Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e
r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Answer:
7.8 Mph
Explanation:
Rate of cycling = 1.1 rev/s
Rear wheel diameter = 26 inches
Diameter of sprocket on pedal = 6 inches
Diameter of sprocket on rear wheel = 4 inches
Circumference of rear wheel = \pi d=26\piπd=26π
Speed would be
\begin{gathered}\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s\end{gathered}Rate of cycling×Diameter of sprocket on rear wheelDiameter of sprocket on pedal×Circumference of rear wheel=1.1×46×26π=134.77432 inches/s
Converting to mph
1\ inch/s=\frac{1}{63360}\times 3600\ mph1 inch/s=633601×3600 mph
134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph134.77432 inches/s=134.77432×633601×3600 mph=7.65763 mph
The Speed of the bicycle is 7.8 mph
Answer:
a) Cr = Co - Fx / D
b) ΔC / Δx = ( CR - Cr ) / ( xR - xRo )
Explanation:
A) Derive an expression for the profile c(r) inside the tissue
F = DΔC / X = D ( Co - Cr ) / X ------ 1
where : F = flux , D = drug diffusion coefficient
X = radial distance between Ro and R
Hence : Cr = Co - Fx / D
B) Express the diffusive flux at outer surface of the balloon
Diffusive flux at outer surface = ΔC / Δx = CR - Cr / xR - xRo
Answer:
Find the answer in the attached in the order
Option a, Option c and Option b
Explanation:
Below is the program to separate odd and even numbers
<u>Explanation</u>:
<u>L1:</u>
mov ah,00
mov al,[BX]
mov dl,al
div dh
cmp ah,00
je EVEN1
mov [DI],dl
add OddAdd,dl
INC DI
INC BX
Loop L1
jmp CAL
<u>EVEN1:</u>
mov [SI],dl
add Even Add,dl
INC SI
INC BX
Loop L1
<u>CAL: </u>
mov ax,0000
mov bx,0000
mov al,OddAdd
mov bl,EvenAdd
MOV ax,4C00h
int 21h
end
The above program separates odd and even numbers from the array using 8086 microprocessor. It has odd numbers in 2000h and even numbers in 3000h.
