6.3 Inlet and Outlet Pipes, Net Filling Rate, Alternating Flow & Tank Leakage Problems

Idealized constant-flow assumptions

These exercises assume constant flow rates independent of water level, additive flows, and no overflow or evaporation. Fill-time formulas start with an empty tank; emptying formulas start with a full one. Equal inlet and drain rates leave the level unchanged. In the alternating-flow diagram, $E_B$ denotes a positive removal magnitude; signed drain rates in the worked equations are negative. If the inlet fills the tank in its first turn, stop there. Otherwise the peak-threshold formula requires positive net cycle progress.

The Hydraulic Analogy to Time & Work

The mathematical framework governing pipes and cisterns mirrors mechanical work, with one critical distinction: pipes possess directional flow.

$$\text{Total Tank Capacity (Volume)} \equiv \text{Total Work } (W)$$ $$\text{Volumetric Flow Rate } (R) \equiv \text{Efficiency } (E)$$ $$\text{Filling or Emptying Duration } (T) \equiv \text{Time Taken } (T)$$ $$\text{Capacity} = \text{Flow Rate} \times \text{Time} \iff W = E \times T$$

Directional Flow Classifications

1. Inlet Pipe (Filling Pipe): Introduces fluid into the cistern. Represents positive work ($+R$). If an inlet fills a cistern in $x$ hours, its rate is $+\frac{1}{x}$ of tank capacity per hour.
2. Outlet Pipe (Waste Pipe / Drain): Evacuates fluid from the cistern. Represents negative work ($-R$). If an outlet empties a full cistern in $y$ hours, its rate is $-\frac{1}{y}$ of tank capacity per hour.

+-----------------------------------------------------------------------------+
|                     INLET AND OUTLET RATE SUPERPOSITION                     |
+-----------------------------------------------------------------------------+
| Inlet Pipe A (Fills in x hours):   Rate = +1/x of tank volume per hour      |
| Outlet Pipe B (Empties in y hours): Rate = -1/y of tank volume per hour     |
|                               │                                             |
|                               ▼                                             |
| Both Pipes Open Simultaneously:                                             |
| Net Flow Rate = (+1/x) + (-1/y) = (y - x) / (xy)                            |
|                               │                                             |
|          ┌────────────────────┴────────────────────┐                        |
|          ▼ (If y > x: Inlet is Faster)             ▼ (If x > y: Drain Faster)│
| Net Rate > 0 -> TANK FILLS                         Net Rate < 0 -> TANK EMPTIES|
| Time to Fill = (x * y) / (y - x)                   Time to Empty = (x * y)/(x - y)|
+-----------------------------------------------------------------------------+

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The Tank Leakage Problem

A universal staple of railway quantitative exams involves a cistern that fills in normal duration $T$, but takes an extended duration $T'$ ($T' > T$) due to a leak or structural fissure at the bottom.

Mathematical Derivation

Let the normal inlet pump rate be $R_{\text{in}} = \frac{1}{T}$, and the leak's emptying rate be $R_{\text{leak}} = \frac{1}{L}$. When both operate simultaneously, the net filling rate is:

$$R_{\text{net}} = R_{\text{in}} - R_{\text{leak}} = \frac{1}{T'}$$ $$\frac{1}{T} - \frac{1}{L} = \frac{1}{T'} \implies \frac{1}{L} = \frac{1}{T} - \frac{1}{T'} = \frac{T' - T}{T \cdot T'}$$

Inverting to find the time $L$ taken by the leak alone to empty a completely full tank:

$$T_{\text{leak}} = \frac{T \cdot T'}{T' - T}$$

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Staged Multi-Pipe Valve Operations & Premature Closures

In yard watering plants, multiple pumps operate concurrently, but one or more valves may be shut down before the tank is completely full to prevent overflow or conserve power.

The Problem Structure

Pipes $A$ and $B$ can fill a tank in $x$ and $y$ minutes respectively. Both pipes are opened simultaneously, but Pipe $A$ is closed $t$ minutes before the tank is full. What is the total time $T$ required to fill the tank?

• Method 1: Direct Work Equation: Pipe $A$ operates for $(T - t)$ minutes; Pipe $B$ operates for the entire $T$ minutes.

$$(T - t) \cdot E_A + T \cdot E_B = \text{Total Capacity } W$$ $$T(E_A + E_B) - t \cdot E_A = W \implies T = \frac{W + t \cdot E_A}{E_A + E_B}$$

• Method 2: Virtual Volume Shortcut: If Pipe $A$ had remained open for those final $t$ minutes, it would have added an extra virtual volume of $t \times E_A$ units. Add this virtual volume to the total tank capacity and divide by their combined rate:

$$T = \frac{\text{Capacity} + (t \times E_A)}{E_A + E_B}$$

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Alternating Pipe Flow & The Peak Threshold Protection

When inlet and drain pipes operate alternately (e.g., Inlet $A$ open in Minute 1, Drain $B$ open in Minute 2), the system behaves identically to the negative work dynamics covered in Section 6.2.

+-----------------------------------------------------------------------------+
|                     ALTERNATING FLOW: THE PEAK THRESHOLD                    |
+-----------------------------------------------------------------------------+
| Inlet A fills in x min (Rate = +E_A) | Drain B empties in y min (Rate = -E_B) |
| Net cycle yield = E_A - E_B units every 2 minutes                           |
|                               │                                             |
|                               ▼                                             |
| CRITICAL RULE: Cistern is FULL the moment water reaches Capacity (C) on    |
| the inlet's turn! Drain B never operates after the tank is filled.          |
|                               │                                             |
|                               ▼                                             |
| 1. Reserve Inlet's single stride: Threshold = C - E_A                       |
| 2. Full 2-min cycles to threshold: k = ceil[(C - E_A) / (E_A - E_B)]         |
| 3. Final fractional minute on inlet turn: dt = (C - Work_k) / E_A           |
+-----------------------------------------------------------------------------+

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Absolute Tank Capacity Determination from Volumetric Telemetry

A high-yield question type on RRB exams asks for the absolute volumetric capacity of the tank in liters. These problems bridge fractional unit rates with physical flow rates given in liters per minute (LPM) or cubic meters per hour ($m^3/h$).

The Systematic Three-Step Solution Framework

1. Step 1: Isolate the Fractional Rate of the Calibrated Pipe: Use the LCM or reciprocal method to calculate the theoretical fractional rate ($R_k = 1/T_k$) of the specific pipe whose absolute physical discharge rate is provided.
2. Step 2: Determine Theoretical Time to Evacuate/Fill Alone: Compute the total time $T_k$ (in minutes or hours) that this specific pipe would take to fill or empty the entire tank alone.
3. Step 3: Multiply Calibrated Discharge Rate by Total Time:

$$\text{Absolute Tank Capacity (Liters)} = \text{Discharge Rate (Liters/minute)} \times T_k \text{ (minutes)}$$

Example: An emptying pipe discharges $40\text{ L/min}$. If calculations prove this emptying pipe alone would evacuate the full tank in $25\text{ minutes}$, the absolute tank capacity is:

$$\text{Capacity} = 40 \frac{\text{liters}}{\text{minute}} \times 25 \text{ minutes} = 1,000 \text{ liters}$$

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Worked Examples

Worked Example 1: Multi-Pump Locomotive Fueling Siding with Staged Closures

Problem: A diesel locomotive fueling siding has three supply lines connected to an underground fuel storage reservoir. Line A can fill the reservoir in 20 hours, Line B in 30 hours, and Line C in 60 hours. All three supply lines are opened simultaneously. However, due to fuel manifold maintenance, Line C is shut off 4 hours before the reservoir is completely filled, and Line B is shut off 2 hours before the reservoir is filled. Line A operates continuously until the reservoir is completely full. What is the total time taken to fill the fuel reservoir?

• Step 1: Determine Total Capacity Units and Pipe Efficiencies:

$$\text{Total Reservoir Capacity } W = \text{LCM}(20, 30, 60) = 60 \text{ units}$$ $$E_A = \frac{60}{20} = 3 \text{ units/hour}, \quad E_B = \frac{60}{30} = 2 \text{ units/hour}, \quad E_C = \frac{60}{60} = 1 \text{ unit/hour}$$

• Step 2: Formulate the Total Time Equation: Let $T$ be the total operating duration in hours.
  • Line A operates for the full duration: $t_A = T$ hours.
  • Line B operates for $t_B = (T - 2)$ hours.
  • Line C operates for $t_C = (T - 4)$ hours.

$$3(T) + 2(T - 2) + 1(T - 4) = 60$$

• Step 3: Solve Algebraically for $T$:

$$3T + 2T - 4 + T - 4 = 60$$ $$6T - 8 = 60 \implies 6T = 68 \implies T = \frac{68}{6} = \frac{34}{3} = 11\frac{1}{3} \text{ hours}$$

• Step 4: Verify via Virtual Volume Shortcut:

$$\text{Virtual Volume Added} = (2 \text{ hrs} \times E_B) + (4 \text{ hrs} \times E_C) = (2 \times 2) + (4 \times 1) = 4 + 4 = 8 \text{ units}$$ $$\text{Total Virtual Capacity} = 60 + 8 = 68 \text{ units}$$ $$\text{Combined Hourly Rate} = E_A + E_B + E_C = 3 + 2 + 1 = 6 \text{ units/hour}$$ $$T = \frac{68}{6} = 11\frac{1}{3} \text{ hours (11 hours 20 minutes)}$$

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Worked Example 2: Overhead Water Tank Leakage & Capacity Derivation

Problem: An overhead coaching service reservoir fills in 6 hours via an electric pump under normal conditions. However, due to an undetected foundation leak at the bottom, it takes 8 hours to fill completely. (a) In how many hours will the leak alone evacuate the entire full reservoir? (b) If an auxiliary washout drainage tap discharging water at 45 liters per minute is opened simultaneously with the normal inlet pump and the active leak, the full reservoir empties in 24 hours. What is the total volumetric capacity of the reservoir in liters?

• Step 1: Calculate Emptying Time of Leak Alone: Using the leakage shortcut:

$$T_{\text{leak}} = \frac{T \cdot T'}{T' - T} = \frac{6 \times 8}{8 - 6} = \frac{48}{2} = 24 \text{ hours}$$

The leak alone empties the reservoir in 24 hours (emptying-rate magnitude $R_{\text{leak}} = \frac{1}{24}$ reservoir/hr).

• Step 2: Determine the Net Emptying Rate with Auxiliary Tap Open: Let $R_{\text{tap}} = \frac{1}{H}$ be the emptying rate of the auxiliary drainage tap. Under all three components, the full reservoir empties in 24 hours (net rate $= -\frac{1}{24}$):

$$R_{\text{in}} - R_{\text{leak}} - R_{\text{tap}} = -\frac{1}{24}$$ $$\frac{1}{6} - \frac{1}{24} - \frac{1}{H} = -\frac{1}{24}$$ $$\frac{1}{6} - \frac{1}{H} = 0 \implies \frac{1}{H} = \frac{1}{6}$$

Therefore, the auxiliary washout tap alone would completely empty the full reservoir in 6 hours.

• Step 3: Convert Time to Minutes and Calculate Absolute Capacity:

$$T_{\text{tap}} = 6 \text{ hours} = 6 \times 60 = 360 \text{ minutes}$$ $$\text{Total Capacity} = \text{Discharge Rate} \times T_{\text{tap}} = 45 \frac{\text{liters}}{\text{minute}} \times 360 \text{ minutes} = 16,200 \text{ liters}$$

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Worked Example 3: Alternating Filling and Drainage Flow with Peak Threshold

Problem: A washing plant reservoir has an inlet pipe A that can fill it in 12 minutes, and a drain pipe B that can empty it in 18 minutes. If pipes A and B are opened alternately for 1 minute each, starting with Inlet Pipe A, in exactly how many minutes will the empty reservoir be completely filled for the first time?

• Step 1: Compute Total Work Units and Minute Rates:

$$\text{Total Capacity } C = \text{LCM}(12, 18) = 36 \text{ units}$$ $$E_A = +\frac{36}{12} = +3 \text{ units/min}, \quad E_B = -\frac{36}{18} = -2 \text{ units/min}$$

• Step 2: Determine Net Progress per 2-Minute Cycle:

$$\text{Work in Minute 1 (Pipe A)} = +3 \text{ units}$$ $$\text{Work in Minute 2 (Pipe B)} = -2 \text{ units}$$ $$\text{Net Yield per 2-minute cycle} = +3 - 2 = +1 \text{ unit every 2 minutes}$$

• Step 3: Establish the Peak Threshold Barrier: Deduct Pipe A's full single-minute stride from total capacity:

$$\text{Threshold Target} = C - E_A = 36 - 3 = 33 \text{ units}$$

• Step 4: Compute Cycles to Reach Threshold:

$$k = \frac{33 \text{ units}}{1 \text{ unit/cycle}} = 33 \text{ complete cycles}$$

• Time elapsed for 33 cycles $= 33 \times 2 = 66 \text{ minutes}$.
• Water level accumulated $= 33 \times 1 = 33 \text{ units}$.
• Step 5: Final Advance in Minute 67: At minute 67, Pipe A opens:

$$\text{Water Added} = +3 \text{ units}$$ $$\text{Cumulative Volume} = 33 + 3 = 36 \text{ units (100\% FULL!)}$$

The reservoir reaches 100% capacity at the end of the 67th minute. Pipe B never operates.

• Conclusion: The reservoir is completely filled in 67 minutes (1 hour 7 minutes).

(Note: Naive division of $36 / 1 \times 2 = 72 \text{ minutes}$ overshoots the true answer by 5 full minutes!)

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Calculation Traps & Tactical Shortcuts

1. Sign Confusion in Net Flow Equations: When an outlet pipe is active, its rate must enter the equation with a minus sign ($-1/y$). Adding drain rates instead of subtracting them is a common blunder under time pressure.
2. The Full Tank Emptying Distinction: Ensure you read whether the problem asks for the time to fill an empty tank or to empty a full tank. When drain rate exceeds inlet rate ($x > y$), the tank will empty, and net time is $\frac{xy}{x - y}$.
3. Unit Consistency in Discharge Conversions: If discharge is given in liters per minute and filling times are given in hours, immediately convert the filling time into minutes ($T \times 60$) before multiplying by discharge rate.