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Nat2105 [25]
3 years ago
7

Is Micah a idiot true or false ooooooooooooooooooooooooooooooooooo

Engineering
1 answer:
Vedmedyk [2.9K]3 years ago
6 0
False because yeah jkdkdlgkdjfkekvkx
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A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a
nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, g_{c} = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

c_{p}= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence v_{a} ^{2} = 0

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} +  \frac{v_{1} ^{2} }{2c_{p} }

T_{1} = 20+460 = 480°R

T_{a}  =480+  \frac{(800)(800}{2(0.240)(25037}= 533.25°R

Pressure at the inlet of compressor at isentropic condition

P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}

P_{a} = (10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}= 14.45 psia

P_{2}= 8P_{a} = 8(14.45) = 115.6 psia

4 0
3 years ago
Read 2 more answers
The parts of a feature control frame are the tolerance value, the datum references, and the
Elan Coil [88]

Answer:

d

Explanation:

4 0
3 years ago
Two previously undeformed rod-shaped specimens of copper are to be plastically deformed by reducing their cross-sectional areas.
mezya [45]
I am not sure I am stuck on this and I have been for 45 min someone please help me and this girl or boy!!
4 0
2 years ago
Which of the following best describes the role of engineers
Fantom [35]

Problem Solvers

Explanation:

Engineers find problems in the world, and then they find solutions for them.

8 0
3 years ago
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
ad-work [718]

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.

5 0
3 years ago
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