Given:
outer radius, R' = 10 m
inner diameter, d = 2 m
inner radius, R =
= 1 m
surface temperature, T' = 
Thermal conductivity of soil, K = 0.52 W/mK
Solution:
To calculate the thermal temperature of conductor, T, we know amount of heat, Q is given by :
Q = 
500 =
T = 68.86 +20 =
Therefore, outside surface temperature is
Answer:
Yes, the flow is turbulent.
Explanation:
Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.
Given:
Diameter of pipe is 10mm.
Velocity of the pipe is 1m/s.
Temperature of water is 200°C.
The kinematic viscosity at temperature 200°C is
m2/s.
Calculation:
Step1
Expression for Reynolds number is given as follows:

Here, v is velocity,
is kinematic viscosity, d is diameter and Re is Reynolds number.
Substitute the values in the above equation as follows:


Re=64226.07579
Thus, the Reynolds number is 64226.07579. This is greater than 2000.
Hence, the given flow is turbulent flow.
Answer:
Temperature at center of apples = 11.2⁰C
Temperature at surface of apples = 2.7⁰C
Amount of Heat transferred = 17.2kJ
Explanation:
The properties of apple are given as:
k = 0.418 W/m.°C
ρ = 840 kg/m³
Cр = 3.81 kJ/kg.°C
α = 1.3*10 ⁻⁷ m²/s
h = 8 W/m².°C
d = 0.09m
r = 0.045m
t = 1 hour = 3600s
<h2>Solution</h2>
Biot number is given as:

The constants λ₁ and A₁ corresponding to Biot number (from the table) are:
λ₁ = 1.476
A₁ = 1.239
Fourier Number is:

As Fourier Number > 0.2 , one term approximates solutions are applicable
The temperature at the center of apples, The temperature at surface of apples and Amount of heat transfer is found in the ATTACHMENT.
Answer: 0.93 mA
Explanation:
In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.
We can express the resistance as follows:
R = ρ.L/A
In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:
A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2
Replacing by the values, we get R as follows:
R = 1.4 1010 Ω
Applying Ohm’s Law, and solving for the current I:
I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA
Answer:
Ps=19.62N
Explanation:
The detailed explanation of answer is given in attached files.