg1'1
ri3i:i
:.i
lil l? '\il rtgiiis t'escir
t'..i
or
U
N)F
*
J
-)
o)
j''l': fulcchanir:a!
-'i-lierlr;.ri *nri i;lr:ii! livsteurs
*
rr
[]rLctir,e i::xiirir Srili.;titns
ri'.r \ l;l lh*ll'.
crrnr
The heat addition in the combustion chamber of the gas cycle is.
O
Q,^:m*(hr-hr)
O
q)
N)
o)
I
The temperatures at those locations is given, thus, we may use the constant specific heat assumption to
-r,
O
calculate the enthalpy change of the air as an ideal gas:
O
O
O
O
O
o
O
O
I
where the subscripts
"1" and'2-
denote the inlet and outlet of the combustion chamber, respectively.
Q,n:fu*,
I
So, the only thing we need is the mass
(l)
offr-r,)
flow rate of air. We can obtain it by performing an enerry
balance on the HRSG:
lRate at
whichl-lRate
at whichl
["n".gy enters ]-[energy
mn
-L
.-l
(-
fu*
c
frgu",in-&g,o*
o-
&o*^,.r*
ft**u..o*-fro*.,in
.
: to*'ug**
7*J
fturo..or,-fto*
.''steam
m*:74l,8OOf x
o
a
.|
h*;n*fusr ho^,6:fu*ft*ro*+&**
o
c-{
exits
(l3sr -82.s)BJu
'tbm
'
:6,203,163
s.2a ryg-x(st7-170)'F
T
o)
5
o)
now, rve insert this in equation
Q,n:6,203'L o:
c)
Sxo
(l)
z+
ffi
x ( t,s6o
-:10
)
oF
:
I -7
42xt
on
?
:
t,to,
U*o&
I
N)
O
.-J
@
-1,
I\)
i\)
(o
!
rv',r'rt. Sl ilv tire ili.. c'oiti
?5
Ccpr.ri;:lrt
i
l'ir
1l
.,\ii rrghts resrrvcd
u
(t
N
1-
*
*i
J=
(t)
O
O
PIr:
ir,!ccl:ini:li .- 'l irt:it:l-:i lilJ f luiii !r siai:,:; .' lli.:i!;'-:,.: ir'il::1r
soLUTtoN 223. THE CORRECT ANSWER
{iri'1si11l;:s
IS (C)
Regenerator
1 bar
G)
T\)
o)
I
4 bar
209 "C
-L
O
O
O
O
I
O
O
O
O
O
I
Air 0.562 kg/s
1 bar
27 0C
-t
<l
.-l
C-
c
o
c-l
o
6
The power required by the compressor is obtained from an enerry balance on the compressor:
W
o-
J
p
m*(hr- h,)=mwc p(7, .r,)
{t n-.: rn *,(h r- h o}: m *c F{T 3- T o)
Setting 14/**s:IV"rt (according CI the probiem statement) we obtain-
(rr-f ,;:1Tr-T) =
CI
o
p=
and sirnilarly, for the H.P. Turbine:
o)
5
*
To:T;(Tz-T t):145"C
At this point, it pays to visualize the process occurring in the turbine on a Mollier diagram (an &-s
diagram):
t\)
O
.-r
@
..J
f\)
i\)
(o
F n=P ot
T
As shown in the figure, the "ideal" process is an itentropic process to the same pressure as the "real"
process. The isentropic efficiency of the turbine is the ratio
lvri'rv.
S
I
;lr.-ii;ef
'7
i,:. c,.rrr:t
{.t
of the real powerdeveloped by the power
i-;i:lr.:,rI:.lit ,ir
:i]iI Aii ri:hi:
r'r:cr-vcrl
=
d,
o
developed if the process were isentropic.
h.-ho
trt- h+,
t;,urr:T--,
O
(r)
t\)
?
O
and after assuming constant specific heat for the gas,
Tr-Tn
il .*b:;
T1 j-'4.
O
O
O
5
o
o
which we can use to solve for
?no,
:
(r.-r')
Tu,=Tr-Ij;;ll
-
?
=
T
q,:921oc-Ozt --7!sYc :7r7.8oC
I
tr
:
P"=
c-
C
A
l
r
Ptl# l--'
\'ll
ltn.s+2a\#
Pa,:4batlffiffi)'n=,-Ot
bar:205kPa
J
c'
o
THE CORRECT ANSWER IS (C)
v)
o_
o)
5
CI
a
:'
9)
l\)
3
@
.-t
N)
l\)
(o
-U
rvrirr.Siar iircPi:. ;.rrl
i1
"
':
C.,1,,r",g.hi.!.:{li: ,rll ri*htslgsen'etl
U
g
N)t-
',',
)F
.J
(=t)
O
O
(.()
I\J
!
--r
(-
regenerator effectiveness is defined as:
u*":P^
(r)
rvhich upon assuming constant specific
heats- can re-cast as:
- -T'-T
.r*_
'
To_T,
4,:mc
o'
(D
o)
8e
n
t^
1 Yin
[,] .-l - )]:=-=
ffiC,
c-{
o-
-o
5
isTr,
n(T
3
*''hich can be obtained from an enerry
-T r)
rxro'+
4s0,000Txo.z+ffi
:825oR =
Iz:
14l5"F
Therefore, the regenerator efftctiveness is (inserting known values into equation (2)):
5
o)
(2)
balance on the combustor:
c
rt)
-
;ir l::ci)l t.tt-,i
With the states labeled as in the figure, the
In the expression above, the only thing rve are missing
-l
S!
SOLUTION 224. THE CORRECT ANSWER IS (D)
o)
I
-t
O
O
O
O
I
O
O
O
o
O
I
rrri.
,,*:ffi:0.9:9aYo
THE CORRECT ANSWER IS: (D)
I
t\)
O
-1,
@
-L
N)
i$
(o
!
i',.,r.*, 5 ar,lheill_..
I
COu":
l:!
t-tp".,li3irl ,r'2UI7 A|! i'igh:s tc;err*il
u
(t
N
f
t'1, \lfc'r.i:rciri
-
!'h,:r:uaj .lrr,i
|liriil
S1,st*rtrs
- ltaciici'
O
O
e
t\)
?
IO
o
l
:
c=
:
the cylinder volume displaced per cycle.
rS: (A)
It is a fictitious
pressure that,
if it acted on the piston during
(l)
where P is the brake power,Vois the total cytinder displaced volume, 11 is the crankshaft rotational
speed (rad/s), and n o is the number of crank revolutions for each power stroke per cylinder
four-stroke cycles and
Calculations are much quicker
Va
fs1
is in in3, and mep is in lbf/in2 we have:
mep[rbr/in']:
"I?]:^"'nu'ooo
olii lN [rp*
1t
?" ,
n*:)
if the above equation is adapted to prescribed unig for all variables- For
example, if power is in hp, N is in rpm,
The torque,
(
n*: I for two-stroke cycles)
(z)
]
power and rotational speed are related by:
(3)
"tnot:Ilry#lll
^ 3.000x110 /^
P:ff:62'8hP
o)
so \ /e can now solve for V ofromequation (2)
nrl,"']:€ffiQ:122.8ini
N)
O
This volume, however, is the total volme of atl four cylinders. The questions asks for the volume of
each cylinder:
N)
(o
:S
!:.cLri.r
Pn*
men:zl^i
o)
-I
@
-.1
N)
P
cycle. It can be shown that mep is given by:
5
-t
o
]o)
lhc
the entire power stroke, would produce the same amount of net work as that produced during the actual
3
c
E
o-
S I a.,
[n an internal combustion engine, the mean effective pressure (mep) is the work per cycle divided by
?
O
5
O
?
\r,,\\:1!
i:.rr;ili Soiirlri>n:
,(- '-='
e SoLUTIoN 22s. THE coRRECT ANSwER
V
o,,:V ol4:30.7in]
THE CORRECT ANSWERIS: (A)
1:i',1:!..: $ 1 11:1 l1gl t l :.
co;ii
7'j
I lllit'i,:irl
"':{lll
,1iJ
f
igiltr ifbL': \.ii
ct
u
N
)F
:F
-_t
3
(t)
O
O
(^)
tv
o)
I
-J,
O
O
O
O
O
O
O
O
O
I
I
{
(-
c
l'l: \lciir;trr,c.ll
ii.le:r't.r: .l:r.i i::r'i.l )',..,u
i,:
Pi
::i'tlii I t'.::l .-riliritrrt-
SOLUTION 226. THE CORRECT ANSWER lS: (C)
The mean effective pressure (mep) is a term frequently used in conjunction with reciprocating engines.
It
is a fictitious pressure that,
if
it acted on the piston during the entire polver stroke, r.vould produce the
same amount of net work as that produced during the actual cycle. The figure below
Otto thermodynamic cycle in P-r, coordinates, [n undergraduate thermo, it was shorvn that the net work
produced is the area inside the cycle. Knor.ving this area, and having fixed the volumes corresponding
to top dead center (TDC) and bottom dead center (BDC} one selects the rnep such that.
W
Iz***
*r:mep(
- I,'-,n)
From this, we can solve for mep:
meo:,-J1ll'
' (Z*.- l'*.)
or, on a per-unit-mass basis:
llto,
t"P= (n'"--
o
r€P
"rJ
c-l
For the Otto cycle, the compression ratio is actually a ratio
of
o-
volurnes:
r:1'*../v.,n, s the expression for mep can
be
5
rwitten as:
o
{t
o)
shots the ideal
v.m
TDC
meo:;ig-' (v*-r,**l ,'f
o)
-t
o
:t
I
*.p=
N)
)
(t)
r,*ff1/r)
O
So we need to figure out the net work and the maximum specific volume. First, to determine the net
@
worlq this can be obtained from the given heat addition and the definition of the cycle thermal
t\)
efficiency:
i\)
.-9-U
=
'?*=1d
4i"
1l',*r: rl,.gi,
and, with the cold-air-standard the thermal efficiency is:
We can plug in known vatues:
r,n*:[l -
8-o'o]x 800 kJ
l7*:1-r(r-el
/kg:451
.8
therefore
u,*,:[l-rit-r)]qr,.
kJ/ kg
Now, for the maximum specific volume, we can use the ideal gas equation of state at the initial state
5ij
i.6r;;',,rrg'iri
.r
'lr-lf
.i.
,1li rrgi:ts;-csctltii
o
C'
N)F
:f
J
3
o)
I'l-, l.lcciriiiriclii
(state
I
- l irei-rlal
:i;ici l:luiii S',,sir::i':s
_ Rr ,_
v-*,:Tr=
N)
o.zszHFxzqo
Now we can use equation
o
O
o
O
(l)
45l.sEIr-. -^
-.
|
xl-l!&l:o2lkPa
iFg
os32r-fr-11
ke\ 8/
O
O
I
:o.s32g
""--kg
to calculate the mep:
*"p:
I
rc
-
A:--lookpu
o)
I
O
O
O
airt-t:
in the Otto cycle drawn in the figure).
O
O
(^)
.-r
rr'rllr jl::: llrti)l:
- liriiiiiil i \ii!-, S*lrili()rrs
l'41
*'l
|
\t
.-l.
-t
(C
o'
c-{
o
U)
c)CI
5
o
=)
o
:'
I
t\)
O
-l
@
t\)
i\)
CO
-U
ivrvtv SIaVthtI)l':.cirnt
{ir:pl,Ligirt C :0
l:. illl
r
i-gitls rcs*t'r'*ij
o
o
N
,(-
r"
)t
J
=)
O
O
O
(.)
This kind of problem is solved with the momentum equation. If a control volume has only one entrance
and one exit, for steady-state flow, the momentum equation is:
ZF:*l/"^-V
I
O
O
O
O
O
O
O
O
I
I ct::r'r
SOLUTION 227. THE CORRECT ANSWER IS: (Bi
O)
-r,
O
* -*' iirii 1lttl'i
^j
This is a vector equation and in general one u,orks with x- and y-components of the equation.
First, define the control volume. We
will
use the volume indicated by the red dashed lines in the figure.
I
We have used blue arrows to indicate velocities and black arrows to indicate forces.
_L
{
(--L
c
o'
{
c
(D
Forces F, and Frare the resultant forces due to the hydrostatic pressure at each location.
ll
ghrA,
Fr:tp
and Fr:tp
ShzA,
a
a
5
o)
o)
o
p
N)
O
oo
-\
t\)
i\)
r,:|oz +$rz
z
! ro
r' rontl--tffi|:r',roo,o,
|'"'_7-|
Similarly, Fz:l,24$Ibf . So, the sum of forces in the x-direction is:
I
F
,: F ,- F ,- F *n": I 1,652 lbf
Now consider the term in the right hand side of the rnomentum equation:
' /{+-+\:
\A, t,l
p
int,v2-t/
'\=
o
n,(I-I\
\A, 'a,l
So, the momentum equation can now be written as:
CO
-u
11 ro
p
I I\\,.rl
v-l;-;l:
ntl
\n2
tl
;,2 pl
v-:
'
L F,
AJ, \-
'-lL
I f ,l
tt1
.t{l
F,
|
i.*1ir,r'iri.i C
l{lil
:1li ricir;s r€s{:}\i:rl
u
q
N)F
{{-
frL, fulccitariicel
- 'l-irtrrnri liii
_)
O
O
I
G)
N)
it2:402,500*
o)
I
O
O
O
O
I
O
O
O
a
O
I
.-r
{
-L
(-
i.rr:i.t i'irtliiirctts
root-''zon'-'lrr,.roo lbf xl32 2lbm ft/srl
ir2:--L
-/ I00 ft'-z}fl.'
62.4lbmlft'l
I I rDt
I
o)
-1,
- l'rrirlti,r
f ii.ri<i 3r'sit:t-:s
s-D
-
i,,
:634.4L
But the problem statement asks for the flow rate in millions of gallons per day, so
/:63a.4t*lgo,+oor
s I day
+sear
l:4r0x
I I lft' I
1*lz
r0'+g!
oay
SOLUTION 228. THE CORRECT ANSWER lS: (D)
Perform an energy balance on the control volume shown with red dashed lines:
c
o'
c-{
T
o
a
o-
o)
:<
o)
-t
c)
at
r*rictrl-[nate
at whichl
I
extts
IRate
fenergy enters
::r
I
fu*ur", h *to,n *
N)
O
Q
r*t:
nt
r, lT ror,rn:f? *.t",
*u,r(hrot,;n
-
&toto,rt
)
@
.-l
N
t\)
(o
-lJ
where ir*",.,
lenergY
-
fr
fu
.*t
ro *
:
|
-
* 9,*,
&*x"r.in)
10,032
I lh I
I
l7.a8gall"
fu,ot h *t'*
*r.r( h,*r.,o,,t
uo,T't
: 60 g at I minx 62.4 tbm / ft I x l-#ll
*
J
+
h
We can now use the energy balance:
Qnu:
o,o.,: 46,000
$
furotctot (Itot,;n
o.+ r
#-
- Ior,*t)r00 "F
-
t'lorr", co.t".( rou,.r,uu,
30,032
\!L
1
.0
#
- I*,""in )
60
oF
: 84,080
?
THE CORRECT ANSV/ER IS: (D)
$'rtrl' 5ia.,,lhr:l!I :.ctm
ri,i
{qrprlrrlll . l,'i7 .i\ll rir:iri: iusclicJ
u
N
*
{-
.J
)=
o)
O
O
(,^)
Therefbre
I,:0.8333 X884 K :736.4 K
Similarly, the density at the tkoat is:
,,:l*]"r-l#,l"#k
o)
I
.-J,
O
O
O
O
O
O
O
O
O
I
-
For
lv{a:l ue find p I p r,:0.6339 , therefore:
I rr ]
o,:lap ulxp o : 0.633e
^..o1r.
"
i
Now we can go back to equation
(l)
I
Vo
:2 6l'*
1045 kPa
6ffi
KgK
\
i
884 K
j
m
J
/
and introduce the known values:
I
.J,
{J.
(-
c
o
-l
c
o
a
o.
o)
5
Alternatite method (faster): Upon finding out that
Ma:|at
I
N)
O
-l
@
.-a
IY
N)
(o
-U
flow
is
the equation
choked in this case. The mass flow rare through the nozzle can be calculatod directly from
for choked flow:
ln
\t+r
.l-t ^ | L lz{x-r\:7. lkg/s
fu"tuu.a:A,f
'oT Rt"o\t+
t
J
o
-t
C)
:t
ttre exit plane, we recognize iltat
(t
5
N
>F
{-
.J
.r
It!: fulechairitai -'l-irlr.:::ll ili:i.i flliii.i Si,'itni:i - ili:tcircc i::tiitn Sttii:ilotls
ir
rr ji.ir liicirl: c(titt
_)
SOLUTTON 240. THE CORRECT ANSWER IS: (C)
O
O
(-^)
across the space' In other
The sensibte heat gain of the space is "picked up" by the supply air as it flows
between the
the sensible heat gain corresponds to the difference in dry bulb temperature
o)
N)
o)
I
-r.
O
O
O
O
(fI
O
O
O
O
t
words,
incoming (supply) air and the exiting (retum) air:
where nz" is the mass flow rate, and co is the specific heat of air. Equation
.oo)
5
(2\
with coe0.24 Btu/lb/'F , and p *:Q.Q 751il ft3 we obtain:
t,, _
v-
c
CI
@
is an energy balance on
Q":pV crLT.
{_I
(-{
c
(l)
instead of mass flow rate:
the air flor,r,ing through the conditioned space. Using volumetric flow rate Iz
-J,
o'
(l)
g,:m"co(7,-T,l
:
oi
62904mrn
o.o75q.o.z+ffi'u"n
THE CORRECT ANSWER IS: (C)
are expressed in cubic
A.lteruate Solution: Typically, in HVAC applications, volumetric air flow rates
l(60min/hr) allows us to use CFMin equation (2)
feet per minute. Defining a new variable
cFM:i
and still obtain
O
in Btu/hr:
(3)
o)
C)
:t
I
N)
O
@
f\)
i\)
(o
-g
heat of dry air
Strictly speaking, the specific heat of humid air is the weighted average of the specific
ratio in pounds of
and water vapor: co:(0.24+0.45:a)etultqaryair)/"Fwherertr is the humidiry
the typical
water per pound of dry air. Because in many conventional air conditioning applications
(in Btu/hr) is approximated by
value of crr is about 0.01 (lbm-water)(lbmdry air) the sensible heat gain
(considering a typical air density of roughly 0-075 lb/ft3):
o.=(0
vvhen A ?"
075$i
.ttlooffl
to
245H+)^r :1 '(cru)
1
s
(4)
we can
is in "F. Since we are given the temperature difference betrveen return and supply,
use equation (4) to solve for the CFM
of supply air:
cFM:*r:##tO:6172
ri'iyii
S
I
ai''il..el)
l:..
coil
.r)
tlor;r rifl-:i
l-,1
.li,'1? r\ll ughls rcfen'e'"i
(t
cr
N*
*
.J
Pl: \ler:l:l:n'c::i
-
"i
irr;.r.;:':i :ll-ri.i i:iili,"l
S\slc:ls * Illiii.'trr:i i.:i.:;tr; liilliltir;r-r:
u.,!''.\'
i :iviitrlll:.. rrr.::
!
3
o)
O
O
(,
t\)
o)
I
-t
O
O
O
O
I
O
O
o
O
O
I
.-t
{"'
--t
e.lr
c
fi
\
(-
-<-
(\:\
^l$
*t
u*-"/
t\
o'
lf
c-{
'5J\
CI
a
\\
j
a-
dl
o)
5
SI-AY*e
o
o
P.E.
t\
tE"-.'
r
Jr-
)n
L
-l
J
"rl
E-
I
_o,
\\\\ry/.
l\)
O
S
I
ar,th eP E.
c-o
m
F
\€
\
t^
--l
-l
\_-"r'
oo
-L
N)
i$
(o
.lJ
r,,'rru,.
S
i;it.ll;tili:
i.nl li
),'i.
i
{ii
t;gh:s :rsr:l.eil