Evaluate the energy loss that would occur if the steel hydraulic tubing described in Example Problem 10.10 is coiled for 4½ revolutions to make a heat exchanger. The inside radius of the bend is the same 200 mm used earlier and the other conditions are the same.

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Accepted Answer

Let’s start by bringing some data from Example Problem 10.10.

r/D = 7.71

[latex] f_T = 0.0108 [/latex]

K = 0.248
v = 4.33 m/s

Now we can compute the value of [latex] K_B[/latex] for the complete coil using Eq. (10–10). Note that each revolution in the coil contains four 90° bends. Then,

n = 4.5 revolutions (4.0 90° bends/rev) = 18

The total bend resistance [latex] K_B[/latex] is

[latex] K_B =\left(n-1\right) \left[0.25\pi f_T\left(r/D\right) +0.5K\right] +K[/latex]

[latex] K_B =\left(18-1\right) \left[0.25\pi \left(0.0108\right) \left(7.71\right) +0.5\left(0.248\right) \right] +0.248[/latex]

[latex] K_B = 3.47[/latex]

Then the energy loss is found from

[latex] h_L = K_B \left(v^2/2g\right) = 3.47\left(4.33\right)^2 / \left[2\left(9.81\right) \right] =3.32N·m/N[/latex]

Question 10.11: Evaluate the energy loss that would occur if the steel hydra...

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