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3 years ago

determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004

Engineering

1 answer:

Afina-wow [57]3 years ago

6 0

Answer:

hello your question is incomplete attached below is the complete question

A) optimum compressor ratio = 9.144

B) specific thrust = 2.155 N.s /kg

C) Thrust specific fuel consumption = 1670.4 kg/N.h

Explanation:

Given data :

Mo = 2.1 , To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k

γ = 1.4

attached below is the detailed solution

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What is the difference between the pressure head at the end of a 150m long pipe of diameter 1m coming from the bottom of a reser

uysha [10]

Answer:

$\frac {p_2- p_1}{\rho g} = 31.06 m$

Explanation:

from bernoulli's theorem we have

$\frac{p_1}{\rho g} + \frac{v_1^{2}}{2g} +z_1 = \frac{p_2}{\rho g} + \frac{v_2^{2}}{2g} +z_2 + h_f$

we need to find pressure head difference i.e.

$\frac {p_2- p_1}{\rho g} = (z_1 - z_2) - h_f$

where h_f id head loss

$h_f = \frac{flv^{2}}{D 2g}$

velocity v = $\frac{1}{n} * R^{2/3} S^{2/3}$

$S = \frac{\delta h}{L} = \frac{40}{150} = 0.267$

hydraulic mean radius R = $\frac{A}{P} = \frac{hw}{2h+w}$

$R = \frac{40*1}{2*40+1} = 0.493 m$

so velocity is = $\frac{1}{0.013} * 0.493^{2/3} 0.267^{1/2}$

v = 24.80 m/s

head loss

$h_f = \frac{0.0019*150*24.80^{2}}{1* 2*9.81}$

$h_f =8.93 m$

pressure difference is

$\frac {p_2- p_1}{\rho g} = 40 - 8.93 = 31.06 m$

$\frac {p_2- p_1}{\rho g} = 31.06 m$

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3 years ago

Technician A says that 18 gauge AWG wire can carry more current flow that 12 gauge AWG wire. Technician B says that metric wire

denpristay [2]

Answer:

Technician B

Explanation:

Both AWG and metric are sized by cross-sectional area.

Technician A is wrong: 12 gauge wire is larger diameter rated for 20 amps in free air. 18 awg is smaller diameter and typically used for speaker wiring, Class II or low voltage and sub-circuits within appliances.

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3 years ago

Which option identifies why Ethan’s skills are valuable to his team in the following scenario?

larisa [96]

Answer:

Explanation:

The options are:

- In an isometric drawing, multiple angles and axes can be shown in one sketch.

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- Computer programs will not be necessary to create the exact dimensions of the design.

Orthographic projections are in either the First or Third Angles but the angles are fixed and do not provide perspective view. Isometric drawings are perspective views from different angles.

So Ethan's skill is valuable because "In an isometric drawing, multiple angles and axes can be shown in one sketch."

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2 years ago

Air modeled as an ideal gas enters a well-insulated diffuser operating at steady state at 270 K with a velocity of 180 m/s and e

ZanzabumX [31]

Answer:

exit temperature 285 K

Explanation:

given data

temperature T1 = 270 K

velocity = 180 m/s

exit velocity = 48.4 m/s

solution

we know here diffuser is insulated so here heat energy is negleted

so we write here energy balance equation that is

0 = m (h1-h2) + m × $(\frac{v1^2}{2}-\frac{v2^2}{2})$ .....................1

so it will be

$h1 + \frac{v1^2}{2} = h2 + \frac{v2^2}{2}$ .....................2

put here value by using ideal gas table

and here for temperature 270K

h1 = 270.11 kJ/kg

$270.11 + \frac{180^2\times \frac{1}{1000}}{2} = h2 + \frac{48.4^2\times \frac{1}{1000}}{2}$

solve it we get

h2 = 285.14 kJ/kg

so by the ideal gas table we get

T2 = 285 K

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3 years ago

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Answer:

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3 years ago