(20 points) An engineer is tasked with developing a model to study a cylindrical heat exchanger in a steam system. The prototype cylinder is 2.54 cm in diameter and the steam properties are: velocity

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An engineer is tasked with developing a model to study a cylindrical heat exchanger in a steam system. The prototype cylinder is 2.54 cm in diameter and the steam properties are: velocity = 30 m/s; density = 0.6 kg/m³; and absolute viscosity = 1.197 X 10⁻⁵ N-s/m², respectively. The model is going to be tested in a water tunnel where the water temperature is 20°C and the velocity is 3 m/s

a) what is the prototype Reynolds number, based on using the cylinder diameter as the characteristic length?

b) what should be the diameter of the model be to ensure dynamic similitude?

a) the prototype Reynolds number is 38195.488

b) 0.01278 m or 1.278 cm should be the diameter of the model be to ensure dynamic similitude

Explanation:

Given that;

1 prototype ; d = 2.54 cm = 0.0254 m, Vp = 30 m/s, Sp = 0.6 kg/m³, Up = 1.197 × 10⁻⁵ N-s/m²

Model{ water at 20°C}; dm = ?, Vm = 3 m/s, Pm = 998.23 kg/m³, Um = 1.002 × 10⁻³ N-s/m²

a) what is the prototype Reynolds number,

to calculate prototype Reynolds number we use the expression;

(Re)p = SpVpdP / Up

we substitute our value

(Re)p =  (0.6 × 30 × 0.0254) / 1.197 × 10⁻⁵

(Re)p = 38195.488

Therefore the prototype Reynolds number is 38195.488

b)

what should be the diameter of the model be to ensure dynamic similitude?

i.e dm = ?

so dynamic similarity [ viscous flow]

(Re)m = (Re)p

[PmVmdm / Um] = [SpVpdP / Up]

we substitute

[998.23 × 3 × dm / 1.002 × 10⁻³] = (0.6 × 30 × 0.0254) / 1.197 × 10⁻⁵

2994.69dm / 1.002 × 10⁻³ = 38195.488

2994.69dm = 38.2718

dm = 38.2718 / 2994.69

dm = 0.01278 m or 1.278 cm

Therefore 0.01278 m or 1.278 cm should be the diameter of the model be to ensure dynamic similitude

gas and liquid