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zubka84 [21]
4 years ago
8

A tank contains liquid nitrogen at -190℃ is suspended in a vacuum shell by three stainless steel rods 0.80 cm in diameter and 3

meters long with a thermal conductivity of 16.3 W/m^2-℃. If the ambient air outside the vacuum shell is 15℃, calculate the magnitude of the conductive heat flow in watts along the support rods. A. 0.168 B. 0.0587 C. 0.182 D. 0.176
Engineering
1 answer:
Svetach [21]4 years ago
7 0

Answer:

C) 0.182 W

Explanation:

The effective conductive section is:

3 * A = 3 * \frac{\pi * d^2}{4} = 3 * \frac{\pi * (0.008 m)^2}{4} = 1.5e-4 m^2

The thermal conductivity of stainless steel is

k = 18 w/(m * K)

Heat conduction in a rod follows this equation:

q = k * A * \frac{(t2 - t1)}{l} = 18 * 1.5e-4 * \frac{(15 - (-190))}{3} = 0.182 W

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During the recovery of a cold-worked material, which of the following statement(s) is (are) true?
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Answer:

Some of the internal strain energy is relieved.

There is some reduction in the number of dislocations.

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The thermal conductivity is recovered to its precold-worked state

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Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m
kodGreya [7K]

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

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