Consider a 25-mm-diameter and 15-m-long smooth tube that is maintained at a constant surface temperature. Fluids enter the tube at 50°C with a mass flow rate of 0.01 kg/s. Determine the tube surface temperatures necessary to heat water, engine oil, and liquid mercury to the desired outlet temperature of 150°C

Answers

Answer 1

Answer: ⚠this is not my answer it from hamzaahmeds

Water: h = 35.53 W/m².k

Engine oil: h = 18.84 W/m².k

Mercury: h = 1.19 W/m².k

Explanation:

Assuming the steady state, one-dimensional heat flow, it is clear that the added to the fluid by tube heat will be equal to the heat transfer through convection outside the tube.

Therefore,

mCΔT = hAΔT

mC = hA

h = mC/A

where,

h = convection coefficient

m = mass flow rate =  0.01 kg/s

C = specific heat capacity of fluid

A = surface area of tube = 2πrL = 2π(0.0125 m)(15 m) = 1.178 m²

FOR WATER:

C = 4186 J/Kg.k

Therefore,

h = (0.01 kg/s)(4186 J/Kg.k)/(1.178 m²)

h = 35.53 W/m².k

FOR ENGINE OIL:

C = 2220 J/Kg.k

Therefore,

h = (0.01 kg/s)(2220 J/Kg.k)/(1.178 m²)

h = 18.84 W/m².k

FOR LIQUID MERCURY:

C = 140 J/Kg.k

Therefore,

h = (0.01 kg/s)(140 J/Kg.k)/(1.178 m²)

h = 1.19 W/m².k


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9514 1404 393

Answer:

  (x, y) = (5.76, 1 5/7)

Explanation:

The location of the centroid in the x-direction is the ratio of the first moment of area about the y-axis to the total area. Similarly, the y-coordinate of the centroid is the first moment of area about the x-axis, divided by the area.

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For k1 = 4/12^3 and k2=12/4^2, these evaluate to ...

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The y-coordinate of the centroid requires we find the distance of the differential of area from the x-axis. We can use (y2 +y1)/2 for that purpose. Then the y-coordinate is ...

  [tex]\displaystyle c_y=\frac{a_y}{a}\\\\a_y=\int_0^{12}{(\frac{y_2+y_1}{2}(y_2-y_1))}\,dx=\frac{1}{2}\int_0^{12}{(\frac{x}{k_2}-(k_1x^3)^2)}\,dx\\\\a_y=\frac{12^2}{4k_2}-\frac{k_1^212^7}{14}=\frac{240}{7}\\\\c_y=\frac{12}{7}\approx1.7143[/tex]

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  (x, y) = (5.76, 1 5/7)

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Explanation:

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