Formula: (V1 D1) + (V2 D2) = Vf Df where Vi = volume of fluid 1 (bbl, gal, etc.) Di = density of fluid 1 (ppg,lb/ft3, etc.)
V2 = volume of fluid 2 (bbl, gal, etc.) D2 = density of fluid 2 (ppg,lb/ft3, etc.)
VF = volume of final fluid mix DF = density of final fluid mix
Example 1: A limit is placed on the desired volume:
Determine the volume of 11.0 ppg mud and 14.0 ppg mud required to build 300 bbl of 11.5 ppg mud:
Given: 400 bbl of 11.0 ppg mud on hand, and 400 bbl of 14.0 ppg mud on hand
Solution: let V1 = bbl of 11.0 ppg mud V2 = bbl of 14.0 ppg mud then a) V1 + V2 = 300 bbl b) (11.0) V1 + (14.0) V2 = (11.5)(300)
Multiply Equation A by the density of the lowest mud weight (D1 = 11.0 ppg) and subtract the result from Equation B:
b) (11.0) (V1 ) + (14.0) (V2 ) = 3450 — a) (11.0) (VL) + (11.0) (V2_) = 3300 0 (3.0) (V2 ) = 150
Therefore: V2 = 50 bbl of 14.0 ppg mud V1 + V2 = 300 bbl V1 = 300 — 50 V1 = 250 bbl of 11.0 ppg mud
Check: V1 = 50 bbl D1 = 14.0 ppg V2 = 150 bbl D2 = 11.0 ppg VF = 300 bbl Df = final density, ppg
(50) (14.0) + (250) (11.0) = 300 Df 700 + 2750 = 300 Df 3450 = 300 Df 3450 - 300 = Df 11.5 ppg = Df
Example 2: No limit is placed on volume:
Determine the density and volume when the two following muds are mixed together:
Given: 400 bbl of 11.0 ppg mud, and 400 bbl of 14.0 ppg mud
Solution:
Formula:
let Vi = bbl of 11.0 ppg mud V2 = bbl of 14.0 ppg mud VF = final volume, bbl
Di = density of 11.0 ppg mud D2 = density of 14.0 ppg mud Df = final density, ppg
(400) (l1.0) + (400) (l4.0) = 800 Df 4400 + 5600 = 800 Df
Therefore: final volume = 800 bbl final density = 12.5 ppg
4. Oil Based Mud Calculations
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