Repeat Problem 3–78E, assuming that a 0.01-in.-thick layer of mineral deposit (k = 0.5 Btu/h · ft · °F) has formed on the inner surface of the pipe.

Question

Repeat Problem 3–78E, assuming that a 0.01-in.-thick layer of mineral deposit ([latex]k = 0.5 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex]) has formed on the inner surface of the pipe.

Accepted Answer

Steam exiting the turbine of a steam power plant at 100°F is to be condensed in a large condenser by cooling water flowing through copper tubes. For specified heat transfer coefficients and 0.01-in thick scale build up on the inner surface, the length of the tube required to condense steam at a rate of 400 lbm/h is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 Heat transfer coefficients are constant and uniform over the surfaces.

Properties The thermal conductivities are given to be [latex] k=223 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex] for copper tube and be [latex]k = 0.5 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex] for the mineral deposit. The heat of vaporization of water at 100°F is given to be 1037 Btu/lbm.

Analysis When a 0.01-in thick layer of deposit forms on the inner surface of the pipe, the inner diameter of the pipe will reduce from 0.4 in to 0.38 in. The individual thermal resistances are

[latex] A_{i}=\pi D_{i} L=\pi(0.4 / 12 ft )(1 ft )=0.105 ft ^{2}[/latex]

[latex] A_{o}=\pi D_{o} L=\pi(0.6 / 12 ft )(1 ft )=0.157 ft ^{2}[/latex]

[latex] R_{i}=\frac{1}{h_{i} A_{i}}=\frac{1}{\left(35 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F ight)\left(0.105 ft ^{2} ight)}=0.2711 h \cdot ^{\circ} F / Btu [/latex]

[latex] R_{ ext {pipe }}=\frac{\ln \left(r_{2} / r_{1} ight)}{2 \pi k L}=\frac{\ln (3 / 2)}{2 \pi\left(223 Btu / h \cdot ft \cdot{ }^{\circ} F ight)(1 ft )}=0.00029 h \cdot ^{\circ} F / Btu[/latex]

[latex] R_{ ext {deposit }}=\frac{\ln \left(r_{1} / r_{ ext {dep }} ight)}{2 \pi k_{2} L}=\frac{\ln (0.2 / 0.19)}{2 \pi\left(0.5 Btu / h \cdot ft \cdot{ }^{\circ} F ight)(1 ft )}=0.01633 h \cdot{ }^{\circ} F / Btu [/latex]

[latex] R_{o}=\frac{1}{h_{o} A_{o}}=\frac{1}{\left(1500 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F ight)\left(0.157 ft ^{2} ight)} = 0.00425 h \cdot ^{\circ} F / Btu[/latex]

[latex] R_{ ext {total }}=R_{i}+R_{ ext {pipe }}+R_{ ext {deposit }}+R_{o} [/latex]

[latex] = 0.27211 + 0.00029 + 0.01633 + 0.00425 = 0.29298 h \cdot ^{\circ} F / Btu[/latex]

The heat transfer rate per ft length of the tube is

[latex] \dot{Q}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{ ext {total }}}=\frac{(100-70)^{\circ} F }{0.29298^{\circ} F / Btu } = 102.40 Btu / h [/latex]

The total rate of heat transfer required to condense steam at a rate of 400 lbm/h and the length of the tube required can be determined to be

[latex] \dot{Q}_{ ext {total }}=\dot{m } h_{f g} = (120 lbm / h )(1037 Btu / lbm ) = 124,440 Btu / h[/latex]

[latex] ext { Tube length } = \frac{\dot{Q}_{ ext {total }}}{\dot{Q}} = \frac{124,440}{102.40} = 1 2 1 5 f t [/latex]

Question 3.79.E: Repeat Problem 3–78E, assuming that a 0.01-in.-thick layer o...

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