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Gala2k [10]
4 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:

Engineering
1 answer:
Lera25 [3.4K]4 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.

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Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb
guajiro [1.7K]

Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = h_{g\ at\ 125KPa} = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

Now, we will find h_{2s}(enthalpy at the outlet for the isentropic process):

h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

Now, the isentropic efficiency of the turbine can be given as follows:

\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%

3 0
3 years ago
A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of
OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0
3 years ago
a buyer can purchase 70 screwdrivers ten 4-inch length twelve 6 inch length twenty 8-inch length are needed. how many heavy 24-i
jasenka [17]

Answer:

28 , 24-inch screwdrivers

Explanation:

The total number of screwdrivers that can be purchased is = 70

4 - inch length screwdrivers = 10

6- inch length screwdrivers = 12

8- inch length screwdrivers = 20

Total = 20 +12 +10 = 42

Remaining = 70-42 = 28

So, heavy 24-inch screwdrivers = 28

3 0
3 years ago
5. (5 points) Select ALL statements that are TRUE A. For flows over a flat plate, in the laminar region, the heat transfer coeff
finlep [7]

Answer:

The following statements are true:

A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction

C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface

E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.

Select ALL statements that are TRUE

B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant

D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed

Explanation:

7 0
3 years ago
A specific internal combustion engine has a displacement volume VD of 5.6 liters. The processes within each cylinder of the engi
Kisachek [45]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0
3 years ago
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