The ductility of a large copper (k = 386 W/mºC, a = must be increased by quenching before it can be extruded. The copper slab is first heated up to have a uniform initial temperature of 1,000 °C. Cold water is then sprayed onto one side of the slab to maintain a constant surface temperature of 20 °C. (a) Draw a simple sketch to show the change of temperature inside the slab with time. (b) Calculate the temperature (in °C) at 10 cm below the surface 20 seconds after the surface temperature is lowered. (c) Calculate the heat transfer flux rate (kW/m2 ) through a plane 8 cm below the surface 15 seconds after the surface temperature is lowered. (d) Calculate the total heat removed by cold water from the surface in the first 5 seconds (J/m2_). 11.234 x10-5 m 2 /s, p = 8.954 g/cm3) slab
The ductility of a large copper (k = 386 W/mºC, a = must be increased by quenching before it can be extruded. The copper slab is first heated up to have a uniform initial temperature of 1,000 °C. Cold water is then sprayed onto one side of the slab to maintain a constant surface temperature of 20 °C. (a) Draw a simple sketch to show the change of temperature inside the slab with time. (b) Calculate the temperature (in °C) at 10 cm below the surface 20 seconds after the surface temperature is lowered. (c) Calculate the heat transfer flux rate (kW/m2 ) through a plane 8 cm below the surface 15 seconds after the surface temperature is lowered. (d) Calculate the total heat removed by cold water from the surface in the first 5 seconds (J/m2_). 11.234 x10-5 m 2 /s, p = 8.954 g/cm3) slab
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Please help and show all steps on how answer was solved. Class is Heat Transfer. Thank you.
![The ductility of a large copper (k= 386 W/m°C, a = 11.234 x10-5 m 2 /s, p = 8.954 g/cm3) slab
must be increased by quenching before it can be extruded. The copper slab is first heated up to
have a uniform initial temperature of 1,000 °C. Cold water is then sprayed onto one side of the
slab to maintain a constant surface temperature of 20 °C.
(a) Draw a simple sketch to show the change of temperature inside the slab with time.
(b) Calculate the temperature (in °C) at 10 cm below the surface 20 seconds after the surface
temperature is lowered.
(c) Calculate the heat transfer flux rate (kW/m2) through a plane 8 cm below the surface 15
seconds after the surface temperature is lowered.
(d) Calculate the total heat removed by cold water from the surface in the first 5 seconds (J/m2).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36997276-e876-4d20-b813-249583ed4792%2Fa7b868db-7364-4ef4-b09f-d1ff08786b59%2Fdx4teb9_processed.png&w=3840&q=75)
Transcribed Image Text:The ductility of a large copper (k= 386 W/m°C, a = 11.234 x10-5 m 2 /s, p = 8.954 g/cm3) slab
must be increased by quenching before it can be extruded. The copper slab is first heated up to
have a uniform initial temperature of 1,000 °C. Cold water is then sprayed onto one side of the
slab to maintain a constant surface temperature of 20 °C.
(a) Draw a simple sketch to show the change of temperature inside the slab with time.
(b) Calculate the temperature (in °C) at 10 cm below the surface 20 seconds after the surface
temperature is lowered.
(c) Calculate the heat transfer flux rate (kW/m2) through a plane 8 cm below the surface 15
seconds after the surface temperature is lowered.
(d) Calculate the total heat removed by cold water from the surface in the first 5 seconds (J/m2).
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