Pycnometer Method for Water Content
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Method of Determining Water Content
1. Oven Drying Method
2. Sand Bath Method
3. Alcohol Method
4. Pycnometer
5. Calcium Carbide Method
6. Radiation Method
7. Torsional Method
Description of this video :
1. SOIL MECHANICS Determination of Water Content by Pycnometer Method
2. For the whole course, visit our website,… This is the excerpt from Lecture 04 of Soil Mechanics www.machenlink.com
3. 4. Pycnometer Method L. 04 Determination of Water Content 900 ml 6 mm Procedure : • It is a quick method in which result is obtained between 10 to 20 minutes. This method is used for only those soil whose specific gravity is known. • Pycnometera is 900 ml with a conical brass top with a 6mm diameter hole at its centre.
4. L. 04 Determination of Water Content • Series of observations to compute water content are as folloWeightweight of the Pycnometer is observed initially (𝒘 𝟏 ). ▪ 200-400 gm moist sample (𝒘 𝟐 )Pycnometerometer. ▪ Empty voluPycnometerometer is filled with water along with simultaneous removal of air present in soil either by constant stirring or by use of vacuum at the tPycnometerometPycnometerometer is being Weightweight after filling the empty volume with water (𝒘 𝟑Pycnometerometer is then completely emptied and again filled with water after cleaning properly (𝒘 𝟒 ). 𝑤1 𝑤2 Moist soil 𝑤3 𝑤4
5. Weight of solid in (b) = WeightWeight of moist soPycnometerometer = 𝑤2 − 𝑤1 weight of water in the moist soil sample 𝑤 𝑤 Weight of water 𝒘 𝒘 = 𝑤 𝑑 𝐺. 𝛾 𝑤 = 𝑤 𝑑 𝛾𝑠 Volume of solid (𝑉𝑑) Volume of water equivalent to volume of solid = 𝑤 𝑑 𝐺. 𝛾 𝑤 Weight of Solid 𝒘 𝒅 = 𝑤2 − 𝑤1 − 𝑤 𝑑 𝐺 = 𝛾𝑠 𝛾 𝑤 𝛾𝑠 = 𝑤 𝑑 𝑉𝑑 𝑤 = 𝑤 𝑤 𝑤 𝑑 ×1Pycnometerometer is being Weightweight after filling the empty volume with water (𝒘 𝟑Pycnometerometer is then completely emptied and again filled with water after cleaning properly (𝒘 𝟒 ). 𝑤1 𝑤2 Moist soil 𝑤3 𝑤4 L. 04 Determination of Water Content (a) (b) (c) (d) 𝛾 𝑑 = 𝑤 𝑑 𝑉
6. 𝑤3 − 𝑤4 = 𝑤 𝑑 𝐺 − 1 𝐺 ⇒ 𝑤 𝑑 = (𝑤3 − 𝑤4) 𝐺 𝐺 − 1 𝑤 𝑤 𝑤 𝑑 𝑤 = × 100 % (𝑤2− 𝑤1) − 𝑤 𝑑 𝑤 𝑑 × 100 %= 𝑤1 𝑤2 Moist soil 𝑤3 𝑤4 (a) (b) (c) (d) If the solids from (c) are replaced with water of equivWeightweight We geWeightweight (𝑤4) from (d). 𝑤4 = 𝑤3 − 𝑤 𝑑 + 𝑤 𝑑 𝐺 𝑤4 = 𝑤3 − 𝑤 𝑑 1 − 1 𝐺 ⇒ = 𝑤 𝑑 𝐺. 𝛾 𝑤 × 𝛾 𝑤 = 𝑤 𝑑 𝐺 EquivWeightweight of water 𝑤 𝑑 𝐺 L. 04 Determination of Water Content
7. 𝑤 𝑤 𝑤 𝑑 𝑤 = × 100 % (𝑤2− 𝑤1) − 𝑤 𝑑 𝑤 𝑑 × 100 %= 𝑤3 − 𝑤4 = 𝑤 𝑑 𝐺 − 1 𝐺 ⇒ 𝑤 𝑑 = (𝑤3 − 𝑤4) 𝐺 𝐺 − 1 Let’Weightweight of solid in (b) = WeightWeight of moist soPycnometerometer = 𝑤2 Weightweight of water in the moist soil sample WeightWeight of water 𝒘 𝒘 = 𝑤 𝑑 𝐺. 𝛾 𝑤 = 𝑤 𝑑 𝛾𝑠 Volume of solid (𝑉𝑑) Volume of water equivalent to volume of solid = 𝑤 𝑑 𝐺. 𝛾 𝑤 Weight of Solid 𝒘 𝒅 = 𝑤2 − 𝑤1 − 𝑤 𝑑 𝑤 = 𝑤 𝑤 𝑤 𝑑 ×100% 𝐺 = 𝛾𝑠 𝛾 𝑤 𝛾𝑠 = 𝑤 𝑑 𝑉𝑑 𝑤1 𝑤2 Moist soil 𝑤3 𝑤4 (a) (b) (c) (d) L. 04 Determination of Water Content If the solids from (c) are replaced with water of equivWeightweight We geWeightweight (𝑤4) from (d). 𝑤4 = 𝑤3 − 𝑤 𝑑 + 𝑤 𝑑 𝐺 𝑤4 = 𝑤3 − 𝑤 𝑑 1 − 1 𝐺 ⇒ = 𝑤 𝑑 𝐺. 𝛾 𝑤 × 𝛾 𝑤 = 𝑤 𝑑 𝐺 Equivalent weight of water 𝑤 𝑑 𝐺 𝑤 = 𝑤2− 𝑤1 𝑤3− 𝑤4 𝐺 − 1 𝐺 − 1 × 100%
8. In order to determine the water content, 370 g of a wet sandy sample was placed in a pycnometerWeightweight oPycnometerometer, sand and water packed to the top of the conical cap was found to be 2148 gWeightweigPycnometerometer full of clean water was 1932 g. Taking G = 2.65, determine the water content of the sample. 04. Example 04 (Water content of moist soil sample using Pycnometer)L. 04 Strategy: Solution: 𝑤 = 𝑤2 − 𝑤1 (𝑤3 − 𝑤4) × 𝐺 − 1 𝐺 − 1 × 100 Step 1: Weight of soil sample is already given separately. Use the formula of water content. Write the given data Weight of soil sample = 𝑤2 − 𝑤1 = 370 gm WeigPycnometerometer, sand and water (𝑤3) = 2148 gm WeigPycnometerometer and water (𝑤4) = 1932 gm
9. Solution: 𝑤 = 𝑤2 − 𝑤1 (𝑤3 − 𝑤4) × 𝐺 − 1 𝐺 − 1 × 100 Step 1:Write the given data Weight of soil sample = 𝑤2 − 𝑤1 = 370 gm WeigPycnometerometer, sand and water (𝑤3) = 2148 gm WeigPycnometerometer and water (𝑤4) = 1932 gm Example 04 (Water content of moist soil sample using Pycnometer)L. 04 𝑤 = 370 2148 − 1932 × 2.65 − 1 2.65 − 1 × 100 w = 6.5 %