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Ultrafiltration

Environmental Engineering · FE Reference Handbook section

Environmental Engineering
7 formulas
10 exam-style examples
~59 min
All Environmental Engineering lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Ultrafiltration — solve for ultrafiltration permeate flux — Ultrafiltration

ultrafiltration membrane permeate flux for drinking water treatment Given transmembrane pressure (deltaP) = 57.0000 kPa; dynamic viscosity (mu) = 0.0011 Pa*s; membrane resistance (R_m) = 910,000,000,000 1/m, determine the ultrafiltration permeate flux (J_w) in m/s.

Given

  • transmembranepressure(deltaP)=57.0000kPatransmembrane pressure (deltaP) = 57.0000 kPa
  • dynamicviscosity(mu)=0.0011Pa∗sdynamic viscosity (mu) = 0.0011 Pa*s
  • membraneresistance(Rm)=910,000,000,0001/mmembrane resistance (R_m) = 910,000,000,000 1/m

Find

ultrafiltration permeate flux (J_w), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except J_w is given, so isolate J_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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for J_w:

    Jw=ΔPμRmJ_{w} = \dfrac{\Delta P}{\mu R_m}
  3. Step 3 — List the givens: transmembrane pressure (deltaP) = 57.0000 kPa, dynamic viscosity (mu) = 0.0011 Pa*s, membrane resistance (R_m) = 910,000,000,000 1/m.

  4. Step 4 — Substitute the given values:

    Jw=ΔP0.0011RmJ_{w} = \dfrac{\Delta P}{0.0011 R_m}
  5. Step 5 — Evaluate:

    Jw=0.0001 m/sJ_{w} = 0.0001\ \text{m/s}
  6. Step 6 — Check: returning J_w = 0.0001 m/s to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
Jw=0.0001 m/sJ_{w} = 0.0001\ \text{m/s}

Why the other options are there

  • 0.0001 — kept a factor of two that cancels in the correct rearrangement.
  • 0.0000 — dropped that same factor in the other direction.
  • 0.0001 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ultrafiltration

Example 2
Ultrafiltration — solve for transmembrane pressure — Ultrafiltration (2)

ultrafiltration flux calculation from transmembrane pressure Given dynamic viscosity (mu) = 0.0018 Pa*s; membrane resistance (R_m) = 567,000,000,000 1/m; ultrafiltration permeate flux (J_w) = 0.0007 m/s, determine the transmembrane pressure (deltaP) in kPa.

Given

  • dynamicviscosity(mu)=0.0018Pa∗sdynamic viscosity (mu) = 0.0018 Pa*s
  • membraneresistance(Rm)=567,000,000,0001/mmembrane resistance (R_m) = 567,000,000,000 1/m
  • ultrafiltrationpermeateflux(Jw)=0.0007m/sultrafiltration permeate flux (J_w) = 0.0007 m/s

Find

transmembrane pressure (deltaP), in kPa

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except deltaP is given, so isolate deltaP 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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for deltaP:

    deltaP=JwμRmdeltaP = J_w \mu R_m
  3. Step 3 — List the givens: dynamic viscosity (mu) = 0.0018 Pa*s, membrane resistance (R_m) = 567,000,000,000 1/m, ultrafiltration permeate flux (J_w) = 0.0007 m/s.

  4. Step 4 — Substitute the given values:

    deltaP=Jw0.0018RmdeltaP = J_w 0.0018 R_m
  5. Step 5 — Evaluate:

    deltaP=689.6 kPadeltaP = 689.6\ \text{kPa}
  6. Step 6 — Check: returning deltaP = 689.6 kPa to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
deltaP=689.6 kPadeltaP = 689.6\ \text{kPa}

Why the other options are there

  • 1,379 — kept a factor of two that cancels in the correct rearrangement.
  • 344.8 — dropped that same factor in the other direction.
  • 758.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ultrafiltration

Example 3
Ultrafiltration — solve for membrane resistance — Ultrafiltration (3)

ultrafiltration membrane resistance and flux relationship Given transmembrane pressure (deltaP) = 200.0 kPa; dynamic viscosity (mu) = 0.0017 Pa*s; ultrafiltration permeate flux (J_w) = 0.0008 m/s, determine the membrane resistance (R_m) in 1/m.

Given

  • transmembranepressure(deltaP)=200.0kPatransmembrane pressure (deltaP) = 200.0 kPa
  • dynamicviscosity(mu)=0.0017Pa∗sdynamic viscosity (mu) = 0.0017 Pa*s
  • ultrafiltrationpermeateflux(Jw)=0.0008m/sultrafiltration permeate flux (J_w) = 0.0008 m/s

Find

membrane resistance (R_m), in 1/m

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except R_m is given, so isolate R_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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for R_m:

    Rm=ΔPμJwR_{m} = \dfrac{\Delta P}{\mu J_w}
  3. Step 3 — List the givens: transmembrane pressure (deltaP) = 200.0 kPa, dynamic viscosity (mu) = 0.0017 Pa*s, ultrafiltration permeate flux (J_w) = 0.0008 m/s.

  4. Step 4 — Substitute the given values:

    Rm=ΔP0.0017JwR_{m} = \dfrac{\Delta P}{0.0017 J_w}
  5. Step 5 — Evaluate:

    Rm=143255187628 1/mR_{m} = 143255187628\ \text{1/m}
  6. Step 6 — Check: returning R_m = 143,255,187,628 1/m to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
Rm=143255187628 1/mR_{m} = 143255187628\ \text{1/m}

Why the other options are there

  • 286,510,375,257 — kept a factor of two that cancels in the correct rearrangement.
  • 71,627,593,814 — dropped that same factor in the other direction.
  • 157,580,706,391 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ultrafiltration

Example 4
Ultrafiltration — solve for ultrafiltration permeate flux (case 2) — Ultrafiltration (4)

ultrafiltration membrane permeate flux for drinking water treatment Given transmembrane pressure (deltaP) = 350.0 kPa; dynamic viscosity (mu) = 0.0013 Pa*s; membrane resistance (R_m) = 940,000,000,000 1/m, determine the ultrafiltration permeate flux (J_w) in m/s.

Given

  • transmembranepressure(deltaP)=350.0kPatransmembrane pressure (deltaP) = 350.0 kPa
  • dynamicviscosity(mu)=0.0013Pa∗sdynamic viscosity (mu) = 0.0013 Pa*s
  • membraneresistance(Rm)=940,000,000,0001/mmembrane resistance (R_m) = 940,000,000,000 1/m

Find

ultrafiltration permeate flux (J_w), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except J_w is given, so isolate J_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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for J_w:

    Jw=ΔPμRmJ_{w} = \dfrac{\Delta P}{\mu R_m}
  3. Step 3 — List the givens: transmembrane pressure (deltaP) = 350.0 kPa, dynamic viscosity (mu) = 0.0013 Pa*s, membrane resistance (R_m) = 940,000,000,000 1/m.

  4. Step 4 — Substitute the given values:

    Jw=ΔP0.0013RmJ_{w} = \dfrac{\Delta P}{0.0013 R_m}
  5. Step 5 — Evaluate:

    Jw=0.0003 m/sJ_{w} = 0.0003\ \text{m/s}
  6. Step 6 — Check: returning J_w = 0.0003 m/s to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
Jw=0.0003 m/sJ_{w} = 0.0003\ \text{m/s}

Why the other options are there

  • 0.0006 — kept a factor of two that cancels in the correct rearrangement.
  • 0.0001 — dropped that same factor in the other direction.
  • 0.0003 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ultrafiltration

Example 5
Ultrafiltration — solve for transmembrane pressure (case 2) — Ultrafiltration (5)

ultrafiltration flux calculation from transmembrane pressure Given dynamic viscosity (mu) = 0.0015 Pa*s; membrane resistance (R_m) = 688,000,000,000 1/m; ultrafiltration permeate flux (J_w) = 0.0002 m/s, determine the transmembrane pressure (deltaP) in kPa.

Given

  • dynamicviscosity(mu)=0.0015Pa∗sdynamic viscosity (mu) = 0.0015 Pa*s
  • membraneresistance(Rm)=688,000,000,0001/mmembrane resistance (R_m) = 688,000,000,000 1/m
  • ultrafiltrationpermeateflux(Jw)=0.0002m/sultrafiltration permeate flux (J_w) = 0.0002 m/s

Find

transmembrane pressure (deltaP), in kPa

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except deltaP is given, so isolate deltaP 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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for deltaP:

    deltaP=JwμRmdeltaP = J_w \mu R_m
  3. Step 3 — List the givens: dynamic viscosity (mu) = 0.0015 Pa*s, membrane resistance (R_m) = 688,000,000,000 1/m, ultrafiltration permeate flux (J_w) = 0.0002 m/s.

  4. Step 4 — Substitute the given values:

    deltaP=Jw0.0015RmdeltaP = J_w 0.0015 R_m
  5. Step 5 — Evaluate:

    deltaP=212.6 kPadeltaP = 212.6\ \text{kPa}
  6. Step 6 — Check: returning deltaP = 212.6 kPa to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
deltaP=212.6 kPadeltaP = 212.6\ \text{kPa}

Why the other options are there

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

Reference: FE Handbook — Ultrafiltration

Example 6
Ultrafiltration — solve for membrane resistance (case 2) — Ultrafiltration (6)

ultrafiltration membrane resistance and flux relationship Given transmembrane pressure (deltaP) = 379.0 kPa; dynamic viscosity (mu) = 0.0011 Pa*s; ultrafiltration permeate flux (J_w) = 0.0002 m/s, determine the membrane resistance (R_m) in 1/m.

Given

  • transmembranepressure(deltaP)=379.0kPatransmembrane pressure (deltaP) = 379.0 kPa
  • dynamicviscosity(mu)=0.0011Pa∗sdynamic viscosity (mu) = 0.0011 Pa*s
  • ultrafiltrationpermeateflux(Jw)=0.0002m/sultrafiltration permeate flux (J_w) = 0.0002 m/s

Find

membrane resistance (R_m), in 1/m

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except R_m is given, so isolate R_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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for R_m:

    Rm=ΔPμJwR_{m} = \dfrac{\Delta P}{\mu J_w}
  3. Step 3 — List the givens: transmembrane pressure (deltaP) = 379.0 kPa, dynamic viscosity (mu) = 0.0011 Pa*s, ultrafiltration permeate flux (J_w) = 0.0002 m/s.

  4. Step 4 — Substitute the given values:

    Rm=ΔP0.0011JwR_{m} = \dfrac{\Delta P}{0.0011 J_w}
  5. Step 5 — Evaluate:

    Rm=1479313036690 1/mR_{m} = 1479313036690\ \text{1/m}
  6. Step 6 — Check: returning R_m = 1,479,313,036,690 1/m to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
Rm=1479313036690 1/mR_{m} = 1479313036690\ \text{1/m}

Why the other options are there

  • 2,958,626,073,380 — kept a factor of two that cancels in the correct rearrangement.
  • 739,656,518,345 — dropped that same factor in the other direction.
  • 1,627,244,340,359 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ultrafiltration

Example 7
Ultrafiltration — solve for ultrafiltration permeate flux (case 3) — Ultrafiltration (7)

ultrafiltration membrane permeate flux for drinking water treatment Given transmembrane pressure (deltaP) = 369.0 kPa; dynamic viscosity (mu) = 0.0013 Pa*s; membrane resistance (R_m) = 923,000,000,000 1/m, determine the ultrafiltration permeate flux (J_w) in m/s.

Given

  • transmembranepressure(deltaP)=369.0kPatransmembrane pressure (deltaP) = 369.0 kPa
  • dynamicviscosity(mu)=0.0013Pa∗sdynamic viscosity (mu) = 0.0013 Pa*s
  • membraneresistance(Rm)=923,000,000,0001/mmembrane resistance (R_m) = 923,000,000,000 1/m

Find

ultrafiltration permeate flux (J_w), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except J_w is given, so isolate J_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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for J_w:

    Jw=ΔPμRmJ_{w} = \dfrac{\Delta P}{\mu R_m}
  3. Step 3 — List the givens: transmembrane pressure (deltaP) = 369.0 kPa, dynamic viscosity (mu) = 0.0013 Pa*s, membrane resistance (R_m) = 923,000,000,000 1/m.

  4. Step 4 — Substitute the given values:

    Jw=ΔP0.0013RmJ_{w} = \dfrac{\Delta P}{0.0013 R_m}
  5. Step 5 — Evaluate:

    Jw=0.0003 m/sJ_{w} = 0.0003\ \text{m/s}
  6. Step 6 — Check: returning J_w = 0.0003 m/s to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
Jw=0.0003 m/sJ_{w} = 0.0003\ \text{m/s}

Why the other options are there

  • 0.0006 — kept a factor of two that cancels in the correct rearrangement.
  • 0.0002 — dropped that same factor in the other direction.
  • 0.0003 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ultrafiltration

Example 8
Ultrafiltration — solve for transmembrane pressure (case 3) — Ultrafiltration (8)

ultrafiltration flux calculation from transmembrane pressure Given dynamic viscosity (mu) = 0.0016 Pa*s; membrane resistance (R_m) = 321,000,000,000 1/m; ultrafiltration permeate flux (J_w) = 0.0006 m/s, determine the transmembrane pressure (deltaP) in kPa.

Given

  • dynamicviscosity(mu)=0.0016Pa∗sdynamic viscosity (mu) = 0.0016 Pa*s
  • membraneresistance(Rm)=321,000,000,0001/mmembrane resistance (R_m) = 321,000,000,000 1/m
  • ultrafiltrationpermeateflux(Jw)=0.0006m/sultrafiltration permeate flux (J_w) = 0.0006 m/s

Find

transmembrane pressure (deltaP), in kPa

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except deltaP is given, so isolate deltaP 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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for deltaP:

    deltaP=JwμRmdeltaP = J_w \mu R_m
  3. Step 3 — List the givens: dynamic viscosity (mu) = 0.0016 Pa*s, membrane resistance (R_m) = 321,000,000,000 1/m, ultrafiltration permeate flux (J_w) = 0.0006 m/s.

  4. Step 4 — Substitute the given values:

    deltaP=Jw0.0016RmdeltaP = J_w 0.0016 R_m
  5. Step 5 — Evaluate:

    deltaP=305.3 kPadeltaP = 305.3\ \text{kPa}
  6. Step 6 — Check: returning deltaP = 305.3 kPa to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
deltaP=305.3 kPadeltaP = 305.3\ \text{kPa}

Why the other options are there

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

Reference: FE Handbook — Ultrafiltration

Example 9
Ultrafiltration — solve for membrane resistance (case 3) — Ultrafiltration (9)

ultrafiltration membrane resistance and flux relationship Given transmembrane pressure (deltaP) = 368.0 kPa; dynamic viscosity (mu) = 0.0012 Pa*s; ultrafiltration permeate flux (J_w) = 0.0001 m/s, determine the membrane resistance (R_m) in 1/m.

Given

  • transmembranepressure(deltaP)=368.0kPatransmembrane pressure (deltaP) = 368.0 kPa
  • dynamicviscosity(mu)=0.0012Pa∗sdynamic viscosity (mu) = 0.0012 Pa*s
  • ultrafiltrationpermeateflux(Jw)=0.0001m/sultrafiltration permeate flux (J_w) = 0.0001 m/s

Find

membrane resistance (R_m), in 1/m

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except R_m is given, so isolate R_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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for R_m:

    Rm=ΔPμJwR_{m} = \dfrac{\Delta P}{\mu J_w}
  3. Step 3 — List the givens: transmembrane pressure (deltaP) = 368.0 kPa, dynamic viscosity (mu) = 0.0012 Pa*s, ultrafiltration permeate flux (J_w) = 0.0001 m/s.

  4. Step 4 — Substitute the given values:

    Rm=ΔP0.0012JwR_{m} = \dfrac{\Delta P}{0.0012 J_w}
  5. Step 5 — Evaluate:

    Rm=3046862063256 1/mR_{m} = 3046862063256\ \text{1/m}
  6. Step 6 — Check: returning R_m = 3,046,862,063,256 1/m to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
Rm=3046862063256 1/mR_{m} = 3046862063256\ \text{1/m}

Why the other options are there

  • 6,093,724,126,511 — kept a factor of two that cancels in the correct rearrangement.
  • 1,523,431,031,628 — dropped that same factor in the other direction.
  • 3,351,548,269,581 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ultrafiltration

Example 10
Ultrafiltration — solve for ultrafiltration permeate flux (case 4) — Ultrafiltration (10)

ultrafiltration membrane permeate flux for drinking water treatment Given transmembrane pressure (deltaP) = 179.0 kPa; dynamic viscosity (mu) = 0.0010 Pa*s; membrane resistance (R_m) = 798,000,000,000 1/m, determine the ultrafiltration permeate flux (J_w) in m/s.

Given

  • transmembranepressure(deltaP)=179.0kPatransmembrane pressure (deltaP) = 179.0 kPa
  • dynamicviscosity(mu)=0.0010Pa∗sdynamic viscosity (mu) = 0.0010 Pa*s
  • membraneresistance(Rm)=798,000,000,0001/mmembrane resistance (R_m) = 798,000,000,000 1/m

Find

ultrafiltration permeate flux (J_w), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Ultrafiltration.
  • Everything except J_w is given, so isolate J_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.
  • Ultrafiltration permeate flux is governed by transmembrane pressure, fluid viscosity, and membrane resistance.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}
  2. Step 2 — Rearrange symbolically for J_w:

    Jw=ΔPμRmJ_{w} = \dfrac{\Delta P}{\mu R_m}
  3. Step 3 — List the givens: transmembrane pressure (deltaP) = 179.0 kPa, dynamic viscosity (mu) = 0.0010 Pa*s, membrane resistance (R_m) = 798,000,000,000 1/m.

  4. Step 4 — Substitute the given values:

    Jw=ΔP0.0010RmJ_{w} = \dfrac{\Delta P}{0.0010 R_m}
  5. Step 5 — Evaluate:

    Jw=0.0002 m/sJ_{w} = 0.0002\ \text{m/s}
  6. Step 6 — Check: returning J_w = 0.0002 m/s to

    Jw=ΔPμRmJ_w = \dfrac{\Delta P}{\mu R_m}

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

Answer:
Jw=0.0002 m/sJ_{w} = 0.0002\ \text{m/s}

Why the other options are there

  • 0.0004 — kept a factor of two that cancels in the correct rearrangement.
  • 0.0001 — dropped that same factor in the other direction.
  • 0.0002 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ultrafiltration

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