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Naddik [55]
3 years ago
5

1. A flywheel is suspended by resting the inside of the rim on a horizontal knife edge so that the wheel can swing in a vertical

plane. The flywheel has a mass of 350Kg. The knife edge is parallel to and making 350mm from the axis of the wheel. The time for making one small oscillation is 1.77 seconds. Assuming that the centre of mass is in the axis of the wheel, determine:
(I).The radius of gyration of the wheel about the axis.
(II). The torque to increase the speed of the flywheel at uniform rate from 240 rev/min to 250 rev/min in 0.75seconds when the flywheel is revolving about its axis.
2. A solid metal cylinder 450mm diameter is suspended with its axis vertical by means of a wire coaxial with the cylinder and rigidly attached to it. The stiffness of the wire is 22Nm per radian of twist. Find the necessary mass of the cylinder so that when it is given a small angular displacement about its axis, it will make 40 vibrations per minute.​
Engineering
1 answer:
sammy [17]3 years ago
8 0

Answer: A fly wheel having a mass of 30kg was allowed to swing as pendulum about a knife edge at inner side of the rim as shown in figure.

Explanation:

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A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni
arlik [135]

Answer:

The population size would be p' = 5000

The yield would be    MaxYield = 2082 \ fishes \ per \ year

Explanation:

So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing

 

    So the current  yield which is mathematically represented as

                               \frac{dN}{dt} =   \frac{2000}{1 \ year }

 Where dN is the change in the number of fish

            and dt is the change in time

So in order to obtain the solution we need to obtain the  rate of growth

    For this we would be making use of the growth rate equation which is

                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

  Where N is the population of the fish which is given as 4,000 fishes

          and  K is the carrying capacity which is given as 10,000 fishes

             r is the growth rate

        Substituting these values into the equation

                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

                      Max Yield  = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by p' would be

                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

7 0
3 years ago
Read 2 more answers
An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

3 0
3 years ago
A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo
Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, e_{min} = 0.51

Void ratio in the loosest state, e_{max} = 0.87

Now,

Dry density, \gamma_d=\frac{\gamma_t}{1+w}

=\frac{18}{1+0.05}

= 17.14 kN/m³

Also,

\gamma_d=\frac{G\gamma_w}{1+e}

here, G = Specific gravity = 2.7 for sand

17.14=\frac{2.7\times9.81}{1+e}

or

e = 0.545

Relative density = \frac{e_{max}-e}{e_{max}-e_{min}}

= \frac{0.87-0.545}{0.87-0.51}

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

or

S = 0.2477

or

S = 0.2477 × 100% = 24.77%

7 0
3 years ago
Which type of engineering design uses an already-existing design?
Arte-miy333 [17]
The answer is D I’m 90% sure
4 0
3 years ago
Read 2 more answers
A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a te
Artyom0805 [142]

Explanation:

a converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures

100% (3 ratings)

A_2 = 0.001 m^2 P_1 = 1 MPa, T_1 = 360 k P_2 = 500 kpa p^gamma - 1/gamma proportional T (1000/500)^1.4 - 1/1.4 = (360/T_2) 2^4/14 = 360/T_2 T_2

3 0
3 years ago
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