Answer:
hello your question has some missing part attached below is the complete question
answer : steady state temperature = 419.713k ≈ 218.7⁰c
Time required to reach a junction ≈ 5 secs
Explanation:
The detailed solution of the given problem is attached below but the solution to the subsequent problem from which the question you asked is referenced to( problem 1 ), is not attached because it was not part of the question you asked
Está constituido por dos bobinas de material conductor, devanadas sobre un núcleo cerrado de material ferromagnético, pero aisladas entre sí eléctricamente. ... Las bobinas o devanados se denominan primario y secundario según correspondan a la entrada o salida del sistema en cuestión, respectivamente.
Answer:
Rate of heat transfer to the room air per meter of pipe length equals 521.99 W/m
Explanation:
Since it is given that the radiation losses from the pipe are negligible thus the only mode of heat transfer will be by convection.
We know that heat transfer by convection is given by

where,
h = heat transfer coefficient = 10.45
(free convection in air)
A = Surface Area of the pipe
Applying the given values in the above formula we get

Answer:
The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.
Solution
Given:
Length = 48 m
Width = 12 m
Depth = 3m
Flow rate = 4 m 3 /s
Water density = 10 3 kg/m 3
Dynamic viscosity = 1.30710 -3 N.sec/m
Now,
At the minimum particular diameter it is stated as follows:
The Reynolds number= 0.1
Thus,
0.1 =ρVTD/μ
VT = Dp² ( ρp- ρ) g/ 10μ²
Where
gn = The case/issue of sedimentation
VT = Terminal velocity
So,
0.1 = Dp³ ( ρp- ρ) g/ 10μ²
This becomes,
0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²
= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)
dp³=3.1343 * 10 ^⁻12
Dp minimum= 1.474 * 10 ^⁻4 meters.