The temperatures of the stainless steel slab and copper slab from the surface.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Water is supplied at 150 ft³/s and 70 psi to a hydraulic turbine through a 3-ft inside-diameter inlet pipe as indicated in the figure below. The turbine discharge pipe has a 4.8-ft inside diameter. The static pressure at section (2), 10 ft below the turbine inlet, is 10 in. Hg vacuum. If the turbine develops 2400 hp, determine the rate of loss of available energy between sections (1) and (2). Section (1) P₁ =70psi Q=150ft³/s D₁ = 3 ft 10 ft Turbine power loss = i P₂ = 10 in. Hg vacuum D₂ =4.8ft Section (2) de hp

This problem studies the response of two single degree of freedom bridge systems shown in Figure 1 under three loading cases. The problem has two parts. Part A and Part B use the same loading cases but the system is modified. Assume the following three loading cases in both Part A and Part B: (a) Harmonic wind load acting on the bridge deck pw(t) = powsin(ωwt) with amplitude pow and forcing circular frequency ωw. (b) Harmonic displacement base excitation acting at the base of the bridge pier ug(t) = ugosin(ωgt) with amplitude ugo and displacement circular frequency ωg. (c) Rectangular pulse load acting on the bridge deck with amplitude pop and pulse duration td. Part A The system includes part of a bridge deck and a bridge pier shown in Figure 1(a). For each loading case find the symbolic expression of the peak shear force in the bridge pier assuming the following: • The bridge deck is rigid and it has a mass m. • The bridge deck is rigidly connected with the bridge pier (i.e.,…

specific speed P #2 Q.2. A Pelton wheel turbine of 1.9 m diameter works under a head of 50 m at 150 rpm. The buckets are exposed to water jet which delivers from a nozzle of 20 cm in diameter. Find the overall efficiency power produced by the wheel if the buckets deflects the jet through an angle of 163°. coefficient of velocity as 0.98 [50 Marks] ·licosply Y and no Take the

Answer 2

Question

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Answer 6

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Answer 7

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]