∫ (𝑢𝑚𝑎𝑥𝑟 𝑅)
2
𝑟𝑑𝑟
𝑅 0
=𝑢𝑚𝑎𝑥2
𝑅2 ∫ 𝑟3𝑑𝑟
𝑅 0
=𝑢𝑚𝑎𝑥2 𝑅2
𝑅4 4
=𝑢𝑚𝑎𝑥2 𝑅2 4
=𝑢𝑚𝑎𝑥2 𝐷2 16
∫ 𝑟2𝑑𝑟 =
𝑅 0
𝑟3 3|
0 𝑅
=𝑅3 3
P5.60 A variable mesh screen produces a linear and axisymmetric velocity profile as indicated in Fig P5.60 in the air flow through a 2 ft diameter circular cross duct. The static pressure upstream and downstream of the screen are 0.2 and 0.15 psi and are uniformly distributed over the flow cross section area. Neglecting the force exerted by the duct wall on the flowing air, calculate the screen drag force.
∑ 𝐹 = 𝑑
𝑑𝑡∫ 𝑉𝜌𝑑∀
𝐶𝑉
+ ∫ 𝑉𝜌𝑉𝑅∙ 𝑛𝑑𝐴
𝐶𝑆
Steady state, fixed CV ie 𝑉𝑠= 0, inlet=1, outlet=2
𝑝1𝐴1− 𝑝2𝐴2− 𝑅𝑥= −𝑉1𝜌𝑉1𝐴1+ ∫ 𝑢2𝜌𝑢22𝜋𝑟𝑑𝑟
𝑅 0
𝑅𝑥= 𝜌𝑉12𝜋𝐷12
4 − 2𝜋𝜌 ∫ (𝑢𝑚𝑎𝑥
𝑟 𝑅)
2
𝑟𝑑𝑟
𝑅 0
+ (𝑝1− 𝑝2)𝜋𝐷12 4
0 = 𝑑
𝑑𝑡∫ 𝜌𝑑∀
𝐶𝑉
+ ∫ 𝜌𝑉𝑅∙ 𝑛𝑑𝐴
𝐶𝑆
0 = ∫ 𝜌𝑉 ∙ 𝑛𝑑𝐴
𝐶𝑆
= −𝜌𝑉1𝐴1+ ∫ 𝜌𝑢22𝜋𝑟𝑑𝑟
𝑅 0
𝜌𝑉1𝜋𝐷12
4 = 𝜌 ∫ (𝑢𝑚𝑎𝑥𝑟 𝑅)
𝑅 0
2𝜋𝑟𝑑𝑟
𝑉1𝐷12
4 =2𝑢𝑚𝑎𝑥
𝑅 ∫ 𝑟2𝑑𝑟
𝑅 0
=2𝑢𝑚𝑎𝑥 𝑅
𝑅3 3 𝑉1𝑅2=2𝑢𝑚𝑎𝑥𝑅2
3 𝑢𝑚𝑎𝑥=3
2𝑉1= 150 𝑓𝑡/𝑠 𝑅𝑥 = 13.3 𝑙𝑏
