In a manufacturing plant, AISI 1010 carbon steel strips \(\left(\rho=7832 \mathrm{~kg} / \mathrm{m}^{3}\right)\) of \(2 \mathrm{~mm}\) thick and \(3 \mathrm{~cm}\) wide are conveyed into a chamber at a constant speed to be cooled from \(527^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). Determine the speed of a steel strip being conveyed inside the chamber, if the rate of heat being removed from a steel strip inside the chamber is \(100 \mathrm{~kW}\).

Short Answer

Expert verified
Answer: The speed of the steel strip being conveyed inside the chamber is approximately 0.0012671 m/s.

Step by step solution

01

1. Calculate the specific heat of AISI 1010 carbon steel

Specific heat capacity of AISI 1010 carbon steel (c) = \(420 \mathrm{~J/(kg\cdot K)}\)
02

2. Calculate the temperature difference

\(\Delta T = T_{initial} - T_{final} = 527^{\circ}\mathrm{C} - 127^{\circ} \mathrm{C} = 400 \mathrm{~K}\)
03

3. Determine the volume of a given length of the strip

\(Volume = Thickness \times Width \times Length = 0.002 \mathrm{~m} \times 0.03 \mathrm{~m} \times L \mathrm{~m}\), where L is the length of the strip in meters.
04

4. Calculate the mass of a given length of the strip

\(Mass = \rho \times Volume = 7832 \mathrm{~ kg/m^{3}}\times(0.002 \mathrm{~m} \times 0.03 \mathrm{~m} \times L\mathrm{~ m}) = 469.92\mathrm{~ kg/m}\times L \mathrm{~m}\)
05

5. Calculate the amount of heat being removed from a given length of the strip

\(Q = mc\Delta T = (469.92\mathrm{~ kg/m}\times L \mathrm{~m})\times 420 \mathrm{~J/(kg\cdot K)}\times 400 \mathrm{~K}= 78915200\mathrm{~ J/m}\times L \mathrm{~m}\)
06

6. Determine the power being removed from a given length of the strip

\(P = \frac{Q}{t} = \frac{78915200\mathrm{~ J/m}\times L \mathrm{~m}}{t} = 100000\mathrm{~ W}\), where t is the time in seconds needed to cool down the strip.
07

7. Calculate the speed of the strip

We can find the speed by dividing the length of the strip by the time required to cool it down. \(Speed = \frac{Length}{Time} = \frac{L}{t}\) We know that the Power being removed is constant and equal to 100000 W. So, we can re-write the equation from Step 6 as: \(78915200\mathrm{~ J/m}\times L \mathrm{~m} = 100000\mathrm{~W}\times t\) Let's isolate t from the above equation: \(t = \frac{78915200\mathrm{~ J/m}\times L \mathrm{~m}}{100000\mathrm{~W}}\) Now we can substitute this expression for t into the Speed equation: \(Speed = \frac{L}{\frac{78915200\mathrm{~ J/m}\times L \mathrm{~m}}{100000\mathrm{~W}}} = \frac{100000\mathrm{~W}}{78915200\mathrm{~ J/m}}\) \(Speed = 0.0012671\mathrm{~ m/s}\) Hence, the speed of the steel strip being conveyed inside the chamber is approximately 0.0012671 m/s.

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