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Paha777 [63]
3 years ago
14

Points puntos points

Engineering
1 answer:
masha68 [24]3 years ago
4 0
Ummmmmmm do u need help with anything ✌?
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3 years ago
A seasonal color change helps animals in the Taiga to
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Answer:

Evade predators

Explanation:

The taiga forest biome which is located in the subartic region bounded by the tundra and the temperate forests in the North and South respectively, extends across the continents of North America, Europe and Asia forms the most enormous land ecosystem of related plants and animals on Earth

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4 years ago
Carbon dioxide gas enters a pipe at 3 MPa and 500 K at a rate of 2 kg/s. CO2 is cooled at constant pressure as it flows in the p
Viktor [21]

Answer:

a) 0.0629 m^{3}/s,ρ1=31.76 kg/m^{3},0.05688 m^{3}/s

b) 0.06171 m^{3}/s,32.40966 kg/m^{3}0.05498 m^{3}/s

c) 0.004%

given data:

n=2 kg/s

R=188.92 J/kgK

T_{1}=500 K

T_{2}=450 K

solution:

a) as we know ideal gas relation

PV=nRT..........(1)

The volume flow rate of inlet is

V_{1}= \frac{nRT_{1} }{P}..................(2)

putting the value in eq(2)

V_{1}  =0.0629 m^{3}/s

to find destiny

ρ1=\frac{n}{V_{1} }...........................................(3)

putting value of n and V1 is eq 3

ρ1=31.76 kg/m^{3}

V_{2} =\frac{nRT_{2} }{P}............(A)

putting value in eq A

    =0.05688 m^{3}/s

to find the rates from the compressibility factors we have to find the reduced pressure and temperature in both cases

P(reduced 1)=\frac{P_{1} }{P_{crit} }=\frac{3.10^{6} }{7.39 . 10^{6}}=0.41

T(reduced 1)= \frac{T_{1} }{T_{crit} }=\frac{500K}{304.25K}=1.64

b) After looking at the chart we obtain Z_{1}=0.98 now the volume flow rate at the inlet is:  

V_{1} =\frac{nRT_{1} Z_{1} }{P} ....................................................(4)

Putting the values in eq(4)      

    =0.06171 m^{3}/s        

Now the density is

ρ=\frac{n}{V_{1} }

  =32.40966 kg/m^{3}

now the reduced pressure and compressibility factor in second case will be

T(reduced 2)=\frac{T_{2} }{T_{crit} }=\frac{450}{304.25}=1.48

P(reduced 2)=P(reduced 1)=0.41

Z_{2}=0.97

V_{2} =\frac{nRT_{2} Z_{2} }{P}............................................(5)

putting value in eq (5)

    =0.05498 m^{3}/s

c) now find the error

Δn/n=modulus(ρ1V_{1}-n/n)*100%.....................(6)

putting the value in eq 6

      =0.004%

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