What does the 6 m difference in reservoir levels equal in this problem?

The sum of entry, friction, enlargement and exit losses

Only the friction loss in both pipes

Only the velocity head in the 30 cm pipe

The entry loss plus the exit loss only

How is the velocity in the 30 cm pipe related to that in the 20 cm pipe?

V2 = V1 x (0.2/0.3)^2 = 0.444 V1

After summing the coefficients (total 2.5329 of V1^2/2g), how is V1 and then Q found?

Set 6 = 2.5329 V1^2/2g, solve V1 = 6.82 m/s, then Q = area of 20 cm pipe x V1

Set 6 = V1^2/2g, then multiply by the 30 cm area

Divide 6 by 2.5329 to get V1 directly

Use the exit loss alone to find V2, then Q = A1 V2

2070 Bhadra · 8 marks

Flow between reservoirs (20 cm and 30 cm pipe)

Short answer

With coefficient of friction f=0.01, friction loss is 4fLV²/2gd. Summing all loss coefficients gives 2.533 equated to 6 m yields V₁=6.82 m/s, discharge Q=0.214 m³/s, and individual losses as tabulated.

Two reservoirs are connected by a pipe line which is 20 cm in diameter for the first 5 m and 30 cm in diameter for the remaining 20 m. The entrance and exit are sharp and change of section is sudden. If the water surface in the upper reservoir is 6 m above that in the lower reservoir, calculate head losses and rate of flow. Take f = 0.01 for all cases.

The total difference in water level (6 m) is used up by all the losses in the pipe: entry loss, friction in both pipes, sudden enlargement loss and exit loss. Since $f = 0.01$ is a coefficient of friction (R.K. Bansal convention), friction loss is $h_f = \dfrac{4fLV^2}{2gd}$ .

Data

Pipe 1: $d_1 = 0.20$ m, $L_1 = 5$ m; Pipe 2: $d_2 = 0.30$ m, $L_2 = 20$ m
$H = 6$ m, $f = 0.01$ , $g = 9.81$ m/s²

 Upper res.
 ~~~~~~~~~|                       H = 6 m
          |==== 20 cm ====|=========== 30 cm ==========|
          |     5 m       |            20 m            |~~~~~
                                               Lower res.

Velocity relation

By continuity, $V_2 = V_1\left(\dfrac{d_1}{d_2}\right)^2 = V_1\left(\dfrac{0.2}{0.3}\right)^2 = 0.4444\,V_1$

Losses in terms of $V_1^2/2g$

Loss	Formula	Coefficient of $V_1^2/2g$
Entrance (sharp)	$0.5V_1^2/2g$	0.5
Friction, pipe 1	$4fL_1V_1^2/(2g d_1)$	$4(0.01)(5)/0.2 = 1.0$
Sudden enlargement	$(V_1-V_2)^2/2g$	$(1-0.4444)^2 = 0.3086$
Friction, pipe 2	$4fL_2V_2^2/(2g d_2)$	$\frac{4(0.01)(20)}{0.3}(0.4444)^2 = 0.5267$
Exit	$V_2^2/2g$	$(0.4444)^2 = 0.1975$
Total		2.5329

Velocity and discharge

\begin{aligned} 6 &= 2.5329\,\frac{V_1^2}{2g} \\ V_1 &= \sqrt{\frac{2 \times 9.81 \times 6}{2.5329}} = 6.817\ \text{m/s} \\ V_2 &= 0.4444 \times 6.817 = 3.030\ \text{m/s} \\ Q &= \frac{\pi}{4}(0.2)^2 \times 6.817 = 0.2142\ \text{m}^3/\text{s} \end{aligned}

Individual head losses

Loss	Value (m)
Entrance	1.184
Friction in 20 cm pipe	2.369
Sudden enlargement	0.731
Friction in 30 cm pipe	1.248
Exit	0.468
Total	6.000

Answer: Discharge $Q \approx 0.214$ m³/s (214 L/s); the losses are as tabulated, adding up to the 6 m head difference.

(If $f = 0.01$ were taken as the Darcy factor, $h_f = fLV^2/2gd$ , the friction terms would be four times smaller and Q would be larger; the coefficient-of-friction reading is the usual one for $f = 0.01$ .)

✓ Answer checked by Claude Opus 5.5 (high effort)

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