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Leno4ka [110]
3 years ago
13

the frequencies 10, 12, 23 and 45 Hz. (a) What is the minimum sampling rate required to avoid aliasing? (b) If you sample at 40

sps, which frequency components will be aliased?
Engineering
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

Sampling rate = 90

23 and 45 Hz

Explanation:

The Nyquist criteria states that the sampling frequency must be at least twice of the highest frequency component.

(a) We have 10, 12, 13, 23, and 45 Hz signals

The highest frequency component in this list is 45 Hz

So minimum sampling frequency needed to avoid aliasing would be

Sampling rate = 2*45 = 90 sps

Hence a sampling rate of 90 samples per second would be required to avoid aliasing.

(b) if a sampling rate of 40 sps is used then

40 = 2*f

f = 40/2 = 20 Hz

Hence 23 and 45 Hz components will be aliased.

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Answer:

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A gas flows through a one-inlet, one-exit control volume operating at steady state. Considering an adiabatic control volume with
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Answer:

b. equal to the specific entropy of the gas at the inlet.

Explanation:

Isentropic process is the process in which the entropy of the system remains unchanged. The word isentropic is formed from the combination of the prefix "iso" which means "equal" and the word entropy.

If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.

3 0
4 years ago
1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn
emmasim [6.3K]

Answer:

1. \dot Q=19600\ W

2. \dot Q=120\ W

Explanation:

1.

Given:

  • height of the window pane, h=2\ m
  • width of the window pane, w=1\ m
  • thickness of the pane, t=5\ mm= 0.005\ m
  • thermal conductivity of the glass pane, k_g=1.4\ W.m^{-1}.K^{-1}
  • temperature of the inner surface, T_i=15^{\circ}C
  • temperature of the outer surface, T_o=-20^{\circ}C

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

\dot Q=k_g.A.\frac{dT}{dx}

here:

A = area through which the heat transfer occurs = 2\times 1=2\ m^2

dT = temperature difference across the thickness of the surface = 35^{\circ}C

dx = t = thickness normal to the surface = 0.005\ m

\dot Q=1.4\times 2\times \frac{35}{0.005}

\dot Q=19600\ W

2.

  • air spacing between two glass panes, dx=0.01\ m
  • area of each glass pane, A=2\times 1=2\ m^2
  • thermal conductivity of air, k_a=0.024\ W.m^{-1}.K^{-1}
  • temperature difference between the surfaces, dT=25^{\circ}C

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

\dot Q=k_a.A.\frac{dT}{dx}

\dot Q=0.024\times 2\times \frac{25}{0.01}

\dot Q=120\ W

5 0
3 years ago
What’s the answer please help
irina1246 [14]

Answer:

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Explanation:

4 0
3 years ago
Gas chromatography separates compounds depending on their__________ . Benzene, m-xylene, and toluene have similar_________ , the
amm1812

Answer and Explanation:

Gas chromatography separates compounds depending on their **polarity and volatility**. Benzene, m-xylene, and toluene have similar **polarities**, therefore, the main basis for separation is **volatility**. The more volatile a component the ** higher its vapor pressure**, hence the more time it spends in the **gaseous mobile phase**, giving it a **shorter** retention time. Therefore, components of a liquid mixture will elute in order of **increasing boiling points/decreasing volatilities/increasing polarities with the stationary phase**.

3 0
3 years ago
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