Air at p=1 atm enters a long tube of length 2.5 m and diameter of 12 mm at an inlet temperature of Tm,i=100oC and mass flowrate
of 300 x 10-6 kg/s. A constant heat flux is applied to the air from the tube surface. If the tube surface temperature at the exit is Ts,o=180oC, determine the heat rate entering the tube. Evaluate properties at T=400 K.
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Answer:
F = 9.11 x 10³ N = 9.11 KN
Explanation:
The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:
ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper
ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)
where,
ΔL = Equivalent Change in Length = ?
ΔT = Change in Temperature = 25°C - 12°C = 13°C
Ls = Length of Steel Segment = 300 mm = 0.3 m
Lb = Length of Brass Segment = 200 mm = 0.2 m
Lc = Length of Copper Segment = 100 mm = 0.1 m
Therefore,
ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)
ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m
ΔL = 123.5 x 10⁻⁶ m ----------------------- equation (1)
Now, we calculate this deflection in terms of an applied force (F):
ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)
ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)
ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)
ΔL = F(13.55 x 10⁻⁹ m/N) --------------------- equation (1)
Comparing equation (1) and equation (2):
123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)
F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)
<u>F = 9.11 x 10³ N = 9.11 KN</u>
