A hot fluid, entering a double piped heat exchanger, at 300
degrees Fahrenheit is to be cooled with cooling water entering at 70 degrees
Fahrenheit. The cooling water can be
heated up to 190 degrees Fahrenheit.
Calculate the mean temperature difference for the following cases:
(1)
Counter-current flow, (2) co-current (parallel) flow, (3)
1-2 multi-pass (1 shell, 2 tube passes), (4) 2-4 multi pass (two shell, 4 tube
passes), (5) 1-1 cross flow, and (6) 1-2 cross flow heat exchangers.
Solution:
A 10-degree approach in the terminal temperatures is
considered as normal industrial practice in the design of heat exchangers. Lets use low case t for temperature of the
cold fluid, upper case T for temperature of the hot fluid. Subscripts 1 and 2 identify incoming and
outgoing conditions. With these
conventions, lower case t1 is 70 degrees Fahrenheit, and lower case
t2 is 190, upper case T2 is 190 plus 10 equals 200
degrees Fahrenheit. Upper case T1
is 300 degrees Fahrenheit.
Case (1) Counter-current, Hot side temperature difference,
Delta Th equals (upper case T1 minus lower case t2)
or (300 minus 190) equals 110 degree Fahrenheit.
Delta Tc equals (upper case T2 minus
lower case t1) or (200 minus 70) equals 130 degree Fahrenheit.
Log mean temperature difference, LMTD, is (130 minus 110)
over log of (130 over 110) equals 119.7 degree Fahrenheit.
Case (2) Co-current, Hot side temperature difference, Delta
Th equals (upper case T1 minus lower case t1)
or (300 minus 70) equals 230 degree Fahrenheit.
Delta Tc equals (upper case T2 minus
lower case t1) or (200 minus 190 equals) 10 degree Fahrenheit.
Log mean temperature difference, LMTD, is (230 minus 10)
over log of (230 over 10) equals 70.2 degree Fahrenheit.
For case (3) through (6), we need to calculate values of
parameters R, and S, in order to use Figure 10-14 of Perry that covers
different arrangements. R is a ratio of
(upper case T1 minus upper case T2) and (lower case t2
minus lower case t1). R is
found to be 0.83. S is a ratio of (lower
case t2 minus lower case t1) and (upper case T1
minus lower case t1). S is
found to be 0.52.
One can read the value of correction factor, FT
for different cases of interest, and multiply it with the LMTD for
counter-current case, to obtain actual mean temperature difference. We can see that
Case (3) 1-2 Multiple Pass - FT equals 0.86, so
delta T equals 0.86 times 119.7 or 102.9 degrees Fahrenheit.
Case (4) 2-4 Multiple Pass - FT equals 0.98, so
delta T equals 0.98 times 119.7 or 117.3 degrees Fahrenheit.
Case (5) 1-1 Cross-Flow - FT equals 0.9, so
delta T equals 0.9 times 119.7 or 107.7 degrees Fahrenheit.
Case (6) 1-2 Cross-Flow - FT equals 0.98, so
delta T equals 0.98 times 119.7 or 117.3 degrees Fahrenheit.