A steel ball of 10 mm diameter at 1000 K is r...
A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in a water environment at 300 K. The convective heat transfer coefficient is 1000 W/m2-K. Thermal conductivity of steel is 40 W/m-K.The time constant for the cooling process τ is 16s. The time required (in s) to reach the final temperature is________
(Important - Enter only the numerical value in the answer)
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The time constant for the cooling process can be calculated using the formula:

τ = (m * Cp) / (h * A)

Where:
τ is the time constant
m is the mass of the steel ball
Cp is the specific heat capacity of steel
h is the convective heat transfer coefficient
A is the surface area of the steel ball

First, we need to calculate the mass of the steel ball:

V = (4/3) * π * (r^3)
V = (4/3) * π * (0.005^3) (converting the diameter to radius)
V = 5.24 x 10^-8 m^3

ρ = m / V (ρ is the density of steel)
m = ρ * V (assuming the density of steel is 7850 kg/m^3)

m = 7850 kg/m^3 * 5.24 x 10^-8 m^3
m = 0.00041 kg

Next, we need to calculate the surface area of the steel ball:

A = 4 * π * r^2
A = 4 * π * (0.005^2)
A = 0.000314 m^2

Now we can calculate the time constant:

τ = (0.00041 kg * Cp) / (1000 W/m^2-K * 0.000314 m^2)
τ = (0.00041 kg * Cp) / 0.314 W/K
τ = (0.00041 kg * Cp) / 0.314

We need to know the specific heat capacity of steel (Cp) to calculate the time constant.
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A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in awater environment at 300 K. The convective heat transfer coefficient is 1000 W/m2-K. Thermalconductivity of steel is 40 W/m-K.The time constant for the cooling process τ is 16s. The timerequired (in s) to reach the final temperature is________(Important - Enter only the numerical value in the answer)Correct answer is '42.22'. Can you explain this answer?
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A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in awater environment at 300 K. The convective heat transfer coefficient is 1000 W/m2-K. Thermalconductivity of steel is 40 W/m-K.The time constant for the cooling process τ is 16s. The timerequired (in s) to reach the final temperature is________(Important - Enter only the numerical value in the answer)Correct answer is '42.22'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in awater environment at 300 K. The convective heat transfer coefficient is 1000 W/m2-K. Thermalconductivity of steel is 40 W/m-K.The time constant for the cooling process τ is 16s. The timerequired (in s) to reach the final temperature is________(Important - Enter only the numerical value in the answer)Correct answer is '42.22'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in awater environment at 300 K. The convective heat transfer coefficient is 1000 W/m2-K. Thermalconductivity of steel is 40 W/m-K.The time constant for the cooling process τ is 16s. The timerequired (in s) to reach the final temperature is________(Important - Enter only the numerical value in the answer)Correct answer is '42.22'. Can you explain this answer?.
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