Answer:crane and engine I guess
Explanation:
Answer:
a) V = 0.354
b) G = 25.34 GPA
Explanation:
Solution:
We first determine Modulus of Elasticity and Modulus of rigidity
Elongation of rod ΔL = 1.4 mm
Normal stress, δ = P/A
Where P = Force acting on the cross-section
A = Area of the cross-section
Using Area, A = π/4 · d²
= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²
δ = 50/3.14 × 10⁻⁴ = 159.155 MPA
E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm
Modulus of Elasticity Е = δ/ε
= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA
Also final diameter d(f) = 19.9837 mm
Initial diameter d(i) = 20 mm
Poisson said that V = Е(elasticity)/Е(long)
= - <u>( 19.9837 - 20 /20)</u>
2.33 × 10⁻³
= 0.354,
∴ v = 0.354
Also G = Е/2. (1+V)
= 68.306 × 10⁹/ 2.(1+ 0.354)
= 25.34 GPA
⇒ G = 25.34 GPA
9514 1404 393
Answer:
13/80
Explanation:
The product is ...
(1 3/10)×(1/8) = (13/10)×(1/8) = (13×1)/(10×8) = 13/80
The exit temperature is 586.18K and compressor input power is 14973.53kW
Data;
- Mass = 50kg/s
- T = 288.2K
- P1 = 1atm
- P2 = 12 atm
<h3>Exit Temperature </h3>
The exit temperature of the gas can be calculated isentropically as

Let's substitute the values into the formula

The exit temperature is 586.18K
<h3>The Compressor input power</h3>
The compressor input power is calculated as

The compressor input power is 14973.53kW
Learn more on exit temperature and compressor input power here;
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