A. This is a RETROFIT problem. All of the design information needed is given (U and area)
B. We must use the
ε
-NTU method here.
C
h
= m
h
c
ph
= 2.2 kg/s * 4.18 kJ/kg-K = 9.196 kW/K
C
c
= m
c
c
pc
= 1.475 kg/s * 3.35 kJ/kg-K = 4.94 kW/K
The cold fluid (solvent) has the lower heat capacity rate, so it will heat more than the hot fluid (water) cools.
q
max
= c
min
(T
h,in
-T
c,in
)
= 4.94 kW/K * (95 – 28
o
C) = 331 kW
Need
ε
to find q. A
o
= N
P
* N
T
* pi * d
o
* L
NTU = UA/c
min
= U
o
A
o
/C
min
= 475 W/m2-K * 4 * 88 * pi * (0.0072 m) * 1.72 m / (4940 W/K)
NTU = 1.317
Need
ε
from NTU. There is an equation in Table 11.4, but it’s much easier to read off of Figure 11.13 (matches: 2 shells, 4
tube passes)
C
r
= C
min
/C
max
= 4.94/9.196 = 0.537
Reading from Figure 11.13,
ε
is approximately 0.61
(somewhere in that area 0.59 to 0.63 is a reasonable estimate)
Thus, q = 0.61 * q
max
= 0.61 * 331 kW = 201.9 kW
q = dU/dt = m
c
c
pc
(T
co
– T
ci
)
201900 W = 4940 (T
co
– 28 C)
T
co
= 68.9 C
(67 – 70, depending on effectiveness value)
C. If saturated steam at 95
o
C were used instead of liquid water, the heat capacity ratio C
min
/C
max
would drop to zero since
C
max
for steam is infinity.
NTU would not change, since U, A and C
min
do not change compared to part B.
So using Figure 11.13, with C
r
= 0, and NTU = 1.317, the effectiveness would increase to approximately 0.73
This can also be easily calculated from Table 11.3- the bottom row has the equation for all heat exchangers where C
r
= 0
(i.e., when one fluid is changing phases). Here,
ε
= 1 – exp (- NTU) = 1 – exp (- 1.317) = 0.732 (about the same as we got
by reading the Figure).
So, using steam at 95 C would cause q = 0.732 * qmax = 0.732 * 331 kW = 242.3 kW
242300 W = 4940 (T
co
– 28 C)
T
co
= 77.0
o
C.
This is somewhat higher than using liquid water. The main difference here is that the fluid temperature on the shell side
retains a higher temperature throughout the exchanger, giving a higher dTlm driving force.

7. Open heart surgery is normally done in hypothermic conditions, where the patient’s blood is cooled before the
surgery and re-warmed afterward. A proposed concentric tube counterflow heat exchanger is designed to have
a length of 0.50 m and inside tube diameter of 55 mm (thin-walled), and has an overall heat transfer coefficient,
U, of 500 W/m
2
-K. Given that the blood enters the heat exchanger at 18
o
C and 0.050 kg/s, and water enters the
annulus at 60
o
C at 0.10 kg/s:
A. What is the temperature of the blood exiting the exchanger?
B. How long would the exchanger need to be to achieve a blood outlet temperature of 37
o
C?
Note:
You may assume the mean temperature of water to be 55
o
C (for determining fluid properties ONLY), and that
the heat capacity of the blood is 3500 J/kg-K.
Part A is a RETROFIT problem, since neither outlet temperature is known. Solve using
ε
-NTU.
Part B is a DESIGN problem, so you can use either
ε
-NTU or LMTD to solve.

8. A large shell and tube heat exchanger has U = 640 W/m
2
-K. There are 200 thin-walled tubes with diameter 2.7 cm and
5.5 m length. They make 4 total passes through the exchanger. There are two shell passes.

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- Fall '10
- Brazel
- Heat, Heat Transfer, DI, K/W