The branding, packaging, and labeling of your product should accomplish all of the following except
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Answer:
The inside temperature, is approximately 248 °C.
Explanation:
The parameters given are;
Temperature of the air = 20°C
Carbon steel surface temperature 250°C
Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²
Convection heat transfer coefficient = 25 W/(m²·K)
Heat lost by radiation = 300 W
Assumption,
Air temperature = 20 °C
Hot plate temperature = 250 °C
Thermal conductivity K = 65.2 W/(m·K)
Steady state heat transfer process
One dimensional heat conduction
We have;
Newton's law of cooling;
q = h×A×( -
) + Heat loss by radiation
= 25×0.325×(250 - 20) + 300
= 2456.25 W
The rate of energy transfer per second is given by the following relation;
Thermal conductivity K = 65.2 W/(m·K)
Therefore;
The inside temperature, = 247.99 °C ≈ 248 °C.
A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.
Explanation:
From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).
To find the resistance of 260 ft (79.25 m) of size 4 AWG,
R= K * L/ A
K = 0.0214 ohm mm²/m
L = 79.25 m
A = 21.2 mm²
R = 0.0214 *
= 0.0214 * 3.738
= 0.0792 ohm.
Thus the resistance of uncoated copper wire is 0.0792 ohm
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ