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vesna_86 [32]
3 years ago
15

For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho

ur c. 2000 km per hour d. 200 km per hour
Engineering
1 answer:
ad-work [718]3 years ago
5 0

Answer:d

Explanation:

Given

Temperature=200^{\circ}\approc 473 K

Also \gamma for air=1.4

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no.=\frac{V}{\sqrt{\gamma RT}}

(a)1000 km/h\approx 277.78 m/s

Mach no.=\frac{277.78}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.63

(b)500 km/h\approx 138.89 m/s

Mach no.=\frac{138.89}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.31

(c)2000 km/h\approx 555.55 m/s

Mach no.=\frac{555.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=1.27

(d)200 km/h\approx 55.55 m/s

Mach no.=\frac{55.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

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Answer:

1) the final temperature is T2 = 876.76°C

2) the final volume is V2 = 24.14 cm³

Explanation:

We can model the gas behaviour as an ideal gas, then

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since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:

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then the work will be

W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)

W = (P1*V1/T1)*(T2-T1)/(1-k)  

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T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C

the final volume is

TV^(k-1)= constant

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3 0
3 years ago
Hello, I have a question, I would be glad if you can help.
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Answer:

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aksik [14]

Answer:

Para x=0:

\phi=1.2 W/m^{2}  

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Explanation

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4 0
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olga_2 [115]

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