Answer:
first on is silicon
Explanation:
Answer:
Silicon
Explanation:
Copper indium dieseline (CIS), cadmium telluride (CdTe), and thin-film silicon are certain polycrystalline thin film materials often used, whereas high-efficiency material such as gallium arsenide (GaAs) often comprise single-crystalline thin film materials3.
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 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
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.