List and explain the major features that the following building designs must have to relate directly to their functions.
1 answer:
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Answer:
Re=100,000⇒Q=275.25
Re=500,000⇒Q=1,757.77
Re=1,000,000⇒Q=3060.36
Explanation:
Given:
For air =25°C ,V=8 m/s
For surface =179°C
L=2.75 m ,b=3 m
We know that for flat plate
⇒Laminar flow
⇒Turbulent flow
<u> Take Re=100,000:</u>
So this is case of laminar flow
From standard air property table at 25°C
Pr= is 0.71 ,K=26.24
So
Nu=187.32 ()
187.32=
⇒h=1.78
heat transfer rate =h
=275.25
<u> Take Re=500,000:</u>
So this is case of turbulent flow
Nu=1196.18 ⇒h=11.14
heat transfer rate =h
=11.14(179-25)
= 1,757.77
<u> Take Re=1,000,000:</u>
So this is case of turbulent flow
Nu=2082.6 ⇒h=19.87
heat transfer rate =h
=19.87(179-25)
= 3060.36
Answer:
V = 30 V
vector (E) = -10*a_p + 17.32*a_Q - 24*a_z
mag(E) = 31.24 V/m
dV / dN = 31.2 V /m
a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z
p_v = 276 pC / m^3
Explanation:
Given:
- The Volt Potential in cylindrical coordinate system is given as:
- The point P is at p = 3 , Q = 60 , z = 2
Find:
values at P for:
a.) V
b.) E
c.) E
d.) dV/dN
e.) aN
f.) rhov in free space
Solution:
a)
Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:
b)
To compute the Electric field from Volt potential we have the following relation:
E = - ∀.V
Where, ∀ is a del function which denoted:
∀ =
Hence,
Plug in the values for point P:
c)
The magnitude of the Electric Field is given by:
E = √((-10)^2 + (17.32)^2 + (24)^2)
E = √975.9824
E = 31.241 N/C
d)
The dV/dN is the Field Strength E in normal to the surface with vector N given by:
dV / dN = | - ∀.V |
dV / dN = | E | = 31.2 N/C
e)
a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:
f)
The charge density in free space:
p_v = E.∈_o
Where, ∈_o is the permittivity of free space = 8.85*10^-12
p_v = 31.24.8.85*10^-12
p_v = 276 pC / m^3