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Maurinko [17]
3 years ago
10

You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from

rest, then it takes 1.2 second for it to reach 63.2% of the steady state speed. You attach a sensor to the motor which measures its angle very accurately. The measured angle is used by the controller whose aim is to provide voltage to the motor and make the angle of the motor same as the desired angle.
Required:
a. What is the transfer function of the motor, from voltage to speed?
b. Draw the block diagram of the system with plant, controller and the feedback path.
c. What is the system type (i.e. the type of GK)?
d. If the desired speed is a constant (say 100 rad/s). What should be range of gain K of the proportional controller, so that steady state error in the speed is less than 10%? (i.e. what number K will ensure that, (desired speed - actual steady state speed)/desired speed < 0.1 ?)
e. What happens the above % error in the steady state speed as the gain K is increased?
Engineering
1 answer:
Aleksandr [31]3 years ago
7 0

Answer:

a) \frac{Ws}{Es}  = \frac{200}{1+1.2s}

b) attached below

c) type zero system

d) k > \frac{g}{200}

e) The gain K increases above % error as the  steady state speed increases

Explanation:

Given data:

Motor voltage  = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; \frac{K1}{1+St} be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= \frac{Ws}{Es}  = \frac{200}{1+1.2s}

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > \frac{g}{200}

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Explanation:

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There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter
weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3  ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o =  8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

                                        u_e = 0.5*e_o * E^2

- Plug in the values given:

                                        u_e = 0.5*8.854 *10^-12 *145^2

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3 years ago
Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba
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Answer:

Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.

Explanation:

The optimum cutting speed for the minimum cost

V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)

Where,

C,n = Taylor equation parameters

T_h =Tool changing time in minutes

C_e=Cost per grinding per edge

C_m= Machine and operator cost per minute

On comparing with the Taylor equation VT^n=C,

Tool life,

T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)

Given that,  

Cost of operator and machine time=\$40/hr=\$0.667/min

Batch setting time = 2 hr

Part handling time: T_h=2.5 min

Part diameter: D=73 mm =73\times 10^{-3} m

Part length: l=250 mm=250\times 10^{-3} m

Feed: f=0.30 mm/rev= 0.3\times 10^{-3} m/rev

Depth of cut: d=3.5 mm

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, C_e= \$20/15+2=\$3.33/edge

Tool changing time, T_t=3 min.

C= 80 m/min

n=0.130

(a) From equation (i), cutting speed for the minimum cost:

V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}

\Rightarrow 47.7 m/min

(b) From equation (ii), the tool life,

T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}

\Rightarrow T=53.4 min

(c) Cycle time: T_c=T_h+T_m+\frac{T_t}{n_p}

where,

T_m= Machining time for one part

n_p= Number of pieces cut in one tool life

T_m= \frac{l}{fN} min, where N=\frac{V_{opt}}{\pi D} is the rpm of the spindle.

\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc

So, the number of parts produced in one tool life

n_p=\frac {T}{T_m}

\Rightarrow n_p=\frac {53.4}{4.01}=13.3

Round it to the lower integer

\Rightarrow n_p=13

So, the cycle time

T_c=2.5+4.01+\frac{3}{13}=6.74 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc

(e) Total time to complete the batch= Sum of setup time and production time for one batch

=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times4.01}{457}=0.4387=43.87\%

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Cost per edge,

C_e= \$8/6=\$1.33/edge

Tool changing time, T_t=1min

C= 650 m/min

n=0.30

(a) Cutting speed for the minimum cost:

V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min [from(i)]

(b) Tool life,

T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min [from(ii)]

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc

n_p=\frac {7}{0.53}=13.2

\Rightarrow n_p=13 [ nearest lower integer]

So, the cycle time

T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc

(e) Total time to complete the batch=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.53}{275.5}=0.0962=9.62\%

Similarly, for the ceramic tool:

C_e= \$10/6=\$1.67/edge

T_t-1min

C= 3500 m/min

n=0.6

(a) Cutting speed:

V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}

\Rightarrow V_{opt}=2105 m/min

(b) Tool life,

T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc

n_p=\frac {2.33}{0.091}=25.6

\Rightarrow n_p=25 pc/tool\; life

So,

T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc

(e) Total time to complete the batch

=2\times60+ {50\times 2.63}=251.5 min=4.19 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.091}{251.5}=0.0181=1.81\%

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2. What is the Function of the Camshaft in an Internal Combustion Engine?
mamaluj [8]

Answer:

camshaft, in internal-combustion engines, rotating shaft with attached disks of irregular shape (the cams), which actuate the intake and exhaust valves of the cylinders.

Explanation:

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3 years ago
If the surface energy of a magnesium oxide - nickel oxide (MgO-NiO) solid solution is 1.05 J/m2 and its elastic modulus is 198 G
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Answer:

The maximum length is 3.897×10^-5 mm

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elastic modulus = 198 GPa = 198×10^9 Pa

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Strain = stress/elastic modulus = 27×10^6/198×10^9 = 1.36×10^-4

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