A double-pipe (shell-and-tube) heat exchanger is constructed of a stainless steel (k=15.1 W / m \cdot K ) inner tube of inner diameter D_{i}=1.5 cm and outer diameter D_{o}=1.9 cm and an outer shell of inner diameter 3.2 cm. The convection heat transfer coefficient is given to be h_{i}=800 W / m ^{2} \cdot K on the inner surface of the tube and h_{o}=1200 W / m ^{2} \cdot K on the outer surface. For a fouling factor of R_{f, i}=0.0004 m ^{2} \cdot K / W on the tube side and R_{f, o}=0.0001 m ^{2} \cdot K / W on the shell side, determine (a) the thermal resistance of the heat exchanger per unit length and (b) the overall heat transfer coefficients, U_{i} and U_{o} based on the inner and outer surface areas of the tube, respectively.
Question : A double-pipe (shell-and-tube) heat exchanger is constructed...
SOLUTION The heat transfer coefficients and the fouling factors on the tube and shell sides of a heat exchanger are given. The thermal resistance and the overall heat transfer coefficients based on the inner and outer areas are to be determined.
Assumptions The heat transfer coefficients and the fouling factors are constant and uniform. Analysis (a) The schematic of the heat exchanger is given in Fig. 11-12. The thermal resistance for an unfinned shell-and-tube heat exchanger with fouling on both heat transfer surfaces is given by Eq. (11-8)
\frac{1}{U A_{s}}=\frac{1}{U_{i} A_{i}}=\frac{1}{U_{o} A_{o}}=R=\frac{1}{h_{i} A_{i}}+\frac{R_{f, i}}{A_{i}}+\frac{\ln \left(D_{o} / D_{i}\right)}{2 \pi k L}+\frac{R_{f, 0}}{A_{o}}+\frac{1}{h_{o} A_{o}}as
R=\frac{1}{U A_{s}}=\frac{1}{U_{i} A_{i}}=\frac{1}{U_{o} A_{0}}=\frac{1}{h_{i} A_{i}}+\frac{R_{f, i}}{A_{i}}+\frac{\ln \left(D_{o} / D_{i}\right)}{2 \pi k L}+\frac{R_{f, 0}}{A_{o}}+\frac{1}{h_{o} A_{o}}where
A_{i}=\pi D_{i} L=\pi(0.015 m )(1 m )=0.0471 m ^{2}A_{o}=\pi D_{o} L=\pi(0.019 m )(1 m )=0.0597 m ^{2}
Substituting, the total thermal resistance is determined to be
R= \frac{1}{\left(800 W / m ^{2} \cdot K \right)\left(0.0471 m ^{2}\right)}+\frac{0.0004 m ^{2} \cdot K / W }{0.0471 m ^{2}}+\frac{\ln (0.019 / 0.015)}{2 \pi(15.1 W / m \cdot K )(1 m )}
+\frac{0.0001 m ^{2} \cdot K / W }{0.0597 m ^{2}}+\frac{1}{\left(1200 W / m ^{2} \cdot K \right)\left(0.0597 m ^{2}\right)}
=(0.02654+0.00849+0.0025+0.00168+0.01396) K / W
= 0.0532^{\circ} C / W
Note that about 19 percent of the total thermal resistance in this case is due to fouling and about 5 percent of it is due to the steel tube separating the two fluids. The rest (76 percent) is due to the convection resistances.
(b) Knowing the total thermal resistance and the heat transfer surface areas, the overall heat transfer coefficients based on the inner and outer surfaces of the tube are
U_{i}=\frac{1}{R A_{i}}=\frac{1}{(0.0532 K / W )\left(0.0471 m ^{2}\right)}=399 W / m ^{2} \cdot Kand
U_{o}=\frac{1}{R A_{0}}=\frac{1}{(0.0532 K / W )\left(0.0597 m ^{2}\right)}=315 W / m ^{2} \cdot KDISCUSSION Note that the two overall heat transfer coefficients differ significantly (by 27 percent) in this case because of the considerable difference between the heat transfer surface areas on the inner and the outer sides of the tube. For tubes of negligible thickness, the difference between the two overall heat transfer coefficients would be negligible.