Answer:
Class of fit:
Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).
Here minimum shaft diameter will be greater than the maximum hole diameter.
Medium Drive Force Fits are FN 2 Fits.
As per standard ANSI B4.1 :
Desired Tolerance: FN 2
Tolerance TZone: H7S6
Max Shaft Diameter: 3.0029
Min Shaft Diameter: 3.0022
Max Hole Diameter:3.0012
Min Hole Diameter: 3.0000
Max Interference: 0.0029
Min Interference: 0.0010
Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.
Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.
Explanation:
Answer:
Absolute pressure=70.72 KPa
Explanation:
Given that Vacuum gauge pressure= 30 KPa
Barometer reading =755 mm Hg
We know that barometer always reads atmospheric pressure at given situation.So atmospheric pressure is equal to 755 mm Hg.
We know that P= ρ g h
Density of 
So P=13600 x 9.81 x 0.755
P=100.72 KPa
We know that
Absolute pressure=atmospheric pressure + gauge pressure
But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.
Absolute pressure=atmospheric pressure + gauge pressure
Absolute pressure=100.72 - 30 KPa
So
Absolute pressure=70.72 KPa
AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.
Explanation:
Answer:
metals, composite, ceramics and polymers.
Explanation:
The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.
i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.
ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.
iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.
iv) Polymers: They have low weight and are poor conductors of electricity and heat
Answer:
1700kJ/h.K
944.4kJ/h.R
944.4kJ/h.°F
Explanation:
Conversions for different temperature units are below:
1K = 1°C + 273K
1R = T(K) * 1.8
= (1°C + 273) * 1.8
1°F = (1°C * 1.8) + 32
Q/delta T = 1700kJ/h.°C
T (K) = 1700kJ/h.°C
= 1700kJ/K
T (R) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8R
= 944.4kJ/h.R
T (°F) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8°F
= 944.4kJ/h.°F
Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C
= 2°C
(273+5) - (273+3)
= 2 K
