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Vinvika [58]
3 years ago
10

Sometimes, we need an amplifier with an accurate gain. Even using 1% tolerance resistors may not be sufficient to provide the re

quired gain tolerance. In these designs, trimmer resistors must be used that can be adjusted after the circuit is built. Each resistor in the circuit can be replaced by a series combination of a fixed resistor and an adjustable resistor. Consider the inverting amplifier below with R1=47kΩ and R2=450kΩ. The source signal internal resistance varies from 0 to Rsmax=1kΩ. Only one trimmer resistor can be added, and it is going to be put in series with the feedback resistor. What are the lowest and highest values R must be able to have so that the gain can be adusted to Av=−10±0.5%? Assume that all resistors have 1% tolerance.

Engineering
1 answer:
vlabodo [156]3 years ago
5 0

Answer:

The value of R_low is 13.1 kΩ while that of R_high is 36.8 kΩ

Explanation:

The diagram is as given as

The circuit is an inverting op amp for which the gain is given as

A_v=\dfrac{R_2+R}{R_1+R_z}

Here R2 is given as 450 kΩ.

R1 is given as 47 kΩ.

The gain is given as

A_v=-10\pm 0.5*\dfrac{10}{100}\\A_v=-10\pm 0.05

For the Lowest value of R the value of the internal resistance rz is 0 and the gain is -10+0.05 so

A_v=-\dfrac{R_2+R}{R_1+R_z}\\-10+0.05=-\dfrac{450+R_l_o_w}{47+0}\\-9.95=-\dfrac{450+R_l_o_w}{47+0}\\R_l_o_w=13.1

So the value of R_low is 13.1 kΩ

For the Highest value of R the value of the internal resistance rz is 1 kΩ and the gain is -10-0.05 so

A_v=-\dfrac{R_2+R}{R_1+R_z}\\-10-0.05=-\dfrac{450+R_{high}}{47+1}\\-10.05=-\dfrac{450+R_{high}}{47+1}\\R_{high}=36.8

So the value of R_high is 36.8 kΩ

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The controller determines if a(n) error exists by calculating the difference between the SP and the PV.

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The Control system is one where it entails if the output is one that has an effect on the input quantity.

So it uses the PV(Process Variable) set against the SP(Setpoint) to know if an error exists.

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3 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
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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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3 years ago
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Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

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