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vfiekz [6]
3 years ago
8

A microwave transmitter has an output of 0.1 W at 2 GHz. Assume that this transmitter is used in a microwave communication syste

m where the transmitting and receiving antennas are parabolas, each 1.2 m in diameter. a. What is the gain of each antenna in decibels
Engineering
1 answer:
ehidna [41]3 years ago
5 0

Answer:

The gain of each antenna in decibels  is 353 .33 dB.

Explanation:

Given:

Frequency = 2 GHz

Diameter  =  1.2 m

To Find:

The gain of each antenna in decibels = ?

Solution:

Relationship between antenna gain and effective area

= \frac{4 \pi f^2 A_e}{c^2}-----------------------------------(1)

Where

f is frequency

A_e  is  effective area

c is the speed of light

A_e = effective area of a parabolic antenna with a face area of A is 0.56A

A_e =  0.56(\pi r^2)

r  is the radius

r = \frac{1.2}{2}

r =  0.6

A_e = 0.56(\pi (0.6)^2)

A_e = 0.56( 0.36 \pi)

A_e = 0.2016 \pi

Substituting the values in eq(1)

= \frac{4 \pi (2 \times 10^9)^2 \times 0.2016\pi}{(3 \times 10^8)^2}

= \frac{4 \pi (4 \times 10^{18} ) \times 0.2016\pi}{(9 \times 10^{16})}

= \frac{4 \pi (4 \times 10^{2} ) \times 0.2016\pi}{9}

= \frac{31.80 \times 10^{2}}{9}

= 353 .33 dB

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Len [333]

This is not a valid question. Please try again.

4 0
3 years ago
A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round
Harlamova29_29 [7]

Answer:

\frac{Q_{square}}{Q_{circle}} =  0.785  

Explanation:

given data

types of drinking straws

  1. square cross-sectional shape
  2. round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square  = perimeter of circle  

4 × S = π × D

here S is length and D is diameter

S = \frac{\pi D}{4}        ....................1

and

ratio of  flow rate through the square and circle is here

\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}  

\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}  

\frac{Q_{square}}{Q_{circle}} =  0.785  

4 0
3 years ago
A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory
Molodets [167]

Answer:

Explanation:

Mean temperature is given by

T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}

Tmean = (Ti + T∞)/2

T_mean = 107.5^{0}

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

\mu = 221.6 × 10⁻⁷N.s/m²

\kappa = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α  = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

              = 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/\mu

Substituting for Pv

Re = 4m/(πD\mu)

     = (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

     = 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0
3 years ago
Differentiate between isohyetal method and arithmetical average method of rainfall​
prohojiy [21]

Answer:

Explained below

Explanation:

The isohyetal method is one used in estimating Rainfall whereby the mean precipitation across an area is gotten by drawing lines that have equal precipitation. This is done by the use of topographic and other data to yield reliable estimates.

Whereas, the arithmetic method is used to calculate true precipitation by the way of getting the arithmetic mean of all the points or arial measurements that will be considered in the analysis.

7 0
3 years ago
What does it mean to say that PEER is a data-driven, consumer-centric, and comprehensive system?
Reika [66]

Answer:

have you heard of gnoogle?

Explanation:have you heard of goongle?

3 0
3 years ago
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