Heat Transfer, Chp:- Heat Exchanger

1. :-

2. Flow arrangement and Temperature
distribution for counter-flow H.T

3. Logarithmic Mean Temperature
Difference (LMTD) for Counter-Flow
Considering an elementary area dA of the heat exchanger.
The rate of flow of heat through this elementary area is given
by :-
Deriving this we get,
Where, θm=LMTD
θ1 = th1
− tc2
θ2 = th2
− tc1
dQ = U ⋅ dA(th − tc) = U ⋅ dA ⋅ Δt …….(i)
Q =
)UA(θ2 − θ1
)ln(θ2 θ1
Since, Q=U ⋅ A ⋅ θm
∴ θm =
θ2 − θ1
)ln(θ2 θ1

4. Effectiveness :-
The heat exchange dQ through an area dA of the heat
exchanger is given by ,
…….(ii)
From expression (ii),we get,
Substituting the value of dQ from expression (i) and
rearranging the equation, we get,
d(th − tc) = −U ⋅ dA(th − tc)
1
Ch
−
1
Cc
dQ = −m
•
hCph
dth = m
•
cCpc
dtc
= −Chdth = Ccdtc
dth =
−dQ
Ch
and dtc =
dQ
Cc
or d(th − tc) = −dQ
1
Ch
−
1
Cc

5. d(th − tc)
(th − tc)
= − U ⋅ dA
1
Ch
−
1
Cc
ln
(th2−tc1)
(th1−tc2)
= −U ⋅ A
1
Ch
−
1
Cc
∴ ln
(th2−tc1)
(th1−tc2)
=
UA
CC
1 −
Cc
Ch
∴
(th2
− tc1
)
(th1
− tc2
)
= exp
UA
CC
1 −
Cc
Ch
Now, From effectiveness equation,
ε=
Q
⋅
Qmax
=
Actual H.T
Max possible H.T
ε =
Ch(th1
− th2
Cmin(th1
− tc1
=
Cc(tc2
− tc1
Cmin(th1
− tc1
∴ 𝑡ℎ2
= 𝑡ℎ1
−
𝜀𝐶min(𝑡ℎ1
− 𝑡 𝑐1
𝐶ℎ

6. ∴ tc2
=
εCmin(th1
− tc1
Cc
+ tc1
Substituting the values of tc2
and th2
, we get
Since,Cmin Cmax=R and UA/Cmin=NTU
ε =
1 − exp −NTU(1 − R)
1 − Rexp −NTU(1 − R)
ε =
1 − exp (−U A Cmin) 1 − Cmin Cmax
1 −
Cmin
Cmax
exp (−U A Cmin) 1 − Cmin Cmax

7. Here, we find that effectiveness of parallel flow and counter-
flow .
ε =
1−exp −NTU(1−R)
1−Rexp − NTU(1−R)
…..Counter flow
ε =
1−exp −NTU(1+R)
1+R
…….(parallel flow)

8. Example:- A Counter-flow heat exchangeris used to cool
0.55 kg/sec with Cp=2.45kJ/kg˚C of Oil from 115˚C to
40˚C by the use of water. The inlet and outlet temp. of
cooling water are 15˚C and 75˚C respectively. The
Overall heat transfer coefficient is 1450W/m2˚C.
Using NTU method, Calculate the following :
1)The mass flow rate of water
2)The effectiveness of Heat exchanger
3)The Surface area required.
Given Data :- th1
= 115˚C , th2
= 40˚C ,
tc1
= 15˚C , tc2
= 75˚C ,
U= 1450W/m2˚C,
mh= 0.55 kg/sec ,
Cph
=2.45kJ/kg˚C ,
Cpc
(water)=4.187kJ/kg˚C

9. i) Q=mCp Δt
∴ m
•
hCph
(th1
− th2
) = m
•
cCpc
(tc2
− tc1
)
∴ 0.55 × 2.45 115 − 40 = m
•
c×4.187(75-15)
∴ m
•
c= 0.40 kg/sec
ii) Ch = m
•
hCph
Cc = m
•
cCpc
=2.45×0.55 =4.187×0.40
=1.3475 =1.6748
Here, Ch<Cc
∴ Cmin = Ch=1.3475
ε =
Ch(th1−th2
Cmin(th1−tc1
=
)1.3475(115−40
)1.3475(115−15
∴ ε = 0.75

10. iii) Heat capacity ratio, R =
Cmin
Cmax
=
1.3475
1.6748
= 0.81
ε =
1 − exp −NTU(1 − R)
1 − Rexp − NTU(1 − R)
∴ 0.75 =
1 − exp −NTU(1 − 0.81)
1 − 0.81exp − NTU(1 − 0.81)
∴ NTU = 2.4736
∴ 2.4736 =
1450 × A
1.3475
NTU =
UA
Cmin
∴ A = 2.2987 m2