Why does my man bun not have its own erodynamics
2 answers:
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Answer:
A) attached below
B) I₁ = 18.1 A , I₂ = 69.39 A
C) V( magnitude) = 454.5 ∠ 5.04° V , Voltage regulation = ≈ -1.2%
Explanation:
A) Schematic diagram attached below
attached below
<u>B) magnitude of primary and secondary winding currents </u>
I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )
VL = Line voltage = 208
cos∅ ( power factor ) = 0.8
P = 20 * 10^3 watts
insert values into equation 1
I₂ = 69.39 A
I₁ ( primary current ) = I₂V2 / V1
I₁ = ( 69.39 * 120 ) / 460 = 18.1 A
<u>C ) Calculate the Primary voltage magnitude and the Voltage regulation</u>
V(magnitude ) = Vp + ( I₁ ∠∅ ) Req ( 1 + j2 = 2.24 ∠63.43° )
= 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )
= 460 + 40.544 ∠ 100.3°
∴ V( magnitude) = 454.5 ∠ 5.04° V
<em>Voltage regulation </em>
= ((Vmag - V1) / V1 )) * 100
= (( 454.5 - 460 / 460 )) * 100
= -1.195 % ≈ -1.2%
Answer:
P_p = 27000 psf
Explanation:
given,
height of the retaining wall = h = 12 ft
internal angle of friction (∅)= 30°
unit weight = 125 pcf
Rankine passive earth pressure = ?
k_p is the coefficient of passive earth pressure
k_p = 3
Passive earth pressure
P_p = 27000 psf
Rankine passive earth pressure on the wall is equal to P_p = 27000 psf
Answer:
110 m or 11,000 cm
Explanation:
- let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
- let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
- let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively
M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s
C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c
<u>Using effectiveness-NUT method</u>
- <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>
C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c
C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c
From the result above cold fluid heat capacity rate is smaller
Dimensionless heat capacity rate, C = minimum capacity/maximum capacity
C= C<em>min</em>/C<em>max</em>
C = 5.016/8.62 = 0.582
.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>
Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)
Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW
.<em>3 Third, we determine the actual heat transfer rate, Q</em>
Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)
Q = (5.016 kW/°c)(80 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW
.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε
ε<em> </em>= Q/Qmax
ε <em>= </em>303.66 kW/702.24 kW
ε = 0.432
.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>
NTU =
NTU =
NTU = 0.661
<em>.6 sixth, we determine Heat Exchanger surface area, As</em>
From the question, the overall heat transfer coefficient U = 640 W/m²
As =
As =
As = 5.18 m²
<em>.7 Finally, we determine the length of the heat exchanger, L</em>
L =
L =
L= 109.91 m
L ≅ 110 m = 11,000 cm