Answer:
maximum allowable electrical power=4.51W/m
critical radius of the insulation=13mm
Explanation:
Hello!
To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)
Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and conduction as shown in the attached image
to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient
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:
hello your question incomplete attached below is the complete question and detailed solution
Answer : Csf = 0.0131
Explanation:
Attached below is the detailed solution
Given data :
ΔTe = 17.1⁰c calculated as ;Ts - Tsat = ( 117.1 - 100 )
Pe = 957.9 kg/m^3
Cp1e = 4217 j/kgk
<em>U</em>e = 279 * 10^-6 n. s / m^2
Pre = 1.76
hfg = 2.257 * 10^6 J/kg
Pv = 0.5955 kg/m^3
б = 0.0589 N/m
q" = 664 * 10^3 w/m^2 ( calculated )
Input these values into equation 1 as contained in the detailed solution
Csf = 0.0131
Answer:
you get electrocuted...........
