Answer: the thermal conductivity of the sample is 22.4 W/m . °C
Explanation:
We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
ASSUMPTIONS
1. Steady operating conditions exist since the temperature readings do not change with time.
2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.
3. The apparatus possess thermal symmetry
ANALYSIS
The electrical power consumed by resistance heater and converted to heat is:
Wₐ = V<em>I</em> = ( 110 V ) ( 0.4 A ) = 44 W
Q = 1/2Wₐ = 1/2 ( 44 A )
Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so
A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²
Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be
Q = kA ( ΔT/L ) → k = QL / AΔT
k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )
k = 22.4 W/m . °C
Answer:
Spray carburetor cleaner on the inside of the bowl and wipe the liquid, dirt, and concentrated fuel off of it. Now take the main jet, spray the cleaner through it and wipe off the dirt. Then take a copper wire, scrub it through the tiny holes in the jet to complete the cleaning process.
Explanation:
Engineer technicians typically hold a two-year associates degree, while engineering technologists likewise hold a bachelors degree.
Answer:
true: the types of building materials that’s should be used in a project does not constitute a conditional for projects’ permits and approvals
Answer:
125 cm³/min
Explanation:
The material rate of removal is usually given by the formula
Material Rate of Removal = Radial Depth of Cut * Axial Depth of Cut * Feed Rate, where
Radial Depth of Cut = 25 mm
Axial depth of cut = 200 mm
Feed rate = 25 mm/min
On multiplying all together, we will then have
MRR = 25 mm * 200 mm * 25 mm/min
MRR = 125000 mm³/min
Or we convert it to cm³/min and have
MRR = 125000 mm³/min ÷ 1000
MRR = 125 cm³/min
