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zhenek [66]
3 years ago
9

Find the compressibility factor Z for oxygen at 3 MPa and 160 K.

Engineering
1 answer:
saveliy_v [14]3 years ago
5 0

Answer:

Z= 0.868

Explanation:

Given that

P= 3 MPa

T = 160 K

We know that

P v= Z R T

P= Pressure

v = specific volume

R= gas constant

T = Absolute temperature

Z=  Compressibility factor

Here specific volume of gas is not given so we assume that specific volume gas

v=0.012\ m^3/kg

We know that for oxygen gas constant

R = 0.259 KJ/kg.K

Now by putting the values

P v = Z R T

3000 x 0.012 = Z x 0.259 x 160

Z= 0.868

So  Compressibility factor is 0.868.

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Answer:

b.

Matrix chain multiplication

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p[] = {5,10,3,12,5,50}

M[0][0] = 0,M[1][1] = 0,M[2][2] = 0,M[3][3] = 0,M[4][4] = 0,M[5][5] = 0,

M[1][2] = M[1][1]+M[2][2]+p[0]*p[1]*p[2] = 0+0+5*10*3 = 150

M[2][3] = M[3][3]+M[2][2]+p[1]*p[2]*p[3] = 0+0+10*3*12 = 360

M[3][4] = M[3][3]+M[4][4]+p[2]*p[3]*p[4] = 0+0+3*12*5 = 180

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M[1][3] = min{M[1][1]+M[2][3]+p[0]*p[1]*p[3] , M[1][2]+M[3][3]+p[0]*p[2]*p[3]}

= {0 + 360 + 600 , 150+0+180} = {960,330} = 330

M[2][4] = min{M[2][2]+M[3][4]+p[1]*p[2]*p[4] , M[2][3]+M[4][4]+p[1]*p[3]*p[4]}

= {0 + 180 + 150 , 360+0+600} = {960,330} = 330

M[3][5] = min{M[3][3]+M[4][5]+p[2]*p[3]*p[5] , M[3][4]+M[5][5]+p[2]*p[4]*p[5]}

= {0 + 3000 + 1800 , 180+0+750} = {4800,930} = 930

M[1][4] = min{M[1][1] + M[2][4] +p[0]*p[1]*p[4] ,M[1][2] + M[3][4] +p[0]*p[2]*p[4] ,

M[1][3] + M[4][4] +p[0]*p[3]*p[4]}

{0+330+250 , 150+180+75 , 330+0+300} = 405

M[2][5] = min{M[2][2] + M[3][5] +p[1]*p[2]*p[5] ,M[2][3] + M[4][5] +p[1]*p[3]*p[5] ,

M[2][4] + M[5][5] +p[1]*p[4]*p[5]}

{0+930+1500 , 360+3000+6000,330+0+2500} = 2430

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{0+2430+2500 , 150+930+750 , 330+3000+3000 , 405+0+1250} = 1655

(a)

MemoizedCutRod(p, n)

r: array(0..n) := (0 => 0, others =>MinInt)

return MemoizedCutRodAux(p, n, r)

MemoizedCutRodAux(p, n, r)

if r(n) = 0 and then n /= 0 then -- check if need to calculate a new solution

q: int := MinInt

for i in 1 .. n loop

q := max(q, p(i) + MemoizedCutRodAux(p, n-i, r))

end loop

end if

r(n) := q

end if

return r(n)

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3 years ago
Air enters a compressor at 100 kPa and 25 ⁰C. It is compressed to 2 MPa and exits the compressor at 540 K. The compressor is at
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(a) The reversible work is 207 kJ/kg

(b) The irreversibility rate is -38.39 kJ/kg

Explanation:

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From air table, h1 =298.18 kJ/kg, s10= 1.69528 kJ/kgK

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h2a= 544.35 kJ/kg,s2a0=2.29906

actual work input to the compressor =wout=h1-h2+Qin

=298.18-544.35+(-150)kJ/kg(- sign indicate heat loss)

=(-246.17)kJ/kg(-ve sign indicates the work is given into the system

a) Reversible work= Win actual - any irreversiblities present

                             =246.17 + irreversibilty

b) irreversibility = T0(Entopy generation Sgen) for air, Sgen

                         =s20-s10-Rln(p2/p1), T0=250C

                         =(25+273)(s2a0-s10-Rlnp2/p1+Qout/Tsurr)

    = 298x[(2.29906-1.69528-0.287kJ/kgK xln(2000kpa/100) + 150 /298]

  = -38.39 kJ/kg

a)Reversible work = Win actual -any irreversiblities present                  

                           =246.17 + irreversibilty

                           =246.17+-38.39

                          =207 kJ/kg

8 0
3 years ago
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