Ask question

Ask question

All categories

3 years ago

5

In a four bar mechanism, L, is a fixed link; L2 is driver crank; L3 is coupler and L4 is follower crank. L=27 cm, L3 =5 cm and L

4=10 mm. Find the length of the link L1 if the linkage can be classified as a double rocker mechanism:

1 answer:

Lera25 [3.4K]3 years ago

6 0

Answer:

$L_1>32 cm$

Explanation:

Given that

$L_2=27 cm$

$L_3=5 cm$

$L_4=10 cm$

Here shortest link is link3

To make the linkage as a double rocker mechanism we have to fix longest link that link is link1

So for double rocker mechanism

$L_3+L_1>L_2+L_4$

Now put the values

$5+L_1>27+10$

$L_1>32 cm$

It means that length of link1 should be greater than 32 cm.

You might be interested in

As a general rule of thumb, in-line engines are easier to work on than the other cylinder arrangements.

siniylev [52]

Answer:

The general rule of thumb is that the SMALLER a substance's atoms and the STRONGER the bonds, the harder the substance. Two of the strongest forms of chemical bonds are the ionic and covalent bonds.

Explanation:

5 0

2 years ago

30 points and brainiest if correct please help A, B, C, D

tatuchka [14]

Answer:

B. to lock the tape into place

Explanation:

the button on the front of the housing locks the tape into place when pressed, preventing the tape from being pulled out further it retracting

4 0

2 years ago

To read signs you need good focal vision

kow [346]

Answer:eyesight

Explanation:

7 0

3 years ago

Read 2 more answers

H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate

Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

$\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0$ (1)

Where:

$\dot m$ - Mass flow, in kilograms per second.

$h_{in}$ , $h_{out}$ - Specific enthalpies at inlet and outlet, in kilojoules per second.

$v_{in}, v_{out}$ - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

$p = 2000\,kPa$

$T = 300\,^{\circ}C$

$h_{in} = 3024.2\,\frac{kJ}{kg}$

$\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$

Where $\nu_{in}$ is the specific volume of water at inlet, in cubic meters per kilogram.

Outlet (Superheated steam)

$p = 600\,kPa$

$T = 160\,^{\circ}C$

$h_{out} = 2758.9\,\frac{kJ}{kg}$

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $h_{in} = 3024.2\,\frac{kJ}{kg}$ , $h_{out} = 2758.9\,\frac{kJ}{kg}$ and $v_{in} = 30\,\frac{m}{s}$ , then the flow speed at outlet is:

$35765-25\cdot v_{out}^{2} = 0$ (2)

$v_{out} \approx 37.823\,\frac{m}{s}$

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius ( $r_{in}$ ), in meters, by this expression:

$\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}}$ (3)

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $v_{in} = 30\,\frac{m}{s}$ and $\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$ , then the inlet radius is:

$r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}$

$r_{in}\approx 0.258\,m$

The inlet radius of the nozzle is approximately 0.258 meters.

7 0

2 years ago

Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.3

Paladinen [302]

Answer:

A) n = 0.6143, c ≈ 640m/min

B) n = 0.6143 , c = 637.53m/min

Explanation:

using the given data

A) A plot of flank wear as a function of time and also A plot for tool when

Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min also for cutting edge speed of 155m/min , time for cutting edge is 10 min

is attached below

calculate for the constant N from the second plot

note : the slope will be negative because cutting speed decreases as time of cutting increase

V1 = 100m/min , V2 = 155m/min, T1 = 20.4 min, T2 = 10 min

= - N = $\frac{In(V2) - In(V1)}{In(T2)-ln(T1)}$

therefore - N = $\frac{5.043 - 4.605}{2.302 -3.015}$

= - 0.6143

THEREFORE ( N ) = 0.6143

Determine for the constant C from the second plot as well

note : C is the intercept on the cutting speed axis in 1 min tool life

connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point

hence C ≈ 640m/min

B) Calculate the values of N and C in the Taylor equation solving simultaneous equations

using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation

Taylor tool life equation = vT = C ------------- equation 1

cutting speed = v = 100m/min and 155m/min

tool life = T = 20.4 min and 10 min

also constant n and c are obtained from the previous plot

back to taylor tool life equation = 100 * 20.4 = C

therefore C = (100)(20.4)^n ---------------- equation 2

also using the second values of v and T

taylor tool life equation = 155 * 10 = C

therefore C = ( 155 )(10)^n ----------------- equation 3

Equate equation 2 and equation 3 and solve simultaneously

(100)(20.4)^n = (155)(10)^n

To find N

take natural log of both sides of the equation

= In ((100)(20.4)^n) = In((155)(10)^n)

= In (100) + nIn(20.4) = In(155) + nIn(10)^n

= n(3.0155) - n (2.3026) = 5.043 - 4.605

= 0.7129 n = 0.438

therefore n = 0.6143

To find C

substitute 0.6143 for n in equation 2

C = (100)(20.4) ^ 0.6143

C = 637.53 m/min

Attached are the two plots for solution A

7 0

3 years ago