Whats is the purpose of the stator winding
1 answer:
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Answer:
0.08kg/s
Explanation:
For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.
The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.
finally you use the two previous equations to make a system and find the mass flows
I attached procedure
Answer:
work done = 48.88 × J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 × J
Answer:
the heat transfer from the pipe will decrease when the insulation is taken off for r₂<
where;
r₂ = outer radius
= critical radius
Explanation:
Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .
The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than critical radius (r₂< ) 0, and reaches a maximum when r₂ =
, and starts to decrease for r₂<
. Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂<
.
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 )