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Eduardwww [97]
3 years ago
10

Methane gas at 25°C, 1 atm enters a reactor operating at steady-state and burns with 80% theoretical air entering at 227°C, 1 at

m. An equilibrium mixture of CO2, CO, H2O(g), H2, and N2 exits at 1427°C, 1 atm. Assume N2 is inert.
Determine, per kmol of methane entering;
(a) the composition of the exiting mixture.
(b) the heat transfer between the reactor and its surroundings, in kJ.
Neglect kinetic and potential energy effects.

Engineering
1 answer:
Damm [24]3 years ago
5 0

Answer:

The composition of existing misture are,

a.

Co2 =0.5679

Co= 0.9326

H2O=1.6326

H2 = 0.3679

N2 = 0.6016

b.

K= 3.3878

Explanation:

Please see attachment for the solving.

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The component of a regenerative vapor power cycle that permits only liquid to pass through to a region of lower pressure is a Valve/trap.

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3 0
1 year ago
Tech A says that the brake pedal uses leverage to multiply foot pressure. Tech B says that when braking hard while moving
Nikolay [14]

Tech- A is correct

Explanation:

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7 0
3 years ago
If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam
Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

\zeta =\dfrac{C}{C_c}

Where C is the damping coefficient and Cc is the critical damping coefficient.

C_c=2\sqrt{mK}

So from above we can say that

\zeta =\dfrac{C}{2\sqrt{mK}}

\zeta \alpha \dfrac{1}{\sqrt K}

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0
3 years ago
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