In the planning process of the product development life cycle what is it important to inventory

What are factor of safety for brittle and ductile material

galben [10]

Explanation:

Step1

Factor of safety is the number that is taken for the safe design of any component. It is the ratio of failure stress to the maximum allowable stress for the material.

Step2

It is an important parameter for design of any component. This factor of safety is taken according to the environment condition, type of material, strength, type of component etc.

Step3

Different material has different failure stress. So, ductile material fails under shear force. Ductile material’s FOS is based on yield stress as failure stress as after yield point ductile material tends to yield. Brittle material’s FOS is based on ultimate stress as failure stress.

The expression for factor of safety for ductile material is given as follows:

$FOS=\frac{\sigma_{yp}}{\sigma_{a}}$

Here, $\sigma_{f}$ is yield stress and $\sigma_{a}$ is allowable stress.

The expression for factor of safety for brittle material is given as follows:

$FOS=\frac{\sigma_{ut}}{\sigma_{a}}$

Here, $\sigma_{ut}$ is ultimate stress and $\sigma_{a}$ is allowable stress.

5 0

3 years ago

while performing a running compression test how should running compression compare to static compression

algol [13]

Answer:

The idle speed of a running compression should be between 50-75 PSI and that is about half of the static compression.

Explanation:

The Running or Dynamic compression is used to determine how well the cylinder in an engine is absorbing air, reserving it for the proper length of time, and releasing it to the exhaust. The static or cranking compression test is used to check the sealing of the cylinder. Before performing the running compression test, the static compression test is first performed to rule out other issues like bent valves.

The standard value for the static compression is given by;

Compression ratio * 14.7 = Manufacturers Specification

The running compression should always be half of the static compression.

5 0

2 years ago

A room has a width of 14.1 feet, a length of 15.5 feet, and a ceiling height of 12.0 ft. The average flow rate for this room's a

prohojiy [21]

Answer:

Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds

ANSWER : 104.904 ft^3/sec

Explanation:

General concept : Average flow rate is the volume of fluid per unit time through an area

Hence the average flow rate of the air conditioning unit of this room

Volume of the room / time taken for the air to cycle the room = v / t

assuming the time taken = 25 seconds

volume of room = width * length * height

= 14.1 * 15.5 * 12 = 2622.6 ft^3

Average flow rate = V/ t

= 2622.6 / 25 = 104.904 ft^3/sec

8 0

3 years ago

A machine used to shred cardboard boxes for composting has a first cost of $10,000, an AOC of $7000 per year, a 3-year life, and

borishaifa [10]

Annual Payment where F is accumulated sum of amount, n is number of years and i is annual rate of interest. The standard notation equation is in the image since i can’t type it-

6 0

2 years ago

Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)

vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0

3 years ago