PROBLEM 11.22
KNOWN: Inlet and outlet temperatures for a shell-and-tube heat exchanger with 10 tubes
making eight passes. Heat transfer coefficient for oil flowing in shell. Mass flow rate of water in
tubes. Tube diameter.
FIND: Oil flow rate required to achieve specified outlet temperature. Tube length required to
achieve specified water heating.
SCHEMATIC:
&h
m
Th,i = 160°C
Th,o = 100°C
Tc,o = 85°C
& c = 2.5 kg / s
m
Shell and tube hx
1 shell pass
8 tube passes
10 tubes
D = 25 mm
ho = 400 W/m2⋅K
Tc,i = 15°C
ASSUMPTIONS: (1) Negligible heat loss to the surroundings, (2) Constant properties, (3)
Negligible tube wall thermal resistance and fouling effects, (4) Fully developed water flow in
tubes.
PROPERTIES: Table A.5, unused engine oil: ( Th = 130°C): cp = 2350 J/kg⋅K. Table A.6,
water ( Tc = 50°C): cp = 4181 J/kg⋅K, μ = 548 × 10-6 N⋅s/m2, k = 0.643 W/m⋅K, Pr = 3.56.
ANALYSIS: From the overall energy balance, Eq. 11.7b, the heat transfer required of the
exchanger is
& cc p,c (Tc,o − Tc,i ) = 2.5 kg / s × 4181 J / kg ⋅ K(85 − 15)°C = 7.317 × 105 W
q=m
Hence from Eq. 11.6b,
&h=
m
q
c p,h (Th,i − Th,o )
=
7.317 × 105 W
= 5.19 kg / s
2350 J / kg ⋅ K(160 − 100)°C
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The required tube length may be obtained using the ε-NTU method. We first calculate the heat
& cc p,c =10,453 W/K. Thus, Cmin = Cc, and
& hc p,h = 12,195 W/K, Cc = m
capacity rates, C h = m
Cr = Cmin/Cmax = 0.857. Then from Eq. 11.21,
ε=
Tc,o − Tc,i
Th,i − Tc,i
=
(85 − 15)°C
= 0.483
(160 − 15)°C
Using Eqs. 11.30b,c for one shell pass and an even number of tube passes, we find
Continued…
PROBLEM 11.22 (Cont.)
E=
2 / ε − (1 + Cr )
(1 + C2r )1/ 2
=
2 / 0.483 − (1 + 0.857)
(1 + 0.8572 )1/ 2
= 1.74
⎛ E −1⎞
2 −1/ 2 ln ⎛ 1.74 − 1 ⎞ = 0.997
NTU = −(1 + C2r ) −1/ 2 ln ⎜
⎟ = −(1 + 0.857 )
⎜
⎟
⎝ E +1⎠
⎝ 1.74 + 1 ⎠
Thus UA = NTU×Cmin = 10,420 W/K. To find the required tube length, we must know the heat
&1 = m
&c/N=
transfer coefficients for the water flow. We calculate the Reynolds number, with m
0.25 kg/s defined as the water flow rate per tube, Eq. 8.6 yields
Re D =
&1
4m
4 × 0.25 kg / s
=
= 23, 234
πDμc π(0.025 m)548 × 10−6 N ⋅ s/m2
Hence the flow is turbulent, and from Eq. 8.60,
Nu D = 0.023Re4D/ 5 Pr 0.4 = 0.023(23, 234)4 / 5 (3.56)0.4 = 119
and
k
0.643 W / m ⋅ K
h c = c Nu D =
119 = 3060 W / m2 ⋅ K
D
0.025 m
Hence U = [1/hc + 1/hh]-1 = 354 W/m2⋅K and we can find the required tube length from
L=
UA
10, 420 W / K
=
= 37.5 m
UNπD 354 W / m2 ⋅ K × 10 × π × 0.025 m
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COMMENTS: (1) With L/D = 1516, the assumption of fully developed conditions throughout
the tube is justified. (2) With eight passes, the shell length is approximately L/8 = 4.7 m.