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Flow between reservoirs (15 cm and 25 cm pipe)

Short answer

Apply Bernoulli equating the 6 m head to entry, friction, sudden enlargement and exit losses. Solving with continuity V₂=0.36V₁ and total loss coefficient 2.950 yields V₁=6.32 m/s, discharge Q=0.112 m³/s.

Two reservoirs are connected by a pipeline which is 15 cm diameter for the first 6 m and 25 cm diameter in remaining 15 m length. The entrance and exit are sharp and change of section is sudden. If water surface in upper reservoir is 6 m above that of lower reservoir, calculate rate of flow. Take friction factor 0.04 for all pipes.

Apply Bernoulli's equation between the two reservoir surfaces. The difference in levels equals the sum of all losses.

Given: pipe 1: d1=0.15d_1 = 0.15 m, L1=6L_1 = 6 m; pipe 2: d2=0.25d_2 = 0.25 m, L2=15L_2 = 15 m; H=6H = 6 m; Darcy f=0.04f = 0.04 (hf=fLV2/2gdh_f = fLV^2/2gd).

 ===== upper reservoir
   |  |                      
   |  | 6 m                   
   |  |______ 15 cm, 6 m ___
   |         |____ 25 cm, 15 m ______  
   ------------------------------- ===== lower

Continuity

V2=V1(d1d2)2=V1(0.150.25)2=0.36 V1V_2 = V_1\left(\frac{d_1}{d_2}\right)^2 = V_1\left(\frac{0.15}{0.25}\right)^2 = 0.36\,V_1

Losses in terms of V12/2gV_1^2/2g

LossFormulaCoefficient of V12/2gV_1^2/2g
Entrance (sharp)0.5V12/2g0.5V_1^2/2g0.5000
Friction, pipe 1fL1d1V122g=0.04×60.15f\frac{L_1}{d_1}\frac{V_1^2}{2g} = 0.04\times\frac{6}{0.15}1.6000
Sudden enlargement(V1−V2)22g=(1−0.36)2\frac{(V_1 - V_2)^2}{2g} = (1 - 0.36)^20.4096
Friction, pipe 20.04×150.25×0.3620.04\times\frac{15}{0.25}\times0.36^20.3110
ExitV222g=0.362\frac{V_2^2}{2g} = 0.36^20.1296
Total2.9502

Velocity and discharge

6=2.9502 V122gV122g=2.0337 mV1=2×9.81×2.0337=6.317 m/sV2=0.36×6.317=2.274 m/sQ=π4(0.15)2×6.317=0.1116 m3/s\begin{aligned} 6 &= 2.9502\,\frac{V_1^2}{2g} \\ \frac{V_1^2}{2g} &= 2.0337\ \text{m} \\ V_1 &= \sqrt{2 \times 9.81 \times 2.0337} = 6.317\ \text{m/s} \\ V_2 &= 0.36 \times 6.317 = 2.274\ \text{m/s} \\ Q &= \frac{\pi}{4}(0.15)^2 \times 6.317 = 0.1116\ \text{m}^3/\text{s} \end{aligned}

Check: losses =1.017+3.254+0.833+0.633+0.264=6.00= 1.017 + 3.254 + 0.833 + 0.633 + 0.264 = 6.00 m.

Answer: rate of flow Q≈0.112Q \approx 0.112 m³/s (about 112 L/s). (If 0.04 were the coefficient in 4fLV2/2gd4fLV^2/2gd, the friction terms would be four times larger and QQ smaller; the Darcy form is used here.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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