For the horizontal series-parallel system of the Figure below, all pipes are 8-cm-diameter asphalted cast iron (ε = 0.12 mm) filled with water at 20°C (ρ = 998 kg/m3, μ = 0.001 kg/ms). The total flow rate is Q = 0.0269 m3/s, the lengths of segments A and C are LA = 250 m and LC = 150 m, and velocity in segment A is VA = 2.06 m/s. Neglect minor losses and find (a) the length of segment B (LB), and (b) the total pressure drop (𝑝1− 𝑝2).
1. Solution (a) Continuity:
𝑄 = 𝑄𝐴+ 𝑄𝐵 =𝑉𝐴(𝜋
4𝑑2)+ 𝑉𝐵(𝜋 4𝑑2) 𝑉𝐵 = 𝑄
𝜋 4 𝑑2
− 𝑉𝐴 = (0.0269) 𝜋
4(0.08)2− (2.06) = 3.29 𝑚/𝑠 𝑉𝐶 = 𝑄
𝜋 4 𝑑2
= (0.0269) 𝜋
4(0.08)2 = 5.35 𝑚/𝑠
For segment A, B and C:
𝜀
𝑑 =(0.12 1000)
(0.08) = 0.0015 (𝑅𝑒𝑑)𝐴 =𝜌𝑉𝐴𝑑
𝜇 =(998)(2.06)(0.08)
(0.001) = 164,470 → 𝑓𝐴 = 0.0229 (𝑅𝑒𝑑)𝐵 =𝜌𝑉𝐵𝑑
𝜇 = (998)(3.29)(0.08)
(0.001) = 262,674 → 𝑓𝐵= 0.0225 (𝑅𝑒𝑑)𝐶 = 𝜌𝑉𝐶𝑑
𝜇 =(998)(5.35)(0.08)
(0.001) = 427,144 → 𝑓𝐶 = 0.0222
For parallel segments A and B the head loss is the same:
(ℎ𝑓)
𝐴 = (ℎ𝑓)
𝐵 → (𝑓𝐿 𝑑
𝑉2 2𝑔)
𝐴
= (𝑓𝐿 𝑑
𝑉2 2𝑔)
𝐵
𝐿𝐵 = 𝐿𝐴𝑓𝐴 𝑓𝐵(𝑉𝐴
𝑉𝐵)
2
= (250)(0.0229)
(0.0225)((2.06) (3.29))
2
= 99.76 𝑚
(b) The energy equation between points (1) and (2) through segment A yields:
(𝑝 𝜌𝑔+𝑉2
2𝑔+ 𝑧)
1
= (𝑝 𝜌𝑔+𝑉2
2𝑔+ 𝑧)
2
+ (ℎ𝑓)
𝐴+ (ℎ𝑓)
𝐶
Since 𝑉1 = 𝑉2 ; 𝑧1 = 𝑧2
(𝑝1− 𝑝2) = 𝜌𝑔 [(ℎ𝑓)
𝐴 + (ℎ𝑓)
𝐶] = 𝜌𝑔 [(𝑓𝐿 𝑑
𝑉2 2𝑔)
𝐴
+ (𝑓𝐿 𝑑
𝑉2 2𝑔)
𝐶
]
(𝑝1− 𝑝2) = (998)(9.81) [(0.0229)(250) (0.08)
(2.06)2
2(9.81)+ (0.0222)(150) (0.08)
(5.35)2
2(9.81)] = 746,052 𝑃𝑎
Alternatively, through segment B:
(𝑝1− 𝑝2) = 𝜌𝑔 [(ℎ𝑓)
𝐵+ (ℎ𝑓)
𝐶] = (998)(9.81) [(0.0225)(99.76) (0.08)
(3.29)2
2(9.81)+ (0.0222)(150) (0.08)
(5.35)2
2(9.81)] = 746,059 𝑃𝑎
