dharm

\M-therm\Th15-3.pm5

HEAT TRANSFER 823

Integrating between inlet and outlet conditions (i.e., from A = 0 to A = A), we get

dθ

1 θ

2

z = –

11

CC 0

UdA

hcA

AA

+

L

N

M

O

Q

P =

=

z.

or ln (θ 2 /θ 1 ) = – UA 11

CChc

+

L

N

M

O

Q

P ...(15.51)

Now, the total heat transfer rate between the two fluids is given by

Q = Ch (–) (–)tt Ctthh ccc 12 = 21 ...(15.52)

or

(^112)
C
tt
h Q
= hh−
...[15.52 (a)]
(^121)
C
tt
c Q
= cc−
...[15.52 (b)]
Substituting the values of^1
Ch
and^1
Cc
into eqn. (15.51), we get
ln (θ 2 /θ 1 ) = – UA
tt
Q
tt
Q
Lhh c c12 21− + −
N
M
O
Q
P

UA
Q
[(tt tthc hc 22 – ) – ( 11 – )] =
UA
Q
(θ 2 – θ 1 )
Q =
UA()
ln ( / )
θθ
θθ
11
21
−
The above equation may be written as
Q = U A θm ...(15.53)
where θm =
θθ
θθ
θθ
θθ
21
21
12
12
− = −
ln ( / ) ln ( / ) ...(15.54)
θm is called the logarithmic mean temperature difference (LMTD).
15.4.4.2. Logarithmic Mean Temperature Difference for “Counter-flow”
Ref Fig. 15.39, which shows the flow arrangement and temperature distribution in a single-
pass counter-flow heat exchanger.
Let us consider an elementary area dA of the heat exchanger. The rate of flow of heat through
this elementary area is given by
dQ = U. dA (th – tc) = U. dA. ∆t ...(15.55)
In this case also, due to heat transfer dQ through the area dA, the hot fluid is cooled down by
dth whereas the cold fluid is heated by dtc. The energy balance over a differential area dA may be
written as
dQ = – m&h. cph. dth = – m&c. cpc. dtc ...(15.56)
In a counter-flow system, the temperatures of both the fluids decrease in the direction of heat
exchanger length, hence the –ve signs.
∴ dth = –
dQ
mc
dQ
&hphhC
=−
and dtc = –
dQ
mc
dQ
&c pccC
=−
∴ dth – dtc = – dQ
11
CChc
−
L
N
M
O
Q
P