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jekas [21]
3 years ago
15

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

f 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 25 mm from the surface
Engineering
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

1791 secs  ≈ 29.85 minutes

Explanation:

( Initial temperature of slab )  T1 = 300° C

temperature of water ( Ts ) = 25°C

T2 ( final temp of slab ) = 50°C

distance between slab and water jet = 25 mm

<u>Determine how long it will take to reach T2</u>

First calculate the thermal diffusivity

∝  = 50 / ( 7800 * 480 ) = 1.34 * 10^-5 m^2/s

<u>next express Temp as a function of time </u>

T( 25 mm , t ) = 50°C

next calculate the time required for the slab to reach 50°C at a distance of 25mm

attached below is the remaining part of the detailed solution

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Answer:

The given grammar is :

S = T V ;

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X = , V | ε

T = float | double

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1.

Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.

From the given grammar,

Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.

No other variables generate variable X or ε.

So, only variable X is nullable.

2.

First of nullable variable X is First (X ) = , and ε (epsilon).

L.H.S.

The first of other varibles are :

First (S) = {float, double }

First (T) = {float, double }

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R.H.S.

First (T V ; ) = {float, double }

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First (, V) = ,

First ( ε ) = ε

First (float) = float

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First (z) = z

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Explanation:

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A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi
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Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

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The flexural strength of a specimen is given by

\sigma = \frac{3 F L}{2 bd^{2} }

\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }

\sigma = 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

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saveliy_v [14]

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Answer:

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The product of x and y can be positive or negative ,so the value of  I_{xy} can be positive as well as negative .

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