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3 years ago

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One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp

osite sides of a rigid, well-insulated container. The partition is free to move and allows conduction from one gas to the other without energy storage in the partition itself. The air and carbon dioxide each behave as ideal gases.
Determine the final equilibrium temperature, in K, and the final pressure, in bar, assuming constant specific heats.

1 answer:

pentagon [3]3 years ago

6 0

Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

$m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}$

1x0.723x $(T_{eq} -350)$ =3x0.780x $(450-T_{eq} )$ ⇒ $T_{eq}$ = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

$V_{a,i} = \frac{mRT_{a,i} }{P_{a,i} }$ = $\frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar}$ = 0.201 $m^{3}$

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

$P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }$

$P_{eq}$ = 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

(0.201+1.275)

= 246.67 KPa = 2.47 bar

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Suppose the working pressure for a boiler is 10 psig, then what is the corresponding absolute pressure?

yanalaym [24]

Answer:

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

Explanation:

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$p_{abs} = p_{atm}+p_{g}$ (1)

Where:

$p_{atm}$ - Atmospheric pressure, measured in pounds per square inch.

$p_{g}$ - Working pressured of the boiler (gauge pressure), measured in pounds per square inch.

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3 years ago

For an Otto cycle, plot the cycle efficiency as a function of compression ratio from 4 to 16.

Elza [17]

Assumptions:

Steady state.
Air as working fluid.
Ideal gas.
Reversible process.
Ideal Otto Cycle.

Explanation:

Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):

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Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

$r =\frac{V_1}{V_2}$

Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

$r = \frac{V_4}{V_3} = \frac{V_1}{V_2}$

Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
Exhaust 1-0: the working fluid is vented to the atmosphere.

If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

$\eta = 1-(\frac{1}{r^{\gamma - 1} } )$

where:

$\gamma = \frac{C_{p} }{C_{v}} : specific heat ratio$

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

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Answer:

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3 years ago

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a_sh-v [17]

Answer:

Outside temperature =88.03°C

Explanation:

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To find temperature we need to balance energy

Heat generation=Heat dissipation

Now find the value

We know that for sphere

$q=\dfrac{2\pi DK}{1-\dfrac{D}{4H}}(T_1-T_2)$

Given that q=500 W

so

$500=\dfrac{2\pi 2\times .6}{1-\dfrac{2}{4\times 10}}(T_1-25)$

By solving that equation we get

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3 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3 × 10-4

Vladimir [108]

Answer:

maximum stress is 2872.28 MPa

Explanation:

given data

radius of curvature = 3 × $10^{-4}$ mm

crack length = 5.5 × $10^{-2}$ mm

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to find out

maximum stress

solution

we know that maximum stress formula that is express as

$\sigma m = 2 ( \sigma o ) \sqrt{\frac{a}{\delta t}}$ ......................1

here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack

so put here all value in equation 1 we get

$\sigma m = 2 ( \sigma o) \sqrt{\frac{a}{\delta t}}$

$\sigma m = 2(150) \sqrt{ \frac{\frac{5.5*10^{-2}}{2}}{3*10^{-4}}}$

σm = 2872.28 MPa

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