2 HOUR HIGH PRESSURE LEAKAGE TEST, WATER MAIN …
2 HOUR HIGH PRESSURE LEAKAGE TEST, WATER MAIN COMPUTATIONS | |
|DATE: |CONTRACT MANAGER: |
|CONTRACT NO.: |INSPECTOR: |
|LOCATION: |LINE & GRADE: |
|CONTRACTOR: |COMPUTED BY: |
|SIZES & TYPES OF WATER MAIN: |CHECKED BY: |
|BOX 1 2 HOUR HIGH PRESSURE TEST |
|High Hydraulic Gradient per contract drawings | | | |
|Minus lowest invert within test section |– | | |
|Total |= | | |
| |X .434 | | |
|Equals the pressure at low point |= | | |
|Lowest Surge Pressure per Standard Specifications |+ | | |
|Total |= PSI | |= Maximum test pressure at lowest invert |
| | | |
|BOX 2 PRESSURE AT LOCATION OF GAUGE FOR 2 HOUR HIGH PRESSURE TEST |
|Difference between lowest invert and gauge |Lowest invert | | | | |
| |Elevation of pressure gauge |– | | | |
| |Difference in elevations |= | |x .434 |= |
| | | | | Pressure Difference |
|Pressure difference plus or minus to maximum test |Maximum test pressure at lowest invert | | | |
|pressure, dependent on location of |(from Box 1) | | |PSI |
|pressure gauge. |+ or – Pressure Difference |+/– | |PSI |
| |Gauge Reading |= | |PSI |
|BOX 3 COMPUTATION FOR ALLOWABLE LEAKAGE USING THE FOLLOWING FORMULA |
|Highest invert | | | | |
|Lowest invert |– | |Maximum test pressure at lowest invert | | |
|Difference |= | |Subtract Pressure Difference |– | |
| |X .434 |Equals Maximum test pressure at highest invert |= | |
|Equals Pressure Difference | | | | |
| | | | | |
|Maximum test pressure at lowest invert | | |PSI | |
|+ Maximum test pressure at highest invert |+ | |PSI | |
| |= | | | |
| |2 |= |PSI |
| | | Average test pressure | |
| | | | | |L = SDvP |
| | | | | |133200 | |
|(S) length of pipe in test section broken | | | | |S = Length of pipe in test section, in|
|down by pipe size(s) | | | | |feet |
| (Multiplied) by (D) diameter of pipe | | | | |D = Nominal diameter of test pipe, in |
| | | | | |inches |
| | | | | | |
|Equals = | | | | | |
|TOTAL (the sum from all of the above row) | |P = Average test pressure, psi |
|Multiplied by square root of average pressure x | |computed by averaging the |
| Equals = | |pressure at the low and high points. |
|(Divided) ÷ |133,200 | |
|Equals MAXIMUM ALLOWABLE Leakage (L) = | |L = Maximum allowable leakage, |
| | |gallons/hour |
| | |2 hour allowable leakage |
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