Steam at 120^ ^ ^ C is continuously passed through a rubber tube that is 50 cm long with inner and outer radii of 1.0 cm and 1.2 cm , respectively. The room temperature is 30^ ^ ^ C . Determine the rate of heat flow

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 233 J s^ -1

The rate of heat flow through a cylindrical wall is given by:

H = \dfrac{2 \pi K L (T_1 - T_2)}{\ln(r_2/r_1)}

Substituting the given values:

K = 0.15 \text{ J s}^{-1} \text{ m}^{-1} ^{{}^{{}^{\circ}}}\textC^{-1}

L = 0.5 \text{ m}

T_1 = 120^{{}^{{}^{\circ}}}\text{C}

,

T_2 = 30^{{}^{{}^{\circ}}}\text{C}

r_1 = 1.0 \text{ cm}

,

r_2 = 1.2 \text{ cm}

H = \dfrac{2 \pi (0.15) (0.5) (120 - 30)}{\ln(1.2/1.0)}

H = \dfrac{0.15 \pi (90)}{\ln(1.2)}

Using

\ln(1.2) \approx 0.1823

and

\pi \approx 3.1416

:

H = \dfrac{13.5 \times 3.1416}{0.1823}

H \approx 232.6 \text{ J s}^{-1}

Rounding to the nearest integer gives

233 \text{ J s}^{-1}

.

Steam at 120^ ^ ^ C is continuously passed through a rubber tube...

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