A student engineer is given a summer job to find the drag force on a new unmaned aerial vehicle that travels at a cruising speed
of 320 km/h in air at 15 C. The student decided to build a 1/50 scale model and test it in a water tunnel at the same temperature.
(a) What conditions are required to ensure the similarity of scale model and prototype.
(b) Find the water speed required to model the prototype.
(c) If the drag force on the model is 5 kN, determine the drag force on the prototype
(a) What conditions are required to ensure the similarity of scale model and prototype.
(b) Find the water speed required to model the prototype.
(c) If the drag force on the model is 5 kN, determine the drag force on the prototype
1 answer:
Answer:
b. 1232.08 km/hr
c. 1.02 kn
Explanation:
a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.
See attachment for the remaining.
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Answer:
Glass: Low-Loss dielectric
α = 8.42*10^-11 Np/m
β = 468.3 rad/m
λ = 1.34 cm
up = 1.34*10^8 m/s
ηc = 168.5 Ω
Tissue: Quasi-Conductor
α = 9.75 Np/m
β = 12.16 rad/m
λ = 51.69 cm
up = 0.52*10^8 m/s
ηc = 39.54 + j 31.72 Ω
Wood: Good conductor
α = 6.3*10^-4 Np/m
β = 6.3*10^-4 Np/m
λ = 10 km
up = 0.1*10^8 m/s
ηc = 6.28*( 1 + j )
Explanation:
Given:
Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz
Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.
Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz
Find:
Determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, up, and ηc:
Solution:
- We need to determine the loss tangent to determine category of the medium as follows:
σ / w*εr*εo
Where, w is the angular speed of wave
εo is the permittivity of free space = 10^-9 / 36*pi
- Now we classify as follows:
- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.
Glass: Low-Loss dielectric
Tissue: Quasi-Conductor
Wood: Good conductor
- Now we will use categorized material base equations from Table 17-1 as follows:
Glass: Low-Loss dielectric
α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)
α = 8.42*10^-11 Np/m
β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)
β = 468.3 rad/m
λ = 2*pi / β = 2*pi / 468.3
λ = 1.34 cm
up = λ*f = 0.0134*10^10
up = 1.34*10^8 m/s
ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)
ηc = 168.5 Ω
Tissue: Quasi-Conductor
α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)
α = 9.75 Np/m
β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)
β = 12.16 rad/m
λ = 2*pi / β = 2*pi / 12.16
λ = 51.69 cm
up = λ*f = 0.5169*100*10^6
up = 0.52*10^8 m/s
ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5
ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5
ηc = 39.54 + j 31.72 Ω
Wood: Good conductor
α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )
β = α = 6.3*10^-4 Np/m
λ = 2*pi / β = 2*pi / 6.3*10^-4
λ = 10 km
up = λ*f = 10,000*1*10^3
up = 0.1*10^8 m/s
ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4
ηc = 6.28*( 1 + j )