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Ilya [14]
2 years ago
11

As cylinder pressure and heat increase due to an increased load condition, the fuel injection management system must ___________

_
Engineering
1 answer:
Temka [501]2 years ago
6 0

possible Answers:

Compensate ⭐⭐⭐⭐⭐

Adjust            ⭐⭐⭐⭐⭐

regulate         ⭐⭐⭐⭐

tune               ⭐⭐⭐

calibrate        ⭐⭐⭐

balance         ⭐⭐

correct           ⭐

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vladimir2022 [97]

Answer:

c

Explanation:

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4 0
2 years ago
Suppose the working pressure for a boiler is 10 psig, then what is the corresponding absolute pressure?
yanalaym [24]

Answer:

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

Explanation:

From Fluid Mechanics, we remember that absolute pressure (p_{abs}), measured in pounds per square inch, is the sum of the atmospheric pressure and the working pressure (gauge pressure). That is:

p_{abs} = p_{atm}+p_{g} (1)

Where:

p_{atm} - Atmospheric pressure, measured in pounds per square inch.

p_{g} - Working pressured of the boiler (gauge pressure), measured in pounds per square inch.

If we suppose that p_{atm} = 14.696\,psi and p_{g} = 10\,psi, then the absolute pressure is:

p_{abs} = 14.696\,psi+10\,psi

p_{abs} = 24.696\,psi

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

8 0
3 years ago
How many power station do we have​
loris [4]

Answer: 9,719

Explanation:

5 0
3 years ago
A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
Kay [80]

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

5 0
3 years ago
Gtjffs
grandymaker [24]

the required documents is 3000

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2 years ago
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