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Virty [35]
3 years ago
13

In what situation you would prefer to use a successive approximation ADC over flash ADC?

Engineering
1 answer:
Levart [38]3 years ago
8 0

Answer and Explanation:

Both both successive approximation normally known SAR type and flash type ADC are used for the conversion of analog signal to digital signal successive approximation type ADC are preferred over flash type when we need fix conversion time and better accuracy and if we have don/t matter with speed of the conversion because flash type has better speed of conversion than SAR type

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Why is it reasonable to say that no system is 100% efficient?​
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A 625 g basketball and a 58.5 g tennis ball are dropped from a height of d = 1.5 m onto the floor. The coefficient of restitutio
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Explanation:

Solution of the problem is given in the attachments.

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A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa
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2 years ago
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Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz
butalik [34]

Answer:

Exit velocity V_2=1472.2 m/s.

Explanation:

Given:

At inlet:

P_1=100 bar,T_=600°C,V_1=35m/s

Properties of steam at 100 bar and 600°C

        h_1=3624.7\frac{KJ}{Kg}

At exit:Lets take exit velocity V_2

We know that if we know only one property inside the dome  then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle  is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

   h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}

 h_2=2541.62\frac{KJ}{Kg}

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

In the case of adiabatic nozzle Q=0,W=0

3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0

V_2=1472.2 m/s

So Exit velocity V_2=1472.2 m/s.

4 0
3 years ago
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