DEBUG_INFO: title=Pipes & Cisterns, type=object, isArray=, length=7
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.
If a leak increases filling time.
- Method: Find Leak Efficiency by subtraction.
- Eff(Leak)=Eff(Net)тИТEff(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=4тЯ╣L=1. Leak Time = 20/1 = 20h. To find Total Liters.
- Logic: Capacity = Rate of Pipe (L/min) ├Ч 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. 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=7.
7 cycles (14h) тЖТ 7 filled.
Next hour A adds 3 тЖТ Full. Total 15h. 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.
Pipes open at different times.
- Method: Calculate partial work done before all open.
- Subtract from Total. Solve rest with Combined Efficiency.
Flow rate depends on pipe Cross-Section Area.
- Rule: Rate тИЭd2 (square of diameter).
- If diameter doubles (dтЖТ2d), Flow becomes 4x. Time becomes 1/4.
Example:
Q: Pipe A (d=1) takes 60m. Pipe B (d=3). Time? Solution: Dia Ratio 1:3. Area Ratio 1:9. Flow Ratio 1:9.
Time Ratio 9:1.
Time B = 60/9=6.66 min.