Page 50 - Design and Operation of Heat Exchangers and their Networks

Basic thermal design theory for heat exchangers 37

The correction factor for the logarithmic mean temperature difference
for various flow arrangements can be calculated with an approximate equa-
tion proposed by Roetzel and Spang (Spang and Roetzel, 1995; Roetzel and
Spang, 2010, 2013, Table 1).
In general, the steady-state fluid temperature distributions in a heat
exchanger can be obtained by solving the “micro” energy equations:
(2.78)
dQ ¼ _m h dh h ¼ _m c dh c
dQ ¼ kt h t c ÞdA (2.79)
ð
ð
t h ¼ f h h h p h Þ, t c ¼ f c h c p c Þ (2.80)
ð
in which h is specific enthalpy and f h and f c are the equation of state for hot
and cold fluids, respectively. Using the “micro” energy equations, we can
solve the steady-state problems with variable overall heat transfer coefficient,
variable heat transfer area along the flow direction, and even evaporation and
condensation analytically or numerically.

Example 2.4 Sizing a counterflow heat exchanger
Consider a counterflow shell-and-tube heat exchanger with one shell pass and
one tube pass. Hot water enters the tube at 100°Cand leaves at 80°C. In the
shell side, cold water is heated from 20°Cto 70°C. The heat duty is expected
to be 350kW. There are totally 53 tubes with the inner diameter of 16mm
and wall thickness of 1mm. The thermal conductivity of the tube wall is
40W/mK. The shell-side heat transfer coefficient can be established as
2
1500W/m K. Calculate the tube length of the heat exchanger.
Solution
We use Eq. (2.68) to size the counterflow heat exchanger:

Q ln t t = t t c 0
0
00
00
c
h
h
ð kAÞ ¼
i ¼ Q
00
0
Δt LM,c t t 00 t t 0
h c h c
ð
½
ð
3 ln 100 70Þ= 80 20ފ
¼ 350 10 ¼ 8087W=K
ð
ð 100 70Þ 80 20Þ
To get the heat exchanger area of tube inside A i , we shall evaluate the
overall heat transfer coefficient based on the area of tube inside. The
properties of water to be used are calculated from the following equations.
Specific isobaric thermal capacity of saturated liquid water (Popiel and
Wojtkowiak, 1998) is as follows:
3 3 1:5
c p,s ¼ 4:2174356 5:6181625 10 t +1:2992528 10 t
4 2
1:1535353 10 t +4:14964
6 2:5
10 t ð kJ=kgKÞ 0°C t 150°CÞ (2.81)
ð
Continued