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julsineya [31]
4 years ago
11

Carbon dioxide (CO2) at 1 bar, 300 K enters a compressor operating at steady state and is compressed adiabatically to an exit st

ate of 10 bar, 520 K. The CO2 is modeled as an ideal gas, and kinetic and potential energy effects are negligible. For the compressor, determine (a) the work input, in kJ per kg of CO2 flowing, (b) the rate of entropy production, in kJ/K per kg of CO2 flowing, and (c) the isentropic compressor efficiency.

Engineering
1 answer:
Zielflug [23.3K]4 years ago
6 0

Answer:

A.) 0.08 kJ/kg.K

B.) 207.8 KJ/Kg

C.) 0.808

Explanation:

From the question, the use of fluids mechanic table will be required. In order to get the compressor processes, the kinetic energy and the potential energy will be negligible while applying the ideal gas model.

Since the steam is a closed system, the carbon dioxide will be compressed adiabatically.

Please find the attached files for the solution and the remaining explanation.

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An urn contains r red, w white, and b black balls. Which has higher entropy, drawing k ~2 balls from the urn with replacement or
svetlana [45]

Answer:

The case with replacement has higher entropy

Explanation:

The complete question is given:

'Drawing with and without replacement. An urn contains r red,  w white  and b black balls. Which has higher entropy, drawing k ≥ 2 balls from the urn with  replacement or without replacement?'

Solution:

- n drawing is the same irrespective of whether there is replacement or not.

-  X to denotes drawing from an urn with r red balls,  w white balls and b black balls. So, n = b + r +  w.

We have:

                                      p_X(cr) = r / n

                                      p_X(cw) = w / n

                                      p_X(cb) = b / n

- Now, if  Xi is the ith drawing with replacement then Xi are independent and p_Xi (x) = pX(x) for x e ( cr , cb , cw ).

- Now, let  Yi be the ith drawing with replacement where Yi are not independent p_Yi (x) = pX(x) for x ∈ X.

- To see this, note  Y1 =  X and assume it is true for  Yi and consider  Yi+1:

    p_Y(i+1) (cr) = p(Y(i+1),Yi).(cr, cr) + p(Y(i+1),Yi).(cr, cw) + p(Y(i+1),Yi).(cr, cb)

= pY(i+1)|Yi  (cr|cr)*pYi  (cr) +pY(i+1)|Yi  (cr|cw)*pYi (cw) + pY(i+1)|Yi  (cr|cb)*pYi (cb)

= r*( r - 1 )/n*(n-1) + w*r/n*(n-1) + b*r/n*(n-1) = r / n =  p_X(cr)

- This means, using the chain rule and the conditioning theore m:

H(Y1, Y2, . . . , Yn) =  H(Y1) +  H(Y2|Y1) +  H(Y3|Y2, Y1) + ... H(Yn|Yn−1, . . . , Y1)

=< SUM H(Yi) = n*H(X) =  H(X1, X2, . . . , Xn)

- with equality if and only if the  Yi were independent:

                          H(Y1, Y2, . . . , Yn) < H(X1, X2, . . . , Xn)

Answer: The case with replacement has higher entropy

   

4 0
3 years ago
Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain
fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

N_{A}  = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1}  } ln(\frac{P_{T} -P_{A2}  }{P_{T} -P_{A1} })

Where

D_{AB} = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

P_{A1} = Pressure at first boundary

P_{A2} = Pressure at the destination boundary

T = System temperature

P_{T} = System pressure

Where P_{T} = 101.3 kPa P_{A2} =0, P_{A1} =y_{A}, P_{T} = 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R =  \frac{kJ}{(kmol)(K)} ,    T = 298 K   and  D_{AB} = 1.18 \frac{cm^{2} }{s} = 1.8×10⁻⁵\frac{m^{2} }{s}

N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65}) = 5.153×10⁻⁴\frac{kmol}{m^{2}s }

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

5 0
3 years ago
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