Corrosion

Materials Science · FE Reference Handbook section

Materials Science

0 formulas

10 exam-style examples

~45 min

All Materials Science lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A table listing the standard electromotive potentials of metals is shown on the previous page.
For corrosion to occur, there must be an anode and a cathode in electrical contact in the presence of an electrolyte.

Core formulas for this FE topic

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

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

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

Example 1

Steady heat flow through a wall — Corrosion

A 1.1 ft thick concrete wall of area 396 ft² has k = 0.9 Btu/(hr·ft·°F) and a 64°F temperature difference. What is the heat flow rate?

Given

k = 0.9 Btu/hr·ft·°F
$A = 396 ft^{2}$
ΔT = 64°F
$t = 1.1 ft$

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 0.9(396)(64)/1.1$
Evaluate
$q = 20,736 Btu/hr$

Answer:

q ≈ 20,736 Btu/hr

Why the other options are there

25,091 Btu/hr (thickness multiplied)
57.6 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion

Example 2

Resistivity of a conductor and capacitance of a parallel plate — Corrosion

A conductor of resistivity 1.00e-6 Ω·m is 2.0 m long with a cross-section of 1.3e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 5.0), plate area 0.0120 m² and spacing 0.00050 m is charged to 143 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 2.0 m, A = 1.3e-5 m^{2}$
$\varepsilon_r = 5.0, A_p = 0.0120 m^{2}, d = 0.00050 m$
$V = 143 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 1.00e-6(2.0)/1.3e-5 = 0.1538 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(5.0)(0.0120)/0.00050 = 1.062e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.062e-9(143) = 1.519e-7 C$

Answer:

R = 0.154 \Omega, C = 1.06e-9 F, Q = 1.52e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 2.12e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion

Example 3

Steady heat flow through a wall — Corrosion (2)

A 1.3 ft thick concrete wall of area 197 ft² has k = 1.9 Btu/(hr·ft·°F) and a 89°F temperature difference. What is the heat flow rate?

Given

k = 1.9 Btu/hr·ft·°F
$A = 197 ft^{2}$
ΔT = 89°F
$t = 1.3 ft$

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.9(197)(89)/1.3$
Evaluate
$q = 25,625 Btu/hr$

Answer:

q ≈ 25,625 Btu/hr

Why the other options are there

43,307 Btu/hr (thickness multiplied)
169.1 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion

Example 4

Resistivity of a conductor and capacitance of a parallel plate — Corrosion (2)

A conductor of resistivity 1.00e-6 Ω·m is 3.5 m long with a cross-section of 1.5e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 2.5), plate area 0.0160 m² and spacing 0.00040 m is charged to 184 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 3.5 m, A = 1.5e-5 m^{2}$
$\varepsilon_r = 2.5, A_p = 0.0160 m^{2}, d = 0.00040 m$
$V = 184 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 1.00e-6(3.5)/1.5e-5 = 0.2333 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(2.5)(0.0160)/0.00040 = 8.854e-10 F$
Formula
$Q = C V$
Substituting
$Q = 8.854e-10(184) = 1.629e-7 C$

Answer:

R = 0.233 \Omega, C = 8.85e-10 F, Q = 1.63e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 3.54e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion

Example 5

Steady heat flow through a wall — Corrosion (3)

A 1.3 ft thick concrete wall of area 158 ft² has k = 1.5 Btu/(hr·ft·°F) and a 67°F temperature difference. What is the heat flow rate?

Given

k = 1.5 Btu/hr·ft·°F
$A = 158 ft^{2}$
ΔT = 67°F
$t = 1.3 ft$

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.5(158)(67)/1.3$
Evaluate
$q = 12,215 Btu/hr$

Answer:

q ≈ 12,215 Btu/hr

Why the other options are there

20,643 Btu/hr (thickness multiplied)
100.5 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion

Example 6

Resistivity of a conductor and capacitance of a parallel plate — Corrosion (3)

A conductor of resistivity 1.00e-6 Ω·m is 3.5 m long with a cross-section of 5.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.5), plate area 0.0160 m² and spacing 0.00030 m is charged to 156 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 3.5 m, A = 5.0e-6 m^{2}$
$\varepsilon_r = 4.5, A_p = 0.0160 m^{2}, d = 0.00030 m$
$V = 156 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 1.00e-6(3.5)/5.0e-6 = 0.7000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.5)(0.0160)/0.00030 = 2.125e-9 F$
Formula
$Q = C V$
Substituting
$Q = 2.125e-9(156) = 3.315e-7 C$

Answer:

R = 0.700 \Omega, C = 2.12e-9 F, Q = 3.31e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 4.72e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion

Example 7

Steady heat flow through a wall — Corrosion (4)

A 0.9 ft thick concrete wall of area 361 ft² has k = 0.8 Btu/(hr·ft·°F) and a 66°F temperature difference. What is the heat flow rate?

Given

k = 0.8 Btu/hr·ft·°F
$A = 361 ft^{2}$
ΔT = 66°F
$t = 0.9 ft$

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 0.8(361)(66)/0.9$
Evaluate
$q = 21,179 Btu/hr$

Answer:

q ≈ 21,179 Btu/hr

Why the other options are there

17,155 Btu/hr (thickness multiplied)
52.8 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion

Example 8

Resistivity of a conductor and capacitance of a parallel plate — Corrosion (4)

A conductor of resistivity 1.00e-6 Ω·m is 1.0 m long with a cross-section of 9.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 2.0), plate area 0.0120 m² and spacing 0.00100 m is charged to 100 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 1.0 m, A = 9.0e-6 m^{2}$
$\varepsilon_r = 2.0, A_p = 0.0120 m^{2}, d = 0.00100 m$
$V = 100 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 1.00e-6(1.0)/9.0e-6 = 0.1111 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(2.0)(0.0120)/0.00100 = 2.125e-10 F$
Formula
$Q = C V$
Substituting
$Q = 2.125e-10(100) = 2.125e-8 C$

Answer:

R = 0.111 \Omega, C = 2.12e-10 F, Q = 2.12e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.06e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion

Example 9

Steady heat flow through a wall — Corrosion (5)

A 1.1 ft thick concrete wall of area 63 ft² has k = 1.7 Btu/(hr·ft·°F) and a 48°F temperature difference. What is the heat flow rate?

Given

k = 1.7 Btu/hr·ft·°F
$A = 63 ft^{2}$
ΔT = 48°F
$t = 1.1 ft$

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.7(63)(48)/1.1$
Evaluate
$q = 4,673 Btu/hr$

Answer:

q ≈ 4,673 Btu/hr

Why the other options are there

5,655 Btu/hr (thickness multiplied)
81.6 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion

Example 10

Resistivity of a conductor and capacitance of a parallel plate — Corrosion (5)

A conductor of resistivity 1.00e-6 Ω·m is 4.0 m long with a cross-section of 1.2e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 5.0), plate area 0.0150 m² and spacing 0.00070 m is charged to 122 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 4.0 m, A = 1.2e-5 m^{2}$
$\varepsilon_r = 5.0, A_p = 0.0150 m^{2}, d = 0.00070 m$
$V = 122 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 1.00e-6(4.0)/1.2e-5 = 0.3333 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(5.0)(0.0150)/0.00070 = 9.486e-10 F$
Formula
$Q = C V$
Substituting
$Q = 9.486e-10(122) = 1.157e-7 C$

Answer:

R = 0.333 \Omega, C = 9.49e-10 F, Q = 1.16e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.90e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Corrosion