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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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One of my tools stopped working, and I found that in order to fix it, I need to replace a broken wire. The broken wire has a dia
professor190 [17]

Answer:

a) I₂ = 2 mA   (The current has decreased)

b) L₂ = 2.4 cm

Explanation:

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a)

Length of wire 2 = L₂ = 15 cm

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V₁ = V₂

using Ohm's Law (V = IR)

I₁R₁ = I₂R₂

Since resistance of wire is given by formula:  R = ρL/A

Therefore,

I₁ρ₁L₁/A₁ = I₂ρ₂L₂/A₂

where,

ρ = resistivity, and it depends upon material of wire and the material of both wires is same and the length of wires is also same.

Hence,    ρ₁ = ρ₂

and      L₁ = L₂

and the equation becomes:

I₁/A₁ = I₂/A₂

I₂ = I₁A₂/A₁

I₂ = (12.5 mA)(3.14 mm²)/(19.63 mm²)

<u>I₂ = 2 mA</u>

<u>Thus, the current has decreased.</u>

<u></u>

b)

In order to have same current the resistance of both wires must be same:

R₁ = R₂

ρ₁L₁/A₁ = ρ₂L₂/A₂

Since,   ρ₁ = ρ₂

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L₂ = L₁A₂/A₁

L₂ = (15 cm)(3.14 mm²)/(19.63 mm²)

<u>L₂ = 2.4 cm</u>

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