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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 water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat
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Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}

Properties: The density of water \delta = 1000 kg/m^3

Conversions:

165 \  ft \  to \  meters  = 50 m  \\ \\7000 \ lbm/s \  to  \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \  to \  kilowatt = 1166 kw \\ \\

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

e_{mech_{in}} - e_{mech_{out}} = gh - 0

Then;

gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}

gh = 0.491 kJ/kg

\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg

= 1559 kW

Therefore; the overall efficiency is:

\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}

= \dfrac{1166 \ kW}{1559 \ kW}

= 0.75

= 75%

b) mechanical efficiency of the turbine:

\eta_{turbine- generator} = \eta_{turbine}\times   \eta_{generator}

thus;

\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%

6 0
3 years ago
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