Steam at 450°F is flowing through a steel pipe (k = 8.7 Btu/h · ft · °F) whose inner and outer diameters are 3.5 in. and 4.0 in., respectively, in an environment at 55°F. The pipe is insulated with 2-in.-thick fiberglass insulation (k = 0.020 Btu/h · ft · °F). If the heat transfer coefficients on

Question

Steam at 450°F is flowing through a steel pipe ([latex]k = 8.7 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex]) whose inner and outer diameters are 3.5 in. and 4.0 in., respectively, in an environment at 55°F. The pipe is insulated with 2-in.-thick fiberglass insulation ([latex]k = 0.020 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex]). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and [latex]5 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F[/latex], respectively, determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations?

Accepted Answer

A steam pipe covered with 2-in thick fiberglass insulation is subjected to convection on its surfaces. The rate of heat loss from the steam per unit length and the error involved in neglecting the thermal resistance of the steel pipe in calculations are 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 center line and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible.

Properties The thermal conductivities are given to be [latex] k=8.7 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex] for steel and [latex] k = 0.020 Btu / h \cdot ft \cdot{ }^{\circ} F [/latex] for fiberglass insulation.

Analysis The inner and outer surface areas of the insulated pipe are

[latex] A_{i}=\pi D_{i} L=\pi(3.5 / 12 ft )(1 ft )=0.916 ft ^{2}[/latex]

[latex] A_{o}=\pi D_{o} L=\pi(8 / 12 ft )(1 ft )=2.094 ft ^{2}[/latex]

The individual resistances are

[latex] R_{i}=\frac{1}{h_{i} A_{i}}=\frac{1}{\left(30 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F ight)\left(0.916 ft ^{2} ight)}=0.036 h \cdot{ }^{\circ} F / Btu [/latex]

[latex] R_{1}=R_{p i p e}=\frac{\ln \left(r_{2} / r_{1} ight)}{2 \pi k_{1} L}=\frac{\ln (2 / 1.75)}{2 \pi\left(8.7 Btu / h \cdot ft \cdot{ }^{\circ} F ight)(1 ft )}=0.002 h \cdot{ }^{\circ} F / Btu [/latex]

[latex] R_{2}=R_{ ext {insulation }}=\frac{\ln \left(r_{3} / r_{2} ight)}{2 \pi k_{2} L}=\frac{\ln (4 / 2)}{2 \pi\left(0.020 Btu / h \cdot ft \cdot ^{\circ} F ight)(1 ft )} = 5.516 h \cdot{ }^{\circ} F / Btu [/latex]

[latex] R_{o}=\frac{1}{h_{o} A_{o}}=\frac{1}{\left(5 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F ight)\left(2.094 ft ^{2} ight)}=0.096 h \cdot{ }^{\circ} F / Btu[/latex]

[latex] R_{ ext {total }} = R_{i}+R_{1}+R_{2}+R_{o}=0.036+0.002+5.516+0.096 = 5.65 h \cdot{ }^{\circ} F / Btu[/latex]

Then the steady rate of heat loss from the steam per ft. pipe length becomes

[latex] \dot{Q}=\frac{T_{\infty 1} - T_{\infty 2}}{R_{ ext {total }}}=\frac{(450-55)^{\circ} F }{5.65 h \cdot ^{\circ} F / Btu }= 6 9 . 9 1 \mathrm { Btu } / \mathrm { h }[/latex]

If the thermal resistance of the steel pipe is neglected, the new value of total thermal resistance will be

[latex] R_{ ext {total }}=R_{i}+R_{2}+R_{o} = 0.036+5.516+0.096 = 5.648 h \cdot ^{\circ} F / Btu [/latex]

Then the percentage error involved in calculations becomes

[latex] ext { error } \%=\frac{(5.65-5.648) h \cdot ^{\circ} F / Btu }{5.65 h \cdot ^{\circ} F / Btu } imes 100 = 0 . 0 3 5 \%[/latex]

which is insignificant.

Question 3.75.E: Steam at 450°F is flowing through a steel pipe (k = 8.7 Btu/...

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