specific gravity for a solids slurry of sludge and water Given fraction solids by weight (W_s) = 0.2660; specific gravity of solids (S_s) = 1.9700; specific gravity of water (S_w) = 1.0060, determine the specific gravity of slurry (SG_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except SG_s is given, so isolate SG_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for SG_s:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.2660, specific gravity of solids (S_s) = 1.9700, specific gravity of water (S_w) = 1.0060.

4. Step 4 — Substitute the given values:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$SG_{s} = 1.1565$$

6. Step 6 — Check: returning SG_s = 1.1565 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 2.3131 — kept a factor of two that cancels in the correct rearrangement.
• 0.5783 — dropped that same factor in the other direction.
• 1.2722 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry

specific gravity calculation for a solids slurry from a thickener Given fraction solids by weight (W_s) = 0.1390; specific gravity of water (S_w) = 0.9980; specific gravity of slurry (SG_s) = 1.1430, determine the specific gravity of solids (S_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except S_s is given, so isolate S_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for S_s:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.1390, specific gravity of water (S_w) = 0.9980, specific gravity of slurry (SG_s) = 1.1430.

4. Step 4 — Substitute the given values:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$S_{s} = 11.4260$$

6. Step 6 — Check: returning S_s = 11.4260 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 22.8521 — kept a factor of two that cancels in the correct rearrangement.
• 5.7130 — dropped that same factor in the other direction.
• 12.5687 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry

slurry specific gravity from percent solids and solids specific gravity Given fraction solids by weight (W_s) = 0.0770; specific gravity of solids (S_s) = 1.2400; specific gravity of water (S_w) = 0.9930, determine the specific gravity of slurry (SG_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except SG_s is given, so isolate SG_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for SG_s:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.0770, specific gravity of solids (S_s) = 1.2400, specific gravity of water (S_w) = 0.9930.

4. Step 4 — Substitute the given values:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$SG_{s} = 1.0085$$

6. Step 6 — Check: returning SG_s = 1.0085 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 2.0169 — kept a factor of two that cancels in the correct rearrangement.
• 0.5042 — dropped that same factor in the other direction.
• 1.1093 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry

specific gravity for a solids slurry of sludge and water Given fraction solids by weight (W_s) = 0.2820; specific gravity of water (S_w) = 1.0100; specific gravity of slurry (SG_s) = 1.2440, determine the specific gravity of solids (S_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except S_s is given, so isolate S_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for S_s:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.2820, specific gravity of water (S_w) = 1.0100, specific gravity of slurry (SG_s) = 1.2440.

4. Step 4 — Substitute the given values:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$S_{s} = 3.0333$$

6. Step 6 — Check: returning S_s = 3.0333 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 6.0666 — kept a factor of two that cancels in the correct rearrangement.
• 1.5167 — dropped that same factor in the other direction.
• 3.3367 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry

specific gravity calculation for a solids slurry from a thickener Given fraction solids by weight (W_s) = 0.0840; specific gravity of solids (S_s) = 1.6100; specific gravity of water (S_w) = 0.9910, determine the specific gravity of slurry (SG_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except SG_s is given, so isolate SG_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for SG_s:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.0840, specific gravity of solids (S_s) = 1.6100, specific gravity of water (S_w) = 0.9910.

4. Step 4 — Substitute the given values:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$SG_{s} = 1.0241$$

6. Step 6 — Check: returning SG_s = 1.0241 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 2.0481 — kept a factor of two that cancels in the correct rearrangement.
• 0.5120 — dropped that same factor in the other direction.
• 1.1265 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry

slurry specific gravity from percent solids and solids specific gravity Given fraction solids by weight (W_s) = 0.2730; specific gravity of water (S_w) = 0.9910; specific gravity of slurry (SG_s) = 1.1710, determine the specific gravity of solids (S_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except S_s is given, so isolate S_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for S_s:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.2730, specific gravity of water (S_w) = 0.9910, specific gravity of slurry (SG_s) = 1.1710.

4. Step 4 — Substitute the given values:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$S_{s} = 2.2680$$

6. Step 6 — Check: returning S_s = 2.2680 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 4.5361 — kept a factor of two that cancels in the correct rearrangement.
• 1.1340 — dropped that same factor in the other direction.
• 2.4948 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry

specific gravity for a solids slurry of sludge and water Given fraction solids by weight (W_s) = 0.1830; specific gravity of solids (S_s) = 1.4900; specific gravity of water (S_w) = 1.0050, determine the specific gravity of slurry (SG_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except SG_s is given, so isolate SG_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for SG_s:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.1830, specific gravity of solids (S_s) = 1.4900, specific gravity of water (S_w) = 1.0050.

4. Step 4 — Substitute the given values:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$SG_{s} = 1.0687$$

6. Step 6 — Check: returning SG_s = 1.0687 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 2.1373 — kept a factor of two that cancels in the correct rearrangement.
• 0.5343 — dropped that same factor in the other direction.
• 1.1755 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry

specific gravity calculation for a solids slurry from a thickener Given fraction solids by weight (W_s) = 0.2950; specific gravity of water (S_w) = 0.9970; specific gravity of slurry (SG_s) = 1.2370, determine the specific gravity of solids (S_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except S_s is given, so isolate S_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for S_s:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.2950, specific gravity of water (S_w) = 0.9970, specific gravity of slurry (SG_s) = 1.2370.

4. Step 4 — Substitute the given values:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$S_{s} = 2.9125$$

6. Step 6 — Check: returning S_s = 2.9125 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 5.8251 — kept a factor of two that cancels in the correct rearrangement.
• 1.4563 — dropped that same factor in the other direction.
• 3.2038 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry

slurry specific gravity from percent solids and solids specific gravity Given fraction solids by weight (W_s) = 0.1420; specific gravity of solids (S_s) = 2.4200; specific gravity of water (S_w) = 1.0080, determine the specific gravity of slurry (SG_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except SG_s is given, so isolate SG_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for SG_s:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.1420, specific gravity of solids (S_s) = 2.4200, specific gravity of water (S_w) = 1.0080.

4. Step 4 — Substitute the given values:

   $$SG_{s} = \dfrac{1}{\dfrac{W_s}{S_s}+\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$SG_{s} = 1.0991$$

6. Step 6 — Check: returning SG_s = 1.0991 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 2.1981 — kept a factor of two that cancels in the correct rearrangement.
• 0.5495 — dropped that same factor in the other direction.
• 1.2090 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry

specific gravity for a solids slurry of sludge and water Given fraction solids by weight (W_s) = 0.1270; specific gravity of water (S_w) = 1.0100; specific gravity of slurry (SG_s) = 1.0790, determine the specific gravity of solids (S_s).

Start with the thinking

• The governing relation printed in this handbook section is Specific Gravity for a Solids Slurry.
• Everything except S_s is given, so isolate S_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Specific gravity for a solids slurry combines the specific gravities of solids and water weighted by solids fraction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

2. Step 2 — Rearrange symbolically for S_s:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

3. Step 3 — List the givens: fraction solids by weight (W_s) = 0.1270, specific gravity of water (S_w) = 1.0100, specific gravity of slurry (SG_s) = 1.0790.

4. Step 4 — Substitute the given values:

   $$S_{s} = \dfrac{W_s}{\dfrac{1}{SG_s}-\dfrac{1-W_s}{S_w}}$$

5. Step 5 — Evaluate:

   $$S_{s} = 2.0344$$

6. Step 6 — Check: returning S_s = 2.0344 to

   $$\dfrac{1}{SG_s} = \dfrac{W_s}{S_s} + \dfrac{W_w}{S_w}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 4.0687 — kept a factor of two that cancels in the correct rearrangement.
• 1.0172 — dropped that same factor in the other direction.
• 2.2378 — rounded an intermediate value before the final step.

Reference: FE Handbook — Specific Gravity for a Solids Slurry