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3 years ago

15

Suppose a large amount of power is required. Which engine would you choose between Otto and Diesel? Why?

1 answer:

Firdavs [7]3 years ago

6 0

Answer:

Otto engine

Explanation:

As we know that

Power = Torque x speed

So we can say that when speed of engine then power of engine also will increases.

The speed of Otto engine is more as compare to Diesel engine so the power of Otto engine is more.But on the other hand torque of Diesel engine is more as compare to Otto engine but the speed is low so the product of speed and torque is more for Otto engine .It means that when requires large amount of power then Otto engine should be use.

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A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle

Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

$X_{s} =1.23(\frac{V_{line}^{2} }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms$

the network voltage phase is

$V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV$

the power transmitted is equal to:

$|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV$

the line induced voltage is

$|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV$

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2 years ago

what are the characteristics of an ideal fluid the general relation between shear stress and velocity gradient

Dafna11 [192]

Answer:

ideal fluid follow Newtonian law

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3 years ago

Using the results of the Arrhenius analysis (Ea=93.1kJ/molEa=93.1kJ/mol and A=4.36×1011M⋅s−1A=4.36×1011M⋅s−1), predict the rate

uysha [10]

Answer:

k = 4.21 * 10⁻³(L/(mol.s))

Explanation:

We know that

k = Ae $^{-E/RT}$ ------------------- euqation (1)

K= rate constant;

A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;

E = activation energy = 93.1kJ/mol;

R= ideal gas constant = 8.314 J/mol.K;

T= temperature = 332 K;

Put values in equation 1.

k = 4.36*10¹¹(M⁻¹s⁻¹)e $^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}$

k = 4.2154 * 10⁻³(M⁻¹s⁻¹)

here M =mol/L

k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)

or

k = 4.21 * 10⁻³((L/mol)s⁻¹)

or

k = 4.21 * 10⁻³(L/(mol.s))

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Answer:

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