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mrs_skeptik [129]
4 years ago
10

Determine the resolution and the quantization error in volts for a 4-bit A/D converter that has a full scale range of EFSR =5 V

. If the desired quantization error should be within 1 mV, how many bits are needed?
Engineering
1 answer:
valentinak56 [21]4 years ago
8 0

Answer:

minimum of 12 bits is required

Explanation:

Full scale range Ef = 5v = L

Resolution R_{ADC} = \frac{L}{2^n -1}

n = 4 bit

R_{ADC} = \frac{5}{2^4 -1} = 0.333 VQuartization error  is[tex] \epsilon = \frac{1}{2} R_{ADC}

               = \frac{1}{2} 0.333 = 0.166 V

we know if error is less than 0.001 V then we have

1/2 R_{ADC} \leq 0.001

R-{ADC} \leq 0.002

0.002\geq \frac{L}{2^n -1}

\frac{L}{2^n -1} \leq 0.002

\frac{L}{0.002} \leq 2^n -1

Solving for n we get

n\geq 11.288

n = 12

Therefore minimum of 12 bits is required

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As the temperature of a thermal radiator is increased Group of answer choices the object appears redder. the object appears blue
Elanso [62]

Answer:

As the temperature of a thermal radiator is increased

Group of answer choices

  • the object appears redder.
  • the object appears bluer.
  • the object emits more power for the same area.
  • the object emits less power for the same area.
  • the object expands to keep the same power per units area.

<em>When the temperature of a thermal radiator increases ;</em>

  • <em>the object emits more power for the same are</em>
  • <em>the object becomes bluer</em>

Explanation:

Thermal radiation involves the transfer of heat between molecules of two substances without direct contact with each other. When a body is heated to a given temperature it begins to emit light which is transferred to nearby objects as thermal radiation. The medium through which the heat is transferred could be liquid, solid, or in a vacuum.

<h3>How temperature affects thermal radiation.</h3>

Temperature determines the amount of heat that is been radiated from a body. An increase in temperature would increase the thermal radiation of the body. The increase in the heat radiation results to increase in the thermal energy of the body. Also when a body is heated it tends to be bluer than a cool object, this is caused by the rapid movement of the molecules.

Therefore When the temperature of a thermal radiator increases ;

  • the object emits more power for the same are
  • the object becomes bluer
4 0
4 years ago
Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra
gladu [14]

Answer:

a) T_{2}=837.2K

b) e=91.3 %

Explanation:

A) First, let's write the energy balance:

W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})  (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 \frac{kJ}{kgK} And its specific R constant is 0.287 \frac{kJ}{kgK}.

The only unknown from the energy balance is T_{2}, so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K

B) The isentropic efficiency (e) is defined as:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}

Where {h_{2s} is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because h_{2}-h_{1} can be obtained from the energy balance  \frac{W}{m}=h_{2}-h_{1}

h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}

An entropy change for an ideal gas with  constant Cp is given by:

s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})

Applying logarithm properties:

ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}

Then,

T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K

So, now it is possible to calculate h_{2s}-h_{1}:

h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}

Finally, the efficiency can be calculated:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3 %

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In some states, repair shops can accept used oil from do-it-yourselfers or ____________.
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Answer:

C. Homeowners

Explanation:

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