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liraira [26]
4 years ago
5

Hot water at an average temperature of 88°C and an average velocity of 0.4 m/s is flowing through a 5-m section of a thin-walled

hot-water pipe that has an outer diameter of 2.5 cm. The pipe passes through the center of a 14-cm-thick wall filled with fiberglass insulation (k = 0.035 W/m·K). The surfaces of the wall are at 18°C.a. Determine the rate of heat transfer from the pipe to the air in the rooms. b. Determine the temperature drop of the hot water as it flows through the 5-m-long section of the wall where the steady rate of heat transfer from the pipe is 39.153 W. For water, use density (rho) 1000 kg/m^3 and specific capacity of heat (cp) =4180 J/kg.°C.

Engineering
1 answer:
SIZIF [17.4K]4 years ago
4 0

Answer:

a) The rate of heat transfer will be 19.58 Watts.

b) The temperature drop of the hot water will be 0.024 Degree Celcius.

Explanation:

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Answer:

Final Length = 11.992 in

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Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

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Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

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Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

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Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

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we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

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