MECHANICAL ENGINEERING: Reliability of a product, failure, failed products, total manufactured

R = 1 - f/m


where:

R = reliability of a product, a measure of the product's probability that it will not fail when used

f = number of failed products

m = number of manufactured products


1. A certain bearing and gear manufacturer has produced a total of 4000 ball bearings and 5000 spur gears. It was later found out that there were 35 ball bearings and 70 spur gears that failed. Determine the Reliability of the class of ball bearings and these kinds of spur gears.


ball bearings:

find:

R = Reliability of these kinds of ball bearings


given:

f = 35

m = 4000


solution:

R = 1 - f/m

R = 1 - 35/4000

R = 1 - 0.0088

R = 0.9912  or 99.12%


spur gears:

find:

R = Reliability of these specific types of spur gears


given:

f = 70

m = 5000


solution:

R = 1 - f/m

R = 1 - 70/5000

R = 1 - 0.014

R = 0.986  or 98.6%

MECHANICAL ENGINEERING: Thermodynamics - Entropy and the Second Law of Thermodynamics

Entropy

- a measure of a system's energy that is unavailable for work

- a measure of the disorder or randomness

- from Greek entropia "a turning towards"



Second law of thermodynamics

- there is no machine with 100% efficiency

- phenomenon of irreversibility in nature

- In a system, a process that occurs will tend to increase the total entropy of the universe

- The entropy of an isolated system which is not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium

- It is impossible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow

- It is impossible to extract an amount of heat from a hot reservoir and use it all to do work. Some amount of heat must be exhausted to a cold reservoir



Few examples of Entropy and the 2nd Law of Thermodynamics

- flow of heat from a hot object to a cold object

- flow of a gas from high pressure to low pressure

- evaporation of water

- melting of ice

- cloud formation in the sky

- adding sugar to a cup of coffee

- carbon dioxide is dissolved in water

- spontaneous (natural) cleaning of a messy house

- shuffling of playing cards

- breaking a mirror or glass



1. Isothermal and reversible process


dS = Q/T


where:

dS = change in entropy, values can be (+)increase or (-)decrease

Q = heat, negative (-) if heat removal, positive (+) if heat addition

T = absolute temperature




2. Temperature not constant


dS = Integral (dQ/T)


dQ = m * c * dT


substituting,

dS = Integral (dQ/T)

dS = Integral (m * c * dT/T)

dS = m * c * Integral (dT/T)


if

T1 = initial temperature

T2 = final temperature


dS = m * c * Integral(T1,T2) (dT/T)


then


dS = m * c * ln T2/T1


where:

m = mass

c = specific heat

dS = change in entropy, values can be (+)increase or (-)decrease

dQ = heat, negative (-) if heat removal, positive (+) if heat addition

T = absolute temperature




1. Find the change in entropy if 1 kg of water at 20 C is changed into ice at 0 C.


find:

dS = total change in entropy


given:

m = 1 kg of water

c of water = 4.2 kJ/kg K

lf = heat of fusion of ice = 334 kJ/kg

T1 = 20 C + 273 = 293 K

T2 = 0 C + 273 = 273 K


solution:

dS = dS1 + dS2


dS1 --> cooling of water


dS1 = m * c * ln T2/T1

dS1 = 1 * 4.2 * ln 273/293

dS1 = - 0.3 kJ/K


dS2 --> freezing of water to ice

dS2 = Q/T


but

Q = - m * lf  --> latent heat


note: Q is negative (-) because of HEAT REMOVAL


substituting,

dS2 = - m * lf/T

dS2 = - 1 * 334/273

dS2 = - 1.22 kj/K


total change in entropy

dS = dS1 + dS2

dS = (- 0.3) + (- 1.22)

dS = - 1.52 kJ/K




2. Calculate the change in entropy when 4 kJ of heat flows from the warm exterior at 40 C into a house at 22 C.


find:

dS = total change in entropy


given:

Q = 4 kJ

T exterior = 40 C + 273 = 313 K

T house = 22 C + 273 = 295 K


solution:

dS = dS exterior + dS house


dS exterior:


The entropy of the exterior decreases, since heat flows out of it (- Q)

dS exterior = -Q/T exterior

dS exterior = -4/313

dS exterior = -0.013 kJ/K



dS house:

The entropy of the house increases since heat flows into it (+ Q)

dS house = Q/T house

dS house = 4/295

dS house = 0.014 kJ/K



total change in entropy

dS = dS exterior + dS house

ds = -0.013 + 0.014

dS = 0.001 kJ/K

MECHANICAL ENGINEERING: Fluid Mechanics and Dynamics - Conservation of Mass, mass flow rate, volume flow rate

Volume = Area * height

V = A * h


if d = h = distance travelled by the cross-sectional area

V = A * d

and the Volume flow rate or the volume per unit time

Vdot = (A * d)/t


but d/t = velocity = v

Vdot = A * v


CONSERVATION OF MASS

The amount of mass that flows past any cross-section is constant

m1 = m2

mass in = mass out


but density D = m/V

m = D * V

substituting in

m1 = m2

D1 * V1 = D2 * V2


if V1 and V2 are volume flow rates and Vdot = A * v

D1 * (A1 * v1) = D2 * (A2 * v2)


the same fluid is flowing so

D1 = D2


and it simplifies to


A1 * v1 = A2 * v2


where:

A1 = cross-sectional area at point1

A2 = cross-sectional area at point2

v1 = velocity of fluid at point1

v2 = velocity of fluid at point2


1. Problem:

Water enters a pipe with diameter of 1 in at 7 fps. Determine the exit velocity of the water if the pipe diameter is 2 in.

find:

v2 = velocity of water at exit


given:

d1 = 1 in

d2 = 2 in

v1 = 7 fps


solution:

convert v1 to in/s

v1 = 7 ft/s * 12 in/ft

v1 = 84 in/s


solving for A1 and A2

A1 = pi/4 * (d1)^2

A1 = 0.785 * (1)^2

A1 = 0.785 * 1

A1 = 0.785 sq. in


A2 = pi/4 * (d2)^2

A2 = 0.785 * (2)^2

A2 = 0.785 * 4

A2 = 3.14 sq. in


by conservation of mass

A1 * v1 = A2 * v2

0.785 * 84 = 3.14 * v2

v2 = 21 in/s


convert v2 to fps

v2 = 21 in/s * 1 ft/12 in

v2 = 21/12

v2 = 1.75 fps


------------------

ALTERNATE SOLUTION:


A1 * v1 = A2 * v2


because (pi/4) and (sq. in) will cancel, there's no need for unit conversion

(d1)^2 * v1 = (d2)^2 * v2

(1)^2 * 7 = (2)^2 * v2

v2 = 7/4

v2 = 1.75 fps

MECHANICAL ENGINEERING: Design of Machine Elements - Rod with a circular cross-section

FS = S/Sa


Sa = F/A


A = pi/4 * d^2

A = 0.785 * d^2

A = pi * r^2

A = 3.14 * r^2


where:

FS = factor of safety

S = strength of the material

Sa = allowable stress

F = Force applied

A = cross-sectional area of material

r = radius of circular rod

d = diameter of circular rod


---


1. Problem: Circular Rod

Determine the safe size of a circular rod to be loaded with an axial force of 3000 lbs considering a factor of safety of 4 and the material strength of the rod 30,000 psi.

find:

d = diameter of circular rod

given:

F = axial force = 3000 lbs

S = material strength = 30,000 psi

FS = factor of safety = 4


solution:

Sa = F/A

Sa = 3000/0.785 d^2 

Sa = 3821.66 d^2  ---> equation1


FS = S/Sa

4 = 30,000/Sa

Sa = 30,000/4

Sa = 7500  ---> equation2


equation2 in equation1

Sa = 3821.66 d^2

7500 = 3821.66 d^2

d^2 = 1.96

d = 1.4 in ---> use 1.5 in