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zlopas [31]
3 years ago
5

Q1-A Lossless transmission line is 80 cm long and operates at a frequency of 500 MHz. Two line parameters are L=0.15μH/m and C=9

0pf/m. Find the characteristic impedance the phase constant, the velocity on the line, and the input impedance for ZL =80 Ω
Engineering
1 answer:
hjlf3 years ago
6 0

The characteristic impedance, the phase constant, the velocity on the line, and the input impedance are respectively;

A) Z_o = 40.82 Ω

B) β = 11.543 rad/m

C) v_p = 2.72 × 10^(8) m/s

D) Z_in = 115.91 Ω

We are given;

Inductance; L = 0.15 μH/m = 0.15 × 10^(-6) H/m

Capacitance; C = 90 pf/m = 90 × 10^(-12) f/m

Frequency; f = 500 MHz = 500 × 10^(6) Hz

Load impedance; Z_l = 80 Ω

Length of transmission line; l = 80cm = 0.8m

A) Formula for characteristic impedance is;

Z_o = √(L/C)

Thus;

Z_o = √((0.15 × 10^(-6))/(90 × 10^(-12)))

Z_o = 40.82 Ω

B) Formula for the phase constant is;

β = ω√(LC)

Where;

ω = 2πf

Thus;

β = (2π × 500 × 10^(6))√(0.15 × 10^(-6) × 90 × 10^(-12))

β = 11.543 rad/m

C) Formula for phase velocity is;

v_p = ω/β

v_p = (2π × 500 × 10^(6))/11.543

v_p = 2.72 × 10^(8) m/s

D) Formula for the input impedance is;

Z_in = Z_o[(Z_l + Z_o*tanβl)/(Z_o + Z_l*tanβl)]

Z_in = 40.82[(80 + 40.82*tan(11.543*0.8))/(40.82 + 80*tan(11.543*0.8))]

Z_in = 40.82(72.1335/25.4031)

Z_in = 115.91 Ω

Read more about impedance at; brainly.com/question/25153504

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Answer:

a) the rate of heat transfer from the pipe to the air is 23.866 watts

b) YES, the rate of heat transfer changes to 3518.61 watt

Explanation:

Given that:

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Bulk fluid temperature of 27°C

we know that

hD/k = 0.028 (Re)^0.8 (Pr)^0.33

Outside diameter of pipe = 6.75 cm

length of the pipe = 34.7 m

velocity of air = 7.6 m/s

Cp of air = 1.005 kJ/Kgk

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thermal conductivity of air = 2.624 × 10⁻⁵ kw/m.k

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hD/k = 0.028 (Re)^0.8 (Pr)^0.33

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a)

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Q = hAΔT

Q = 0.0414 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 23.866 watts

therefore the rate of heat transfer from the pipe to the air is 23.866 watts

b)

Now the flow direction changes to parallel flow, then

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([34.7 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

h = 6.1036 w/m².k

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so to find the rate of heat transfer Q

Q = hAΔT

Q = 6.1036 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 3518.61 watt

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Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
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Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

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Answer:

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Explanation:

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Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

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