ST E P H E N z A K A N Y C z E
E
‘
A
A
E
NOMOGRAPHS -
~~~
-
-~
~
~~
~
-
FOR U N S T E A D Y STATE HEAT TRANSFER
e u r equations, derived for different situations involving unsteady state heat transfer to agitated batches of liquid, can be readily solved using four nomographs. Two types of equipment are covered:
r
-Heating or cooling of an agitated batch using a jacketed vessel arrangement or a coil in a tank with a nonisothermal heating or cooling medium. -Heating or cooling of an agitated batch using a counterflow external exchanger with a nonisothermal heating or cooling medium. Ecluations for these cases have been developed by D. Q. Kern, in his book “Process Heat Transfer.” We have maintained the equation numbers used in his work, and have used the same nomenclature, except for the addition of subscripts to indicate batch, heating medium, or cooling medium. The equations (given on the next page) depend on these assumptions: -Over-all heat transfer coefficient, u, is constant for the process and constant over the entire surface flow rates are constanl
--Liquid
-Specific
heats of the liquids are constant
-The heating or cooling medium has temperature
Batch heating and batch cooh‘ng problems can
-Agitation
-No
be solved rapidb
-Heat
a
constant inlet
produces a uniform botch liquid temperature
partial phase changes occur losses ore negligible
(Cmrrimud on next pngc)
VOL 55
NO. 1 J A N U A R Y 1 9 6 3
27
Equations-to
solve them, unfold the nomographs
COIL IN T A N K O R JACKETED VESSEL
EXTERNAL HEAT EXCHANGER (COUNTERFLOW)
Nonisothermal heating medium
Nonisothermal heating medium
1n---T i - t i - (WC), T1 - t~ (Mc)b
K1
(
- 1
~
K1
)e Nonisothermal cooling medium
Nonisothermal cooling medium
U
The charts on the following pages permit rapid and accurate determination of heat transfer area, heat transfer time, over-all coefficient of heat transfer, initial or final batch temperature, or batch charge; providing the other variables are known at the desired conditions. A nomograph for the cooling of a liquid batch with a nonisothermal cooling medium has been previously presented by Seco (2) from the equations of Troupe ( 3 ) . N o treatment of the actual mathematical relationships used to develop the nomographs is given here. For this information one is referred to the original thesis ( 4 ) which is summarized in this article.
W W
8
Subscripts
= =
t
=
K1, K2\ KI, K 4 J = M T = TI
= = =
Te t
= =
tl t2
REFERENCES (1) Kern, D. Q., "Process Heat Transfer," Chap. 18, McGra\vHill, New York, 1930. (2) Seco, M. F.; "Nomograph for Easy Batch Cooling Calcillatiom:'$ Chem. Eng., 172-3 (December 1953). (3) Troupe, R. A., "How to Tailor Exchanger Area to Fit Batch Cooling Time," Chem. Eng., 128-31 (September 1952). (4) Zakanycz, S., M.S.Ch.E. thesis, Newark College of Enginerring, Newark, N. J., 1959.
F.
100 to 1000
12 tl
F. F.
...
t2
F.
IWClh B.t.u./lhr.) iWCIa B.t.u./(hr.) lwc), B.t.u./(hr.)
F. F.
B.t.u./lhr.)
F.
F.
( M c ) ~ B.t.u./O F. 1MC)b B.t.u./O F. U
B.t.u./hr.
A 0
Ft.2 Hr.
28
AUTHOR J . J . Salamone is a Professor of Chemical Engi-
neering at the Newark College of Engineering. Stephen Zakanycz, emflloyed by fVationa1 Lead Co. of Ohio while this article was written, is now with the Aeronautical System Division, Wright-Patterson Air Force Base, Ohio. The authors wish to thank Joseph Kojolt and Clyde Kearn.r, of the Ohio State Unioersity, for their counselling in dewlofling the nomographic form f o r the solutions to the equations.
RANGE OF VARIABLES W H I C H CAM BE SOLVED Equation 78.9 Equation 18.7 1 Equation 18.16
L7nit
11
(WC)b
= =
h 1
heat transfer area, sq. ft. specific heat of hot batch or heating medium, B.t.u./(lb.)( O F.) specific heat of cold batch or cooling medium: B.t.u./(lb.)(" F.) constants in batch heating and cooling equations, dimensionless weight of batch liquid, lb. temperature of hot batch at any time or heating medium at any temperature, F. initial temperature, O F. final temperature, F. temperature of cold batch at any time or cooling medium at any temperature, F. initial temperature, ' F. final temperature, F.
Variable
batch chaige cooling medium = heating medium = inlet temperature of heating or cooling medium
h C
NOMENCLATURE A C
ox-er-all coefficient of heat transfer, B.t.u./(hr.)(ft.2) ( " F.1 = weight flow of hot batch or heating medium, Ib. per hr. = weight flow of cold batch or cooling medium, lb. perlhr. = time, hr. =
(ft.*) 1'
F.)
70 t o 990 71 to 999 (5 to 60) x 103 ... ... ... 240 to 2 x 104 ... 10 to 1000 10 t o 1000 0.04 to 30
80 to 999 71 to 990 70 to 990
... /
.
.
... 15 t o 60)
x
... ...
103
240 t o 2 x 104 10 to 1000 10 t o 1000 0.04to 30
INDUSTRIAL A N D ENGINEERING CHEMISTRY
100 to 1000
...
70 t o 990 71 to 999 (2 to 61 X l o 4 ... ... 1 0 3 t o 104 400 t o 2 X lo5 ... 1 t o 100 10 t o 1000 0.4 t o 150
Equation 18.17
80 to 999 71 t o 990 70 to 990
... 103 to 104 ( 2 t o 6) X l o 4
... 400 to 2 X lo4 1 t o 100 10 t o 1000 0.4 to 150
I
Y
I .0 0.0 0.6
- --
--
- - - -_ -.
0.4-
-
0.2
t
0.08
-0.010 --0.008
--0.006 1.2
I
/
I
loo0
/
/
500 400
/
,
10
/5
6
8
l x 105
UA
/ /
4
1.5 1.8 2
104
/
50
20
100
200
100
200
500
loo0
U
10
20
50
A
xxx)
FIGURE 18.11 Nomograph Solution From ( T z t l ) = 25 vertically to line ( T I
An agitated tank containing a charge of 10,000 pounds of liquid A is to be cooled from 270° to 95' F. Available as a cooling medium is 20,000 pounds per hour of liquid B a t a temperature of 70' F. Also available is a n internal submerged coil, 5 square feet in area, which yields a U of 400 B.t.u./(hr.)(ft.2)(o F.) for the flow rate above. Specific heat of liquid A is 0.5, of liquid B also 0.5. Calculate the time it will take to cool liquid
-
A. Analytical Solution
Kz In
=
~
~
4
~
0
~
~
~
= ~ ~1.2214 ~ 2 ~
~
0
270 - 70 = (20,000(0.5))( 1.2214 - 1 95 - 70 10,000(0.5) 1.2214 8 = 5.73 hr.
0
0
~
~
-
t l ) = 200. Now horizontally to R1 reference line, then vertically to Rz curve. Now horizontally to A' scale. Connect this point with (MC),/(wc), = 0.50 and intersection of this line with RS line gives reference point on R3. Align U = 400 with A = 5 on Uand A lincs to give UA = 2000. Proceed vertically from this point to (wc), = 10,000. From this point move horizontally to point on Y scale. Align this point with reference point on Ra scale, extend line, and read answer. 0 ~ 6 ~ 0 = 5.8 hr.
% error
=
5.80
- 5.73 5.73
x
100
=
1.22
.
.
X
A'
Key
,
1
4 B
I
10.4
I I
Y
I
:I=?
F F
-1.0
-:o.s ~
@
0.6
-- 0.2
-0.10
-0.08
-- 0.06.
20.04
I I
-0.04
I I
TI
, 4
-
1
T
105
10
20
50
200
103
\
5y
A
1MM
.
\
\
I
IOW
103
200
XK)
20
50
10
U
FIGURE 18.9 An agitated tank containing 5000 pounds of liquid A is to be heated from 350' to 590' F. Available as a heating medium is 40,000 pounds per hour of a gas B at a temperature of 600' F. Also available is an internal submerged coil, 500 square feet in area, which yields a U of 20 B.t.u./ (hr.)(ft.2)(0F.) for the flow rate above. Specific heat of liquid A is 1.0, for gas B, 0.5. Calculate the time it will take for heating the batch. Analytical Solution
K1 = In
e(2o)(soo)/(40,0Oo)(O.~)
600 - 350 - 40,000(0.5) 600 - 590 5000(1)
e
=
2.044 hr.
=
1.6487
1.6487 1.6487
(
'>
e
Nomograph Solution From 600 - 350 = 250 on T I
- t l scale go horizontally to T I - tz, = 10, then vertically to R1 curve. Now horizontally to A' scale (value 1.40). Connect this point with (Mc),/(WC), = 0.25 on B scale. Intersection of this line with Rz scale gives reference point on Rz scale. Align U = 20 with A = 500 on U and A lines to give UA = lo4. Proceed vertically from this point to (WC), = 20,000. From this point move horizontally to point on Y scale. Align this point with reference point on Rz scale, extend line and read answer.
e
=
2.08
yo error 2.08 - 2.044 x 2.044
100
=
1.76
X
I
'3
T30
0.8
0.6
-2 1.4
-1.0 Ln 0 4
).2
e
-0.8 -0.6
_-0.4 -0.2
1.10 1.08 -0.10
1.06
-0.08
-0.06
-0.04
tl
t
5.10
f
J
Y
0.8
0.06
c 0.04
ooo f2-fl
10.4
B
R3
Y
"1
0.4.-
0.2-
0.04
U A
1x102
1x105
\ I
1
\
,
v
I
I
4 \
I
I
I , , I
I
6 8 1 0
I
20
I
I
40
I
I
I I I I
6080103
u
,
FIGURE 18.17 A batch composed of 2000 pounds of liquid A is to be cooled from 800' F. to 500' F. by a cooling medium of 40,000 pounds per hour of gas B entering at 200' F. and flowing through a counterflow exchanger with 5 square feet of surface area. A pump connected to the tank circulates the liquid through the exchanger at a rate of 2000 pounds per hour. A U of 20 B.t.u./ (hr.) (ft.2)(' F.) may be expected. Specific heats of liquid A and gas B are both 0.5. Calculate the time required to cool the batch.
1'
800 - 200 500 - 200 (1.0997 - 1) (2000) (40,000)(0.5) 0 2000 [1.0997(40,000)(0.5) - 2000(0.5)] 7.386 hr.
1
Connect this point with ( M C ) , - -
[(
wc>b
= 0.95.
Intersection of this line with RB
(WC) E
scale gives reference point on R3 scale. Align U = 20 with A = 5 (or see method above) to give UA = 100. Proceed vertically from this point to
In
=
scale.
L&h r
Analytical Solution
6
Nomograph Solution
From 500 - 200 = 300 on Tz - tl scale go vertically to line (TI - t l ) = 600. Now horizontally to RI line (TI - tl = lOOO), then vertically to Rz curve. Now horizontally to A'
d
-
] = 0.00095. (wc>,-~ 4
I
1
From this
point move horizontally to point on Y scale. Actually this point is right on Y scale and no horizontal movement is necessary for this illustration. Now align this point with reference point on Rg scale, extend line, and read answer. 0 = 7.0 hr. 7.386 - 7.0 % error X'100 = 5.2 7.386
i6 4
2 ~
1.o
0.8
0.6
0.4
.
.
.
.
LL-U
1M0
em
B
i
0'41
5
i
0.06
0.04
1
I
1
I
2
I
I
4
1
1
1
1
1
1
1
I
.-- ,
I, I
E 10 ,'20 -
40
I
I
I
j
l
60 EO 103
U
I I
J
920
I
I
m
I
i 200
I
I
100
l
l
80
1
I
60
I .
/ I
I
40
I
I
I
A
10
20
FIGURE 18.16 A batch composed of 40,000 pounds of liquid A a t a temperature of 100°F. is to be heated to 550°F. by a heating medium of 50,000 pounds per hour of gas B entering a t 600' F. and flowing through a counterflow exchanger with 50 square feet of surface area. A pump connected to the tank circulates the liquid through the exchanger at a rate of 3000 pounds per hour. A U of 20 B.t.u./(hr.)(ft.2)(" F.) may be expected. Specific heats of liquid A and gas B are 1.0 and 0.6, respectively. Calculate the time required for heating the batch.
Nomograph Solution
From 600 - 100 = 500 on T I - tl scale go horizontally to TI - t 2 = 50, then vertically to R1 curve. Now horizontally to A' scale (value 1.0).
L] = 12 on B scale.
1
r I )
7
= 0.0003.
-1 -
1
From this
point move horizontally to point on Y scale (value 0.26). Align this point with reference point on Rz scale, extend line, and read answer.
= 1.3495,
(3000)(50,000) (0.60) (1.3499 - 1) 40,000 [1.3499(50,000)(0.60) - 3000(1)] B = 109.5 hr.
-
Intersection of this line (WC), with l i a scale gives reference point on Rs scale. Align U = 20 with A = 50 on U and A lines to give UA = 1000. Proceed vertically from this
Analytical Solution
K S = e ( 2 0 ) ( 6 0 ) [%&) 600 - 100 In 600 - 500
[Gk
Connect this point with ( M C ) ~
0 = 108 hr.
a
% error
=
109.5 - 108 X 100 = 1.37 109.5
