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

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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.

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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.

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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.

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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.

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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)]