biomass as an energy source for a waste-to-energy power plant Given biomass mass processed (m) = 17,750 kg/day; higher heating value of biomass (HHV) = 10,180 kJ/kg; conversion efficiency (eta) = 0.5000, determine the usable energy output (E) in kJ/day.

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except E is given, so isolate E 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for E:

   $$E = m\,HHV\,\eta$$

3. Step 3 — List the givens: biomass mass processed (m) = 17,750 kg/day, higher heating value of biomass (HHV) = 10,180 kJ/kg, conversion efficiency (eta) = 0.5000.

4. Step 4 — Substitute the given values:

   $$E = 17750\,10180\,0.5000$$

5. Step 5 — Evaluate:

   $$E = 90347500\ \text{kJ/day}$$

6. Step 6 — Check: returning E = 90,347,500 kJ/day to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 180,695,000 — kept a factor of two that cancels in the correct rearrangement.
• 45,173,750 — dropped that same factor in the other direction.
• 99,382,250 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source

biomass as an energy source conversion efficiency calculation Given higher heating value of biomass (HHV) = 13,920 kJ/kg; conversion efficiency (eta) = 0.4700; usable energy output (E) = 529,056,000 kJ/day, determine the biomass mass processed (m) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except m is given, so isolate m 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for m:

   $$m = \dfrac{E}{HHV\,\eta}$$

3. Step 3 — List the givens: higher heating value of biomass (HHV) = 13,920 kJ/kg, conversion efficiency (eta) = 0.4700, usable energy output (E) = 529,056,000 kJ/day.

4. Step 4 — Substitute the given values:

   $$m = \dfrac{529056000}{13920\,0.4700}$$

5. Step 5 — Evaluate:

   $$m = 80866\ \text{kg/day}$$

6. Step 6 — Check: returning m = 80,866 kg/day to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 161,731 — kept a factor of two that cancels in the correct rearrangement.
• 40,433 — dropped that same factor in the other direction.
• 88,952 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source

biomass as an energy source from agricultural waste feedstock Given biomass mass processed (m) = 14,860 kg/day; higher heating value of biomass (HHV) = 19,540 kJ/kg; usable energy output (E) = 778,023,000 kJ/day, determine the conversion efficiency (eta).

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except eta is given, so isolate eta 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for eta:

   $$\eta = \dfrac{E}{m\,HHV}$$

3. Step 3 — List the givens: biomass mass processed (m) = 14,860 kg/day, higher heating value of biomass (HHV) = 19,540 kJ/kg, usable energy output (E) = 778,023,000 kJ/day.

4. Step 4 — Substitute the given values:

   $$\eta = \dfrac{778023000}{14860\,19540}$$

5. Step 5 — Evaluate:

   $$\eta = 2.6795$$

6. Step 6 — Check: returning eta = 2.6795 to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 5.3589 — kept a factor of two that cancels in the correct rearrangement.
• 1.3397 — dropped that same factor in the other direction.
• 2.9474 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source

biomass as an energy source for a waste-to-energy power plant Given biomass mass processed (m) = 90,200 kg/day; higher heating value of biomass (HHV) = 10,120 kJ/kg; conversion efficiency (eta) = 0.5400, determine the usable energy output (E) in kJ/day.

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except E is given, so isolate E 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for E:

   $$E = m\,HHV\,\eta$$

3. Step 3 — List the givens: biomass mass processed (m) = 90,200 kg/day, higher heating value of biomass (HHV) = 10,120 kJ/kg, conversion efficiency (eta) = 0.5400.

4. Step 4 — Substitute the given values:

   $$E = 90200\,10120\,0.5400$$

5. Step 5 — Evaluate:

   $$E = 492924960\ \text{kJ/day}$$

6. Step 6 — Check: returning E = 492,924,960 kJ/day to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 985,849,920 — kept a factor of two that cancels in the correct rearrangement.
• 246,462,480 — dropped that same factor in the other direction.
• 542,217,456 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source

biomass as an energy source conversion efficiency calculation Given higher heating value of biomass (HHV) = 10,990 kJ/kg; conversion efficiency (eta) = 0.2600; usable energy output (E) = 160,678,000 kJ/day, determine the biomass mass processed (m) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except m is given, so isolate m 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for m:

   $$m = \dfrac{E}{HHV\,\eta}$$

3. Step 3 — List the givens: higher heating value of biomass (HHV) = 10,990 kJ/kg, conversion efficiency (eta) = 0.2600, usable energy output (E) = 160,678,000 kJ/day.

4. Step 4 — Substitute the given values:

   $$m = \dfrac{160678000}{10990\,0.2600}$$

5. Step 5 — Evaluate:

   $$m = 56232\ \text{kg/day}$$

6. Step 6 — Check: returning m = 56,232 kg/day to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 112,464 — kept a factor of two that cancels in the correct rearrangement.
• 28,116 — dropped that same factor in the other direction.
• 61,855 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source

biomass as an energy source from agricultural waste feedstock Given biomass mass processed (m) = 75,350 kg/day; higher heating value of biomass (HHV) = 10,380 kJ/kg; usable energy output (E) = 927,281,000 kJ/day, determine the conversion efficiency (eta).

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except eta is given, so isolate eta 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for eta:

   $$\eta = \dfrac{E}{m\,HHV}$$

3. Step 3 — List the givens: biomass mass processed (m) = 75,350 kg/day, higher heating value of biomass (HHV) = 10,380 kJ/kg, usable energy output (E) = 927,281,000 kJ/day.

4. Step 4 — Substitute the given values:

   $$\eta = \dfrac{927281000}{75350\,10380}$$

5. Step 5 — Evaluate:

   $$\eta = 1.1856$$

6. Step 6 — Check: returning eta = 1.1856 to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 2.3712 — kept a factor of two that cancels in the correct rearrangement.
• 0.5928 — dropped that same factor in the other direction.
• 1.3041 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source

biomass as an energy source for a waste-to-energy power plant Given biomass mass processed (m) = 90,310 kg/day; higher heating value of biomass (HHV) = 12,590 kJ/kg; conversion efficiency (eta) = 0.4300, determine the usable energy output (E) in kJ/day.

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except E is given, so isolate E 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for E:

   $$E = m\,HHV\,\eta$$

3. Step 3 — List the givens: biomass mass processed (m) = 90,310 kg/day, higher heating value of biomass (HHV) = 12,590 kJ/kg, conversion efficiency (eta) = 0.4300.

4. Step 4 — Substitute the given values:

   $$E = 90310\,12590\,0.4300$$

5. Step 5 — Evaluate:

   $$E = 488911247\ \text{kJ/day}$$

6. Step 6 — Check: returning E = 488,911,247 kJ/day to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 977,822,494 — kept a factor of two that cancels in the correct rearrangement.
• 244,455,624 — dropped that same factor in the other direction.
• 537,802,372 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source

biomass as an energy source conversion efficiency calculation Given higher heating value of biomass (HHV) = 12,460 kJ/kg; conversion efficiency (eta) = 0.3300; usable energy output (E) = 768,394,000 kJ/day, determine the biomass mass processed (m) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except m is given, so isolate m 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for m:

   $$m = \dfrac{E}{HHV\,\eta}$$

3. Step 3 — List the givens: higher heating value of biomass (HHV) = 12,460 kJ/kg, conversion efficiency (eta) = 0.3300, usable energy output (E) = 768,394,000 kJ/day.

4. Step 4 — Substitute the given values:

   $$m = \dfrac{768394000}{12460\,0.3300}$$

5. Step 5 — Evaluate:

   $$m = 186875\ \text{kg/day}$$

6. Step 6 — Check: returning m = 186,875 kg/day to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 373,751 — kept a factor of two that cancels in the correct rearrangement.
• 93,438 — dropped that same factor in the other direction.
• 205,563 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source

biomass as an energy source from agricultural waste feedstock Given biomass mass processed (m) = 22,820 kg/day; higher heating value of biomass (HHV) = 11,500 kJ/kg; usable energy output (E) = 236,882,000 kJ/day, determine the conversion efficiency (eta).

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except eta is given, so isolate eta 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for eta:

   $$\eta = \dfrac{E}{m\,HHV}$$

3. Step 3 — List the givens: biomass mass processed (m) = 22,820 kg/day, higher heating value of biomass (HHV) = 11,500 kJ/kg, usable energy output (E) = 236,882,000 kJ/day.

4. Step 4 — Substitute the given values:

   $$\eta = \dfrac{236882000}{22820\,11500}$$

5. Step 5 — Evaluate:

   $$\eta = 0.9026$$

6. Step 6 — Check: returning eta = 0.9026 to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 1.8053 — kept a factor of two that cancels in the correct rearrangement.
• 0.4513 — dropped that same factor in the other direction.
• 0.9929 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source

biomass as an energy source for a waste-to-energy power plant Given biomass mass processed (m) = 3,000 kg/day; higher heating value of biomass (HHV) = 15,810 kJ/kg; conversion efficiency (eta) = 0.2500, determine the usable energy output (E) in kJ/day.

Start with the thinking

• The governing relation printed in this handbook section is Biomass as an Energy Source.
• Everything except E is given, so isolate E 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.
• Biomass as an energy source converts the chemical energy of organic feedstock into usable energy at a given conversion efficiency.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = m \times HHV \times \eta$$

2. Step 2 — Rearrange symbolically for E:

   $$E = m\,HHV\,\eta$$

3. Step 3 — List the givens: biomass mass processed (m) = 3,000 kg/day, higher heating value of biomass (HHV) = 15,810 kJ/kg, conversion efficiency (eta) = 0.2500.

4. Step 4 — Substitute the given values:

   $$E = 3000\,15810\,0.2500$$

5. Step 5 — Evaluate:

   $$E = 11857500\ \text{kJ/day}$$

6. Step 6 — Check: returning E = 11,857,500 kJ/day to

   $$E = m \times HHV \times \eta$$

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

Why the other options are there

• 23,715,000 — kept a factor of two that cancels in the correct rearrangement.
• 5,928,750 — dropped that same factor in the other direction.
• 13,043,250 — rounded an intermediate value before the final step.

Reference: FE Handbook — Biomass as an Energy Source