Pipes & Cisterns

Similar to Time & Work, but with a twist: Negative Work. Master Inlet (Filling) vs Outlet (Emptying) logic and 'Leakage' problems.

1. Inlet vs Outlet (Sign Convention)

Direction matters.

  • Inlet (Fill): Positive Work (+).
  • Outlet (Empty): Negative Work (-).
  • Net Rate: Rate(In) - Rate(Out).

Example:

Q: A fills in 10h, B empties in 15h. Together?
Solution: Total=30. A=+3, B=-2. Net=+1. Time = 30/1 = 30h.

2. Leak at Bottom

If a leak increases filling time.

  • Method: Find Leak Efficiency by subtraction.
  • Eff(Leak)=Eff(Net)тИТEff(Pipe)Eff(Leak) = Eff(Net) - Eff(Pipe). Usually Net is slower.

Example:

Q: Pipe A fills in 4h. Due to leak, takes 5h.
Solution: TW=20. A=+5. Net(A-L)=+4. Why? Slower.
5тИТL=4тАЕтАКтЯ╣тАЕтАКL=15 - L = 4 \implies L = 1. Leak Time = 20/1 = 20h.

3. Capacity Calculation

To find Total Liters.

  • Logic: Capacity = Rate of Pipe (L/min) ├Ч\times Time taken by THAT Pipe (min).
  • First find Time of Pipe X using Ratios, then multiply by its tangible rate.

Example:

Q: Leak empties in 10h. Inlet adds 4L/min. Full tank empties in 15h.
Solution: Leak=-3, Net=-2. Inlet=+1. Inlet Time=30h. Cap=30├Ч60├Ч4=7200L\text{Cap} = 30\times60\times4 = 7200L.

4. Alternating Pipes

Positive and Negative cycles.

  • Trap: Tank fills only at the END of a positive cycle. Do not divide directly.
  • Hack: Subtract last positive work from Total. Divide rest by Cycle. Then add last step manually.

Example:

Q: A(+3) 1h, B(-2) 1h. TW=10.
Solution: Cycle(2h) = +1. Target 10тИТ3=710-3 = 7.
7 cycles (14h) тЖТ\to 7 filled.
Next hour A adds 3 тЖТ\to Full. Total 15h.

5. Multiple Pipes Logic

Complex combinations.

  • Rule: Sum all efficiencies algebraically.
  • If Net Eff is negative, tank will EMPTY (Total Work = Initial Fills).

Example:

Q: A(+5), B(+4), C(-10). Full tank. Open all.
Solution: Net = 5+4-10 = -1.
Time = Capacity / 1. Tank empties.

6. Time Lag (Opening Differently)

Pipes open at different times.

  • Method: Calculate partial work done before all open.
  • Subtract from Total. Solve rest with Combined Efficiency.

7. Diameter & Flow Rule

Flow rate depends on pipe Cross-Section Area.

  • Rule: Rate тИЭd2\propto d^2 (square of diameter).
  • If diameter doubles (dтЖТ2dd \to 2d), Flow becomes 4x4x. Time becomes 1/41/4.

Example:

Q: Pipe A (d=1d=1) takes 60m. Pipe B (d=3d=3). Time?
Solution: Dia Ratio 1:3. Area Ratio 1:9. Flow Ratio 1:9.
Time Ratio 9:1.
Time B = 60/9=6.6660/9 = 6.66 min.