Answer:
pls I am sorry dis is not a reasonable question
Determine size of a standard square key made of 1045 hot rolled steel for a 2 inch DIA shaft transmitting 100 HP at 500 rpm with factor of safety 2.5 for yield in direct shear. W____inch L_____inch
Answer:
Width = Length = 1.148 inches
Explanation:
We have been given the following data:
D = diameter = 2 inch = 0.0508m
P = Power = 100 HP = 74570 W
N = 500 rpm
Safety Factor = 2.5
Step 1:
We need to find yield strength which is represented by σ(y).
σ(y) for 1045 hot rolled steel = 330MPa
Step 2:Find Shear Strength. Formula is given:
τ(y) = σ(y) / 2
τ(y) = 330 / 2
τ(y) = 165 MPa
τ(y) = 165 × 10⁶ Pa
τ(y) = 165 × 10⁶ kg.m⁻¹.s⁻²
Step 3:
Find Torque. Formula is given:
T = 60P / 2πN
T= (60)(74570) / 2π(500)
T = 1424.9 Nm
Step 4:Find Shear Force. Formula is given:
F = 2T/d
F = 2(1424.9)/0.0508
F = 56098.43 N
F = 56098.43 kg.m.s⁻²
Step 5:Find length by the given formula:
F/L² = τ(y)/Safety factor
Rearrange for L:
L = √(F· Safety factor / τ(y))
Substitute the values found in previous steps to calculate L.
L = 0.02915 meters
Convert it into inches:
L = 1.148 in
As it si a square key:
L = W
Width = 1.148 in
A piston-cylinder device contains 0.1 kg of hydrogen gas (PG model: cv=10.18, k = 1.4, R= 4.12 kJ/kg-K) at 1000 kPa and 300 K. The gas undergoes an expansion process and the final conditions are 500 kPa, 270 K. If 10 kJ of heat is transferred into the gas from the surroundings at 300 K, determine (a) the boundary work (Wb), and (b) the entropy generated (Sgen) during the process
Answer:
(a) 151.84 kJ
(b) 2.922 kJ/K
Explanation:
(a) The parameters given are;
Mass of hydrogen gas, H₂ = 0.1 kg = 100 g
Molar mass of H₂ = 2.016 g/mol
Number of moles of H₂ = 100/2.016 = 49.6 moles
V₁ = mRT/P = 0.1×4.12×300/1000 = 0.1236 m³
P₁/P₂ = (V₂/V₁)^k
V₂ = (P₁/P₂)^(1/k)×V₁ =0.1236 × (1000/500)^(1/1.4) = 0.3262 m³
Boundary work done = (V₂ - V₁)(P₂ + P₁)/2 = (0.3262 - 0.1236)*(500 + 1000)/2 = 151.84 kJ
(b) Entropy generated ΔS = Cv · ㏑(T₂/T₁) + R ·㏑(v₂/v₁)
=10.18 × ㏑(270/300) + 4.12 ·㏑(0.3262/0.1236) = 2.922 kJ/K.
An example of matching and converting would be finding a new market in which to see a product, such as online or internationally.
True
False
Answer:
True
Explanation:
Matching involves putting compatible strength and opportunities together to acquire comparative advantage over the competition
Converting involves the transformation of perceived prospects or threats and weakness or lack of capacity into a chance or opening and strength
Therefore, since finding a new market in which to see a product such as online or internationally can help match the strength of the searcher to the opportunities available internationally and convert weaknesses finding a new market is an example of matching and converting
Explain working mechanism of safety valve.
Answer:
Safety valves are part of the safety features placed in the system to prevent the system from experiencing over pressures
Explanation:
A safety valve is a valve that is actuated automatically when the pressure on the inlet portion rises to a specified value to allow the outflow or discharge of fluid such as liquid, steam or gas out of the system to prevent the system pressure limit from being exceeded. The design of safety valves is such that the valve closes the emergency outlet port again once the system pressure returns to normal.
A prototype boat is 30 meters long and is designed to cruise at 9 m/s. Its drag is to be simulated by a 0.5-meter-long model pulled in a tow tank. For Froude scaling find a)-The tow speed b)-The ratio of model to prototype drag c)-The ratio of model to prototype power.
Answer:
a) 1.16 m/s
b) 1/216000
c) (√15)/6480000
Explanation:
The parameters given are;
Length of boat prototype, lp = 30 m
Speed of boat prototype = 9 m/s
Length of boat model, lm= 0.5 m
a) lm/lp = 0.5/30 = 1/60 = ∝
(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)
vm = 9 × (1/60)^(1/2) = 1.16 m/s
b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;
Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³
Fm/Fp = (1/60)³ = 1/216000
c) The ratio of the model to prototype power pm/p[tex]_p[/tex] = (Fm/Fp) × (vm/vp) = ∝³×√∝
The ratio of the model to prototype power pm/p[tex]_p[/tex] = √(1/60) × (1/60)³
pm/p[tex]_p[/tex] = √(1/60) × (1/60)³ = (√15)/6480000
Describe how you could achieve a high quality finish on your work
Answer:
efort
Explanation:
try your hardest