Question

The Reynolds number for condensate flow is defined as $\operatorname{Re}=4 \dot{m} / p \mu_{l}\(, where \)p$ is the wetted perimeter. Obtain simplified relations for the Reynolds number by expressing \(p\) and \(\dot{m}\) by their equivalence for the following geometries: \((a)\) a vertical plate of height \(L\) and width \(w,(b)\) a tilted plate of height \(L\) and width \(w\) inclined at an angle \(u\) from the vertical, (c) a vertical cylinder of length \(L\) and diameter \(D,(d)\) a horizontal cylinder of length \(L\) and diameter \(D\), and (e) a sphere of diameter \(D\).

Short answer

Answer: The expression for the Reynolds number for a vertical plate is: \(\operatorname{Re}= \frac{4 \rho_l wL}{(2(w+L))\mu_{l}}\). 2. How does the angle of inclination (u) affect the Reynolds number for a tilted plate? Answer: The angle of inclination (u) affects the Reynolds number for a tilted plate by changing the wetted perimeter in the expression: \(\operatorname{Re}= \frac{4 \rho_l wL}{(2(w+L\cos{u}))\mu_{l}}\). The greater the angle, the smaller the contribution of the height to the wetted perimeter, and thus the larger the Reynolds number. 3. How does the expression for the Reynolds number differ between a vertical cylinder and a horizontal cylinder? Answer: For a vertical cylinder, the expression for the Reynolds number is \(\operatorname{Re}=\frac{4L\rho_l}{\mu_l}\), while for a horizontal cylinder, it is \(\operatorname{Re}=\frac{8L\rho_l}{\mu_l}\). The difference is a factor of 2, due to the difference in wetted perimeter for the horizontal cylinder, which is only half the circumference. 4. What is the expression for the Reynolds number for a sphere in terms of its diameter (D), condensate density (\(\rho_l\)), and liquid viscosity (\(\mu_l\))? Answer: The expression for the Reynolds number for a sphere is: \(\operatorname{Re}=\frac{16\rho_l D}{9\mu_l}\).

Step-by-step solution

Geometry (a): Vertical Plate

For a vertical plate of height \(L\) and width \(w\), we can consider the height \(L\) as the 'wetted length' and the width \(w\) as the 'wetted width'. Therefore, the wetted perimeter is given by: \[p = 2(w + L)\] The mass flow rate \(\dot{m}\) depends on the condensate density \(\rho_l\) and the plate's surface area. In this case, the mass flow rate is: \[\dot{m} = \rho_l wL\] Substituting these expressions for \(p\) and \(\dot{m}\) into the Reynolds number equation, we get: \[\operatorname{Re}= \frac{4 \rho_l wL}{(2(w+L))\mu_{l}}\]

Geometry (b): Tilted Plate

For a tilted plate of height \(L\) and width \(w\) inclined at an angle \(u\) from the vertical, we can consider the projected height as \(L \cos{u}\). Hence, the wetted perimeter is given by: \[p = 2(w + L\cos{u})\] The mass flow rate for the tilted plate is the same as for the vertical plate: \[\dot{m} = \rho_l wL\] Substituting these expressions into the Reynolds number equation, we get: \[\operatorname{Re}= \frac{4 \rho_l wL}{(2(w+L\cos{u}))\mu_{l}}\]

Geometry (c): Vertical Cylinder

For a vertical cylinder of length \(L\) and diameter \(D\), the wetted perimeter is the circumference of the cylinder, which is given by: \[p = \pi D\] The mass flow rate depends on the condensate density \(\rho_l\) and the cylinder's outer surface area. So, the mass flow rate is: \[\dot{m} = \rho_l \pi LD\] Substituting these expressions into the Reynolds number equation, we get: \[\operatorname{Re}= \frac{4 \rho_l \pi LD}{\pi D\mu_{l}} = \frac{4L\rho_l}{\mu_l}\]

Geometry (d): Horizontal Cylinder

For a horizontal cylinder of length \(L\) and diameter \(D\), the wetted perimeter is half the circumference of the cylinder because the condensate only clings to the bottom half. So, the wetted perimeter is: \[p = \frac{1}{2} \pi D\] The mass flow rate for a horizontal cylinder is the same as for a vertical cylinder: \[\dot{m} = \rho_l \pi LD\] Substituting these expressions into the Reynolds number equation, we get: \[\operatorname{Re}= \frac{4 \rho_l \pi LD}{(\frac{1}{2}\pi D\mu_{l})} = \frac{8L\rho_l}{\mu_l}\]

Geometry (e): Sphere

For a sphere of diameter \(D\), the wetted perimeter can be considered as the circumference of the sphere's great circle, which is given by: \[p = \pi D\] The mass flow rate for a sphere depends on the condensate density \(\rho_l\) and the sphere's surface area. So, the mass flow rate is: \[\dot{m} = \rho_l \frac{4}{3} \pi \left(\frac{D}{2}\right)^2\] Substituting these expressions into the Reynolds number equation, we get: \[\operatorname{Re}= \frac{4 \rho_l (\frac{4}{3} \pi (\frac{D}{2})^2)}{\pi D\mu_{l}} = \frac{16\rho_l D}{9\mu_l}\]