A 50-m-long section of a steam pipe whose outer diameter is 10 cm passes through an open space at 15°C. The average temperature of the outer surface of the pipe is measured to be 150°C. If the combined heat transfer coefficient on the outer surface of the pipe is 20 W/m^2 · °C, determine (a) the

Question

A 50-m-long section of a steam pipe whose outer diameter is 10 cm passes through an open space at 15°C. The average temperature of the outer surface of the pipe is measured to be 150°C. If the combined heat transfer coefficient on the outer surface of the pipe is [latex]20 W / m ^{2} \cdot{ }^{\circ} C[/latex], determine (a) the rate of heat loss from the steam pipe, (b) the annual cost of this energy lost if steam is generated in a natural gas furnace that has an efficiency of 75 percent and the price of natural gas is $0.52/therm (1 therm = 105,500 kJ), and (c) the thickness of fiberglass insulation ([latex]k=0.035 W / m \cdot{ }^{\circ} C[/latex]) needed in order to save 90 percent of the heat lost. Assume the pipe temperature to remain constant at 150°C.

Accepted Answer

A 50-m long section of a steam pipe passes through an open space at 15°C. The rate of heat loss from the steam pipe, the annual cost of this heat loss, and the thickness of fiberglass insulation needed to save 90 percent of the heat lost 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 conductivity is constant. 4 The thermal contact resistance at the interface is negligible. 5 The pipe temperature remains constant at about 150°C with or without insulation. 6 The combined heat transfer coefficient on the outer surface remains constant even after the pipe is insulated.

Properties The thermal conductivity of fiberglass insulation is given to be [latex]k=0.035 W / m \cdot{ }^{\circ} C[/latex]

Analysis (a) The rate of heat loss from the steam pipe is

[latex] A_{o}=\pi D L=\pi(0.1 m )(50 m ) = 15.71 m ^{2}[/latex]

[latex] \dot{Q}_{ ext {bare }}=h_{o} A\left(T_{s}-T_{ ext {air }} ight)=\left(20 W / m ^{2} \cdot{ }^{\circ} C ight)\left(15.71 m ^{2} ight)(150 - 15){ }^{\circ} C = 4 2 , 4 1 2 W[/latex]

(b) The amount of heat loss per year is

[latex] Q=\dot{Q} \Delta t=(42.412 kJ / s )(365 imes 24 imes 3600 s / yr )=1.337 imes 10^{9} kJ / yr[/latex]

The amount of gas consumption from the natural gas furnace that has an efficiency of 75% is

[latex] Q_{ ext {gas }}=\frac{1.337 imes 10^{9} kJ / yr }{0.75}\left(\frac{1 ext { therm }}{105,500 kJ } ight)=16,903 ext { therms } / yr[/latex]

The annual cost of this energy lost is

Energy cost = (Energy used)(Unit cost of energy)

[latex] =(16,903 ext { therms } / yr )(\$ 0.52 / ext { therm })=\$ 8790 / yr[/latex]

(c) In order to save 90% of the heat loss and thus to reduce it to [latex] 0.1 imes 42,412 = 4241 W[/latex], the thickness of insulation needed is determined from

[latex] \dot{Q}_{ ext {insulated }}=\frac{T_{s}-T_{ ext {air }}}{R_{o}+R_{ ext {insulation }}}=\frac{T_{s}-T_{ ext {air }}}{\frac{1}{h_{o} A_{o}}+\frac{\ln \left(r_{2} / r_{1} ight)}{2 \pi k L}} [/latex]

Substituting and solving for [latex]r_{2}[/latex], we get

[latex] 4241 W =\frac{(150-15){ }^{\circ} C }{\frac{1}{\left(20 W / m ^{2} \cdot{ }^{\circ} C ight)\left[\left(2 \pi r_{2}(50 m ) ight] ight.}+\frac{\ln \left(r_{2} / 0.05 ight)}{2 \pi\left(0.035 W / m \cdot{ }^{\circ} C ight)(50 m )}} \longrightarrow r_{2}=0.0692 m[/latex]

Then the thickness of insulation becomes

[latex] t_{ ext {insulation }} = r_{2} - r_{1} = 6.92 - 5 = 1 . 9 2 c m[/latex]

Question 3.70: A 50-m-long section of a steam pipe whose outer diameter is ...

Related Answered Questions