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The wall of an oven used to cure plastic parts is of thickness L = 0.05 m and is exposed to large surroundings and air at its ou

Umnica [9.8K]

Answer: the temperature of the inner surface is 599.55 K

Explanation:

first, let Am² represent area of the oven surface

now heat transfer rate due to convection

qconv = hAΔT

where ΔT is temperature difference in kelvin (To -Ts) temperature of outer surface - temperature of surroundings

ΔT = 400K - 300k = 100k

and h = 20 W/m².K

so we substitute

qconv = 20 × A × 100 = 2000A W

Now heat transfer due to emission

qemi = ∈α(To⁴ - Ts⁴)A

∈ is emissivity = 0.8, α is boltzman constant = 5.67×10⁻⁸ W/m⁻²K⁻⁴

so we substitute

qemi = 0.8 × 5.67×10⁻⁸ (400⁴ - 300⁴)A

qemi = 4.536×10⁻⁸ × 1.75×10¹⁰ × A

qemi = 793.8A W

NOW total heat loss rate = Qconv + Qemi

2000A + 793.8A = 2793.8A W

so since outer surface is to be maintained at a temp of 400K then heat must be transferred from inner surface to outer surface at the same rate.

Therefore

Heat transfer rate to outer surface by conduction = kA(Ti-To)/L

k is thermal conductivity 0.7 W/m.K, Ti is temperature of inner surface, L is the thickness of plate 0.05m

so Heat transfer rate = 0.7× A × ( Ti - 400 ) / 0.05

Now

This heat transfer rate equals heat loss rate

so 0.7× A × ( Ti - 400 ) / 0.05 = 2793.8 × A

divide both side by A

THEREFORE

Ti = 400 + 199.55

Ti = 599.55 K

∴ the temperature of the inner surface is 599.55 K

5 0

3 years ago

Increase time to respond to threats through position and speed adjustments is risk management.

marysya [2.9K]

I believe that this would be a true accusation.

5 0

3 years ago

Shane's 100-watt radio draws 7 amps of current on a 120-volt circuit. What is the resistance in the radio?

Elis [28]

Answer:

Resistance, R = 2.04 Ohms.

Explanation:

Given the following data;

Current = 7 amps

Circuit voltage = 120 Volts

Power = 100 Watts

To find the resistance;

First of all, we would calculate the potential difference (voltage level) used by the radio.

Power = current * voltage

100 = 7 * voltage

Voltage = 100/7

Voltage = 14.29 Volts

Next, we solve for the resistance of the radio;

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

$V = IR$

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

Substituting into the formula, we have;

$14.29 = 7*R$

$R = \frac {14.29}{7}$

Resistance, R = 2.04 Ohms.

7 0

3 years ago

What do you do when two parts of a transfer line move at different rates?

Tatiana [17]

Answer:Buffer

Explanation:

4 0

3 years ago

A hypothetical metal alloy has a grain diameter of 1.7 102 mm. After a heat treatment at 450C for 250 min, the grain diameter ha

Romashka-Z-Leto [24]

Answer:

the required time for the specimen is 1109.4 min

Explanation:

Given that;

diameter of metal alloy d₀ = 1.7 × 10² mm

Temperature of heat treatment T = 450°C = 450 + 273 = 723 K

Time period of heat treatment t = 250 min

Increased grain diameter 4.5 × 10² mm

grain diameter exponent n = 2.1

First we calculate the time independent constant K

dⁿ - d₀ⁿ = Kt

K = (dⁿ - d₀ⁿ) / t

we substitute

K = (( 4.5 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 250

K = (373032.163378 - 48299.511117) / 250

K = 1298.9306 mm²/min

Now, we calculate the time required for the specimen to achieve the given grain diameter ( 8.7 × 10² mm )

dⁿ - d₀ⁿ = Kt

t = (dⁿ - d₀ⁿ) / K

t = (( 8.7 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 1298.9306

t = ( 1489328.26061158 - 48299.511117) / 1298.9306

t = 1441028.74949458 / 1298.9306

t = 1109.4 min

Therefore, the required time for the specimen is 1109.4 min

4 0

3 years ago