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zhannawk [14.2K]
4 years ago
10

What advantage might there be to having the encoder located on the motor side of the gearhead instead of at the output shaft of

the gearhead (in other words so that it would rotate at the speed of the armature rather than at the speed of the output shaft of the gearhead)?
Engineering
1 answer:
pickupchik [31]4 years ago
7 0

The most accurate answer to that process is definitely precision. The Rotary encoder is an electro-mechanical device that converts the angular position or motion of a shaft or axle to analog or digital output signals. The efficiency of these devices is subject to the position and angle of the axis in front of the encoder.

Most cars use reduction systems in their gearboxes that convert a certain signal input into an output. Mechanically for example, a 20: 1 reduction box already infers that if there is a revolution in the input at the output there are 20. That same transferred to the encoder pulses would imply greater precision.

For example a decoder with 50 holes would have to read 1000 pulses (50 * 20) which is basically a degree of accuracy of 0.36 degrees. In this way it is possible to conclude that if the assembly of the encoder is carried out next to the motor and not at the output, it can be provided with greater precision at the time of reading.

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A hollow steel shaft with and outside diameter of (do)-420 mm and an inside diameter of (di) 350 mm is subjected to a torque of
-Dominant- [34]

Answer:

a.  \tau=51.55 MPa

b.\tau=42.95MPa

c.\theta=7.67\times 10^{-3} rad.

Explanation:

Given: D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa

We know that

\dfrac{\tau}{J}=\dfrac{T}{r}=\dfrac{G\theta}{L}

J for hollow shaft J=\dfrac{\pi (D_o^4-D_i^4)}{64}

(a)

 Maximum shear stress \tau =\dfrac{16T}{\pi Do^3(1-K^4)}

      K=\dfrac{D_i}{D_o}⇒K=0.83

\tau =\dfrac{16\times 300\times 1000}{\pi\times 0.42^3(1-.88^4)}

   \tau=51.55 MPa

(b)

We know that \tau \alpha r

So \dfrac{\tau_{max}}{\tau}=\dfrac{R_o}{r}

\dfrac{51.55}{\tau}=\dfrac{210}{175}

\tau=42.95MPa

(c)

\dfrac{\tau_{max}}{R_{max}}=\dfrac{G\theta }{L}

\dfrac{51.55}{210}=\dfrac{80\times 10^3\theta }{2500}

\theta=7.67\times 10^{-3} rad.

8 0
4 years ago
When a technician is ohming out a load in a circuit, what reading should the technician get if the load is good
xz_007 [3.2K]

When a technician is ohming out a load in a circuit, the reading that the technician should get if the load is good is some numbers.

<h3>What does Ohming out a circuit mean?</h3>

The term “Ohming out a motor” is known to be the act of measuring the electrical resistance that is present in the motor windings and comparing it with the normal values.

Note that When a technician is ohming out a load in a circuit, the reading that the technician should get if the load is good is some numbers as it will tell if there is a measure of comparison or not.

Learn more about technician from

brainly.com/question/2328085

#SPJ1

6 0
2 years ago
Refrigerant 134a at p1 = 30 lbf/in2, T1 = 40oF enters a compressor operating at steady state with a mass flow rate of 200 lb/h a
Mazyrski [523]

Answer:

a) \mathbf{Q_c = -3730.8684 \ Btu/hr}

b) \mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}

Explanation:

From the properties of Super-heated Refrigerant 134a Vapor at T_1 = 40^0 F, P_1 = 30  \ lbf/in^2 ; we obtain the following properties for specific enthalpy and specific entropy.

So; specific enthalpy h_1 = 109.12 \ Btu/lb

specific entropy s_1 = 0.2315 \ Btu/lb.^0R

Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at P_2 = 160 \ lbf/in^2 ; we obtain the following properties:

h_2  = 115.91 \ Btu/lb\\\\ s_2 = 0.2157 \ Btu/lb.^0R

Given that the power input to the compressor is 2 hp;

Then converting to  Btu/hr ;we known that since 1 hp = 2544.4342 Btu/hr

2 hp = 2 × 2544.4342 Btu/hr

2 hp = 5088.8684 Btu/hr

The steady state energy for a compressor can be expressed by the formula:

0 = Q_c -W_c+m((h_1-h_e) + \dfrac{v_i^2-v_e^2}{2}+g(\bar \omega_i - \bar \omega_e)

By neglecting kinetic and potential energy effects; we have:

0 = Q_c -W_c+m(h_1-h_2) \\ \\ Q_c = -W_c+m(h_2-h_1)

Q_c = -5088.8684 \ Btu/hr +200 \ lb/hr( 115.91 -109.12) Btu/lb  \\ \\

\mathbf{Q_c = -3730.8684 \ Btu/hr}

b)  To determine the entropy generation; we employ the formula:

\dfrac{dS}{dt} =\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c

In a steady state condition \dfrac{dS}{dt} =0

Hence;

0=\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c

\sigma _c = m( s_1 -s_2)  - \dfrac{Qc}{T}

\sigma _c = [200 \ lb/hr (0.2157 -0.2315) \ Btu/lb .^0R  - \dfrac{(-3730.8684 \ Btu/hr)}{(40^0 + 459.67^0)^0R}]

\sigma _c = [(-3.16 ) \ Btu/hr .^0R  + (7.4667 ) Btu/hr ^0R}]

\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}

5 0
3 years ago
Thermoplastics burn upon heating. a)-True b)- false?
miv72 [106K]

Yes, It is indeed true that Thermoplastics and thermosettingplastics burn upon heating.

4 0
3 years ago
Consider the function f(n) = n
alexira [117]

get off this

Explanation:

4 0
2 years ago
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