compressors used to raise natural gas pipeline pressure Given specific heat ratio (k) = 1.3900; gas constant (R) = 291.0 J/kg*K; inlet temperature (T_1) = 296.0 K; discharge pressure (p_2) = 540.0 kPa; inlet pressure (p_1) = 102.0 kPa, determine the specific work (w_s) in kJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Compressors (isentropic work).
• Everything except w_s is given, so isolate w_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.
• Compressors do isentropic work on a gas raising its pressure from inlet to discharge conditions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

2. Step 2 — Rearrange the relation so that w_s stands alone on the left-hand side.

3. Step 3 — List the givens: specific heat ratio (k) = 1.3900, gas constant (R) = 291.0 J/kg*K, inlet temperature (T_1) = 296.0 K, discharge pressure (p_2) = 540.0 kPa, inlet pressure (p_1) = 102.0 kPa.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$w_{s} = 183.0\ \text{kJ/kg}$$

6. Step 6 — Check: returning w_s = 183.0 kJ/kg to

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

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

Why the other options are there

• 366.0 — kept a factor of two that cancels in the correct rearrangement.
• 91.5112 — dropped that same factor in the other direction.
• 201.3 — rounded an intermediate value before the final step.

Reference: FE Handbook — Compressors

a centrifugal blower on a dust-collection duct Given volumetric flow (Q) = 9.8000 m^3/s; pressure rise (\Delta p) = 9,100 Pa; efficiency (\eta) = 0.5700, determine the shaft power (W) in W.

Start with the thinking

• The governing relation printed in this handbook section is Blower / compressor fluid power.
• Everything except W is given, so isolate W 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.
• A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W = \dfrac{Q \Delta p}{\eta}$$

2. Step 2 — Rearrange symbolically for W:

   $$W = \dfrac{Q \Delta p}{\eta}$$

3. Step 3 — List the givens: volumetric flow (Q) = 9.8000 m^3/s, pressure rise (\Delta p) = 9,100 Pa, efficiency (\eta) = 0.5700.

4. Step 4 — Substitute the given values:

   $$W = \dfrac{9.8000 9100}{0.5700}$$

5. Step 5 — Evaluate:

   $$W = 156456\ \text{W}$$

6. Step 6 — Check: returning W = 156,456 W to

   $$W = \dfrac{Q \Delta p}{\eta}$$

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

Why the other options are there

• 312,912 — kept a factor of two that cancels in the correct rearrangement.
• 78,228 — dropped that same factor in the other direction.
• 172,102 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

compressors compressing air in an isentropic process Given specific heat ratio (k) = 1.3100; gas constant (R) = 259.0 J/kg*K; discharge pressure (p_2) = 350.0 kPa; inlet pressure (p_1) = 137.0 kPa; specific work (w_s) = 337.0 kJ/kg, determine the inlet temperature (T_1) in K.

Start with the thinking

• The governing relation printed in this handbook section is Compressors (isentropic work).
• Everything except T_1 is given, so isolate T_1 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.
• Compressors do isentropic work on a gas raising its pressure from inlet to discharge conditions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

2. Step 2 — Rearrange the relation so that T_1 stands alone on the left-hand side.

3. Step 3 — List the givens: specific heat ratio (k) = 1.3100, gas constant (R) = 259.0 J/kg*K, discharge pressure (p_2) = 350.0 kPa, inlet pressure (p_1) = 137.0 kPa, specific work (w_s) = 337.0 kJ/kg.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$T_{1} = 1239\ \text{K}$$

6. Step 6 — Check: returning T_1 = 1,239 K to

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

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

Why the other options are there

• 2,478 — kept a factor of two that cancels in the correct rearrangement.
• 619.5 — dropped that same factor in the other direction.
• 1,363 — rounded an intermediate value before the final step.

Reference: FE Handbook — Compressors

a compressor supplying pneumatic construction tools Given shaft power (W) = 99,537 W; pressure rise (\Delta p) = 36,400 Pa; efficiency (\eta) = 0.6300, determine the volumetric flow (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Blower / compressor fluid power.
• Everything except Q is given, so isolate Q 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.
• A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W = \dfrac{Q \Delta p}{\eta}$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{W \eta}{\Delta p}$$

3. Step 3 — List the givens: shaft power (W) = 99,537 W, pressure rise (\Delta p) = 36,400 Pa, efficiency (\eta) = 0.6300.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{99537 0.6300}{36400}$$

5. Step 5 — Evaluate:

   Q = 1.7228\ \text{m^3/s}

6. Step 6 — Check: returning Q = 1.7228 m^3/s to

   $$W = \dfrac{Q \Delta p}{\eta}$$

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

Why the other options are there

• 3.4455 — kept a factor of two that cancels in the correct rearrangement.
• 0.8614 — dropped that same factor in the other direction.
• 1.8950 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

compressors in a refrigeration cycle raising refrigerant pressure Given specific heat ratio (k) = 1.3100; gas constant (R) = 258.0 J/kg*K; inlet temperature (T_1) = 286.0 K; discharge pressure (p_2) = 420.0 kPa; inlet pressure (p_1) = 143.0 kPa, determine the specific work (w_s) in kJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Compressors (isentropic work).
• Everything except w_s is given, so isolate w_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.
• Compressors do isentropic work on a gas raising its pressure from inlet to discharge conditions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

2. Step 2 — Rearrange the relation so that w_s stands alone on the left-hand side.

3. Step 3 — List the givens: specific heat ratio (k) = 1.3100, gas constant (R) = 258.0 J/kg*K, inlet temperature (T_1) = 286.0 K, discharge pressure (p_2) = 420.0 kPa, inlet pressure (p_1) = 143.0 kPa.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$w_{s} = 90.5537\ \text{kJ/kg}$$

6. Step 6 — Check: returning w_s = 90.5537 kJ/kg to

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

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

Why the other options are there

• 181.1 — kept a factor of two that cancels in the correct rearrangement.
• 45.2769 — dropped that same factor in the other direction.
• 99.6091 — rounded an intermediate value before the final step.

Reference: FE Handbook — Compressors

a positive-displacement blower feeding an aeration basin Given shaft power (W) = 81,133 W; volumetric flow (Q) = 6.7000 m^3/s; pressure rise (\Delta p) = 13,400 Pa, determine the efficiency (\eta).

Start with the thinking

• The governing relation printed in this handbook section is Blower / compressor fluid power.
• 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.
• A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W = \dfrac{Q \Delta p}{\eta}$$

2. Step 2 — Rearrange symbolically for \eta:

   $$\eta = \dfrac{Q \Delta p}{W}$$

3. Step 3 — List the givens: shaft power (W) = 81,133 W, volumetric flow (Q) = 6.7000 m^3/s, pressure rise (\Delta p) = 13,400 Pa.

4. Step 4 — Substitute the given values:

   $$\eta = \dfrac{6.7000 13400}{81133}$$

5. Step 5 — Evaluate:

   $$\eta = 1.1066$$

6. Step 6 — Check: returning \eta = 1.1066 to

   $$W = \dfrac{Q \Delta p}{\eta}$$

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

Why the other options are there

• 2.2132 — kept a factor of two that cancels in the correct rearrangement.
• 0.5533 — dropped that same factor in the other direction.
• 1.2172 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

compressors used to raise natural gas pipeline pressure Given specific heat ratio (k) = 1.3600; gas constant (R) = 294.0 J/kg*K; discharge pressure (p_2) = 460.0 kPa; inlet pressure (p_1) = 95.0000 kPa; specific work (w_s) = 294.0 kJ/kg, determine the inlet temperature (T_1) in K.

Start with the thinking

• The governing relation printed in this handbook section is Compressors (isentropic work).
• Everything except T_1 is given, so isolate T_1 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.
• Compressors do isentropic work on a gas raising its pressure from inlet to discharge conditions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

2. Step 2 — Rearrange the relation so that T_1 stands alone on the left-hand side.

3. Step 3 — List the givens: specific heat ratio (k) = 1.3600, gas constant (R) = 294.0 J/kg*K, discharge pressure (p_2) = 460.0 kPa, inlet pressure (p_1) = 95.0000 kPa, specific work (w_s) = 294.0 kJ/kg.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$T_{1} = 510.8\ \text{K}$$

6. Step 6 — Check: returning T_1 = 510.8 K to

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

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

Why the other options are there

• 1,022 — kept a factor of two that cancels in the correct rearrangement.
• 255.4 — dropped that same factor in the other direction.
• 561.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Compressors

a centrifugal blower on a dust-collection duct Given volumetric flow (Q) = 6.7000 m^3/s; pressure rise (\Delta p) = 41,200 Pa; efficiency (\eta) = 0.8200, determine the shaft power (W) in W.

Start with the thinking

• The governing relation printed in this handbook section is Blower / compressor fluid power.
• Everything except W is given, so isolate W 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.
• A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W = \dfrac{Q \Delta p}{\eta}$$

2. Step 2 — Rearrange symbolically for W:

   $$W = \dfrac{Q \Delta p}{\eta}$$

3. Step 3 — List the givens: volumetric flow (Q) = 6.7000 m^3/s, pressure rise (\Delta p) = 41,200 Pa, efficiency (\eta) = 0.8200.

4. Step 4 — Substitute the given values:

   $$W = \dfrac{6.7000 41200}{0.8200}$$

5. Step 5 — Evaluate:

   $$W = 336634\ \text{W}$$

6. Step 6 — Check: returning W = 336,634 W to

   $$W = \dfrac{Q \Delta p}{\eta}$$

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

Why the other options are there

• 673,268 — kept a factor of two that cancels in the correct rearrangement.
• 168,317 — dropped that same factor in the other direction.
• 370,298 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

compressors compressing air in an isentropic process Given specific heat ratio (k) = 1.3500; gas constant (R) = 279.0 J/kg*K; inlet temperature (T_1) = 303.0 K; discharge pressure (p_2) = 530.0 kPa; inlet pressure (p_1) = 150.0 kPa, determine the specific work (w_s) in kJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Compressors (isentropic work).
• Everything except w_s is given, so isolate w_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.
• Compressors do isentropic work on a gas raising its pressure from inlet to discharge conditions.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

2. Step 2 — Rearrange the relation so that w_s stands alone on the left-hand side.

3. Step 3 — List the givens: specific heat ratio (k) = 1.3500, gas constant (R) = 279.0 J/kg*K, inlet temperature (T_1) = 303.0 K, discharge pressure (p_2) = 530.0 kPa, inlet pressure (p_1) = 150.0 kPa.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$w_{s} = 126.2\ \text{kJ/kg}$$

6. Step 6 — Check: returning w_s = 126.2 kJ/kg to

   $$w_s = \dfrac{k}{k-1} R T_1 \left[ \left(\dfrac{p_2}{p_1}\right)^{(k-1)/k} - 1 \right]$$

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

Why the other options are there

• 252.5 — kept a factor of two that cancels in the correct rearrangement.
• 63.1185 — dropped that same factor in the other direction.
• 138.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Compressors

a compressor supplying pneumatic construction tools Given shaft power (W) = 43,557 W; pressure rise (\Delta p) = 26,300 Pa; efficiency (\eta) = 0.6200, determine the volumetric flow (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Blower / compressor fluid power.
• Everything except Q is given, so isolate Q 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.
• A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W = \dfrac{Q \Delta p}{\eta}$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{W \eta}{\Delta p}$$

3. Step 3 — List the givens: shaft power (W) = 43,557 W, pressure rise (\Delta p) = 26,300 Pa, efficiency (\eta) = 0.6200.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{43557 0.6200}{26300}$$

5. Step 5 — Evaluate:

   Q = 1.0268\ \text{m^3/s}

6. Step 6 — Check: returning Q = 1.0268 m^3/s to

   $$W = \dfrac{Q \Delta p}{\eta}$$

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

Why the other options are there

• 2.0536 — kept a factor of two that cancels in the correct rearrangement.
• 0.5134 — dropped that same factor in the other direction.
• 1.1295 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors