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3 years ago

What is the IMA of this pulley belt system if the diameter of the input

Engineering

1 answer:

Stella [2.4K]3 years ago

8 0

Answer:

2.8

Explanation:

The ideal mechanical advantage of the pulley IMA = D'/D where D' = diameter of output pulley = 7 inches and D = diameter of input pulley = 2.5 inches

So, IMA = D'/D

= 7/2.5

= 2.8

So, the ideal mechanical advantage of the pulley IMA = 2.8

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An incremental encoder is rotating at 15 rpm. On the wheel there are 40 holes. How many degrees of rotation would 1 pulse be?

elena-s [515]

Answer:

1 pulse rotate = 9 degree

Explanation:

given data

incremental encoder rotating = 15 rpm

wheel holes = 40

solution

we get here first 1 revolution time

as 15 revolution take = 60 second

so 1 revolution take = $\frac{60}{15}$

1 revolution take = 4 seconds

and

40 pulse are there for 1 revolution

40 pulse for 360 degree

so 1 pulse rotate is = $\frac{360}{40}$

1 pulse rotate = 9 degree

3 0

3 years ago

Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an

ser-zykov [4K]

Answer:

volumetric flow rate = $0.0251 m^3/s$

Velocity in pipe section 1 = $6.513m/s$

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the volume will be mass/density

25/997 = $0.0251 m^3/s$

volumetric flow rate = $0.0251 m^3/s$

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s$

<em>Pipe section B:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s$

7 0

3 years ago

The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc

madreJ [45]

Answer:

Option B

$116 ft/s^{2}$

Explanation:

$\theta=2 rev=2(2\pi)=4\pi$

$\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-4^{2})$

$\omega_f=8.14 rads/s$

$v=r\omega=1.75*8.14=14.245 ft/s$

Centripetal acceleration $=\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}$

Tangential component=dr=2*1.75=3.5

$Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}$

5 0

3 years ago

The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (

mr Goodwill [35]

Answer:

a) the output voltage is 50 V

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V $_s$ /( 1 - D )

given that; V $_s$ = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

- the average inductor current

$I_L$ = V $_s$ / ( 1 - D )²R

given that R = 12.5 Ω, V $_s$ = 20 V, D = 0.6

we substitute

$I_L$ = 20 / (( 1 - 0.6 )² × 12.5)

$I_L$ = 20 / (( 0.4)² × 12.5)

$I_L$ = 20 / ( 0.16 × 12.5 )

$I_L$ = 20 / 2

$I_L$ = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

$I_{Lmax$ = [V $_s$ / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V $_s$ = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

$I_{Lmax$ = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

$I_{Lmax$ = [20 / 2 ] + [ 60 / 20 ]

$I_{Lmax$ = 10 + 3

$I_{Lmax$ = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

$I_{Lmax$ = [V $_s$ / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V $_s$ = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

$I_{Lmin$ = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

$I_{Lmin$ = [20 / 2 ] -[ 60 / 20 ]

$I_{Lmin$ = 10 - 3

$I_{Lmin$ = 7 A

Therefore, the maximum inductor current is 7 A

c) the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

$I_D$ = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

$I_D$ = 50 / 12.5

$I_D$ = 4 A

Therefore, the average current in the diode under ideal components is 4 A

7 0

3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago