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Ilia_Sergeevich [38]
3 years ago
11

A fuel gas containing 45.00 mole% methane and the balance ethane is burned completely with pure oxygen at 25.00 degree C, and th

e products are cooled to 25.00 degree C. Suppose the reactor is continuous. Take a basis of calculation of 1.000 mol/s of the fuel gas, assume some value for the percent excess oxygen fed to the reactor (the value you choose will not affect the results), and calculate - Q(kW), the rate at which heat must be transferred from the reactor if the water vapor condenses before leaving the reactor and if the water remains as a vapor. Now suppose the combustion takes place in a constant-volume batch reactor. Take a basis of calculation 1.000 mol of the fuel gas charged into the reactor, assume any percent excess oxygen, and calculate -Q(kJ) for the cases of liquid water and water vapor as products.
Engineering
1 answer:
Dennis_Churaev [7]3 years ago
7 0

Answer:

A)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

Explanation:

Given data :

45.00 % mole of methane

55.00 % of ethane

attached below is a detailed solution

A) calculate - Q(kw)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B ) calculate  - Q ( KJ )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

since combustion takes place in a constant-volume batch reactor

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Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

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Explanation:

Given that:

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24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

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\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

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Answer:

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. The flexure strength test was performed on a concrete beam having a cross section of 0.15m by 0.15m and a span of 0.45m. If th
ivann1987 [24]

Answer:

σ =5.39Mpa

Explanation:

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Where

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F= Failure load (in N).

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step two:

Given data

F=40.45 kN= 40450N

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