dharm

\M-therm\Th15-3.pm5

HEAT TRANSFER 825

Substituting the values of

1

Ch

and

1

Cc

into eqn. (15.58), we get

ln (θ 2 /θ 1 ) = – U A

tt

Q

tt

Q

Lhh c c12 21− − −

N

M

O

Q

P

= –

UA

Q^ [(tt tthc hc^12 – ) – ( 2 1– )] = –

UA

Q (θ^1 – θ^2 ) =

UA

Q (θ^2 – θ^1 )

or Q =

UA()

ln ( / )

θθ

θθ

21

21

−

...(15.61)

Since Q = U A θm

∴θm =

θθ

θθ

θθ

θθ

21

21

12

12

−

−
ln ( / ) ln ( / )
...(15.61(a))
A special case arises when θ 1 = θ 2 = θ in case of a counter-flow heat exchanger. In such a case,
we have

θm =

θθ

θθ

− =

ln ( / )

0
0
This value is indeterminate. The value of θm for such can be found by applying L′ Hospital’s
rule :

lim

θθln ( / )

θθ

21 θθ

21

→ 21

−

= (/)θθlim ln ( / )

θ θ

θ

211 θθ

1 2

1

21

1

→

−

L

N

M

O

Q

P

Let (θ 2 /θ 1 ) = R. Therefore, the above expression can be written as

lim

()

R ln ( )

R

→ R

−

1

θ 1

Differentiating the numerator and denominator with respect to R and taking limits, we get

lim

()R→ (/ )R

=

(^11)
θ θ
Hence when θ 1 = θ 2 eqn. (15.61) becomes
Q = U A θ
θm (LMTD) for a counter-flow unit is always greater than that for a parallel-flow unit ; hence
counter-flow heat exchanger can transfer more heat than parallel-flow one ; in other words a
counter-flow heat exchanger needs a smaller heating surface for the same rate of heat transfer. For
this reason, the counter-flow arrangement is usually used.
When the temperature variations of the fluids are relatively small, then temperature variation
curves are approximately straight lines and adequately accurate results are obtained by taking the
arithmetic mean temperature difference (AMTD).
AMTD =
tt tt tt tthh cc hc hc12 12 11 22
22 2 2

• − + =()( )−+ −=θθ 12 + ...(15.62)
  However, practical considerations suggest that the logarithmic mean temperature difference
  (θm) should be invariably used when
  θ
  θ
  1
  2

  
  

  1.7.
  Example 15.18. The flow rates of hot and cold water streams running through a parallel-flow
  heat exchanger are 0.2 kg/s and 0.5 kg/s respectively. The inlet temperatures on the hot and cold
  sides are 75°C and 20°C respectively. The exit temperature of hot water is 45°C. If the individual heat
  transfer coefficients on both sides are 650 W/m^2 °C, calculate the area of the heat exchanger.