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3 years ago

1. advantages of 2 pulley system

Engineering

1 answer:

n200080 [17]3 years ago

5 0

Answer:

Advantages

The main advantage in the use of pulleys is that the effort becomes less as compared to the normal lifting of the weights. In other words, it reduces the amount of actual force required to lift heavy objects. It also changes the direction of the force applied. These two advantages in the use pulleys make them an important tool for heavy lifting. It also provides a mechanical advantage.

The other advantage in the use of pulleys is that the distance between the operator and weight. There is a safe distance between them which avoids any disaster. Pulleys are easy to assemble and cost-effective. The combination of different directional pulleys can change the position of the load with little effort. Though there are moving parts in the pulley system they require less or no lubrication after installation.

Disadvantages

Apart from the above-said advantages while using pulley systems, there are several disadvantages in their use. The main disadvantage in the use of the pulley system is that it requires large space to install and operate. The mechanical advantage of pulleys can go to higher values but need more space to install them.

In some cases, the ropes/belts move over the wheel with no grooves, the chances of the slip of ropes/belts from the wheel are inevitable. If the system is installed to use for a long time, they require maintenance and regular check-up of ropes/cables as the friction between the wheels and cables/ropes occur causing wear and tear to them. Continuous use of the system makes the ropes weak. The rope may break while using the system causing damages to the operator, surrounding place and the load which is being lifted.

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Select the statement that is false.

ra1l [238]

Answer:

Explanation:

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3 years ago

A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60

Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

T_2 = T_1 * r^(k-1/k)

T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

T_7 = T_6 * r^(-k-1/k)

T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

flow(m) = 7.941 lbm/s

- The heat input is as follows:

Q_in = c_p*(T_6 - T_5)

Q_in = 0.24*(1400 - 1022.84)

Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

Q_out = c_p*(T_10 - T_1)

Q_out = 0.24*(711.744 - 520)

Q_out = 56.01856 Btu/lbm

5 0

3 years ago

A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.

sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c) $Stress\ ratio =-0.65$

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

$\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}$

$2\sigma_m={\sigma_1+\sigma_2}$

2 x 46.2 = σ₁ + σ₂

σ₁ + σ₂ = 92.4 MPa --------1

The amplitude stress given as

$\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}$

$2\sigma_a={\sigma_1-\sigma_2}$

2 x 219 = σ₁ - σ₂

σ₁ - σ₂ = 438 MPa --------2

By adding the above equation

2 σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

$Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}$

$Stress\ ratio =\dfrac{-172.8}{265.2}$

$Stress\ ratio =-0.65$

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0

3 years ago

A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu

krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

$\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}$

$10*2 = \sqrt{1 + 8f^2 - 1$

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

$\frac{V_1}{\sqrt{g*y_1}} = 7.416$

$V_1 = 7.416 *\sqrt{32.2*1}$

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

$V_2 = \frac{3666.66}{80*10}$

= 4.208 ft/sec

Froude number, F2 =

$\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}$

F2 = 0.235

d) $El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}$

$El = \frac{(10-1)^3}{4*1*10}$

$= \frac{9^3}{40}$

= 18.225ft

e) for critical depth, we use :

$y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3$

= 3.80 ft

7 0

3 years ago

Read 2 more answers

In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yie

Snezhnost [94]

Answer:

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

Explanation:

(a)

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.

Given:

Rake angle is 12°.

Chip thickness before cut is 0.32 mm.

Chip thickness is 0.65 mm.

Calculation:

Step1

Chip reduction ratio is calculated as follows:

$r=\frac{t}{t_{c}}$

$r=\frac{0.32}{0.65}$

r = 0.4923

Step2

Shear angle is calculated as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

Here, $\phi$ is shear plane angle, r is chip reduction ratio and $\alpha$ is rake angle.

Substitute all the values in the above equation as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

$tan\phi=\frac{0.4923cos12^{\circ}}{1-0.4923sin12^{\circ}}$

$tan\phi=\frac{0.48155}{0.8976}$

$\phi=28.21^{\circ}$

Thus, the shear plane angle is 28.21°.

(b)

Step3

Shears train is calculated as follows:

$\gamma=cot\phi+tan(\phi-\alpha)$

$\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})$ $\gamma = 2.155$ .

Thus, the shear strain rate is 2.155.

6 0

3 years ago