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4 years ago

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Explain how use of EGR is effective in reducing NOx emissions 4. In most locations throughout the U.S., the octane number of reg

ular grade gasoline [defined as (R+M)/2] is 87. However, in certain high elevation locations (parts of CO, ID, MT, UT, and WY) regular gasoline may have a lower octane value of 85. Explain how a lower octane fuel can be used in these areas without it leading to unacceptable knocking

1 answer:

TiliK225 [7]4 years ago

3 0

Answer:please see attached file

Explanation:

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The combustion chamber has different shapes depending on the make and model of the engine. True or false

Alisiya [41]

Answer:

svdsdfdfsdfssdf

Explanation:

fsdsdfsdffsdsfdsdf

5 0

3 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$

Where t is given by

$1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec$

So,

$\Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2$

$$ \Delta \delta = 6.75 \: elec \: deg$

The change in torque in rpm/s is given by

$\Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ }$

$\Delta \delta =28.12 \: \: rpm/s$

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t$

Where P is the number of poles of the turbo-generator.

$Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2$

$Rotor \: speed = 1500 + 5.62$

$$ Rotor \: speed = 1505.62 \:\: rpm$

4 0

3 years ago

) A brick of clay is being dried in a batch dryer using constant drying conditions. You have been asked to determine the amount

dimaraw [331]

Answer:

The drying time is calculated as shown

Explanation:

Data:

Let the moisture content be = 0.6

the free moisture content be = 0.08

total moisture of the clay = 0.64

total drying time for the period = 8 hrs

then if the final dry and wet masses are calculated, it follows that

t = (X0+ Xc)/Rc) + (Xc/Rc)* ln (Xc/X)

= 31.3 min.

4 0

3 years ago

Which one of the following types of gas is flammable? a. Neon b. Peroxide. c. Acetylene d. Carbon Dioxide

Delvig [45]

Answer:

D Carbon Dioxide

Explanation:

3 0

3 years ago

Read 2 more answers

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha

Anton [14]

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

$\sigma=\frac{F}{A}$

$F=\sigma*A$

Convert given values to base units:

275 MPa = $275*10^{6}$ Pa

325 $mm^{2}$ = 0.000325 $m^{2}$

Substituting in given values:

F = $(275*10^{6})*(0.000325)=89375$ N

3 0

3 years ago