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alexdok [17]
3 years ago
14

A thick steel slab ( 7800 kg/m3, 480 J/kg·K, 50 W/m·K) is initially at 300°C and is cooled by water jets impinging on one of its

surfaces. The temperature of the water is 25°C, and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling, how long will it take for the temperature to reach 50°C at a distance of 30 mm from the surface?

Engineering
2 answers:
AleksandrR [38]3 years ago
4 0

Answer: 67.392s

Explanation: detailed calculation is shown below

dlinn [17]3 years ago
3 0

Answer:

Time(t) = 2592.91 seconds

Explanation:

From the question, we have;

ρ = 7800 kg/m

c =480 J/kg⋅K

k = 50 W/m⋅K

Ti = 300°C

Ts = 25°C

x = 25 mm or in meters; = 0.025 m

From formula, we know that;

(T(x,t) - T(s))/(T(i) - T(s)) = erf(x/(2√(αt))

Where erf is error function

And α = k/(ρc)

So, solving we have;

(50 - 25)/(300 - 25) = erf(x/(2√(αt))

erf(x/(2√(αt)) = 0.0909

From the error function table which i attached, 0.0909 will give us approximately 0.0806 upon interpolation.

Thus, (x/(2√(αt)) = 0.0806

Let's make "t" the subject of the formula;

x² = 0.0806²(2²)(αt)

t = x²/0.026α

Let's find α

From earlier, we saw that;

α = k/(ρc)

Thus; α = 50/(7800 x 480) = 1.335 x 10^(-5) m²/s

Also, from the question, x = 30mm or 0.03m

So, t = 0.03²/(0.026 x 1.335 x 10^(-5)) = 2592.91 seconds

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Option A

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Answer:

Below is the required code.

Explanation:

%% Newton Raphson Method

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clc;

x0=input('Initial guess:\n');

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cc=input('Condition of convergence:\n');

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Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

>>

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

>>

(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

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Guess x=1.75:

Initial guess:

1.75

Condition of convergence:

0.01

x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

x = 216.970438 and error = 0.267241 at iteration = 6.000000

x = 216.971251 and error = 0.000813 at iteration = 7.000000

The solution x = 216.971251

>>

Guess x=3.0:

Initial guess:

3

Condition of convergence:

0.01

x = 3.309861 and error = 0.309861 at iteration = 1.000000

x = 3.305651 and error = 0.004210 at iteration = 2.000000

The solution x = 3.305651

>>

Guess x=4.7:

Initial guess:

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0.01

x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

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The solution x = 0.451732

>>

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

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