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Inga [223]
3 years ago
7

Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass

ratio of 0.06. The heat released during combustion is 4.5 x 10^8 ft-lbf per slug of fuel (or 18,000 BTU/lbm).
1. Assuming one-dimensional frictionless flow with γ = 1.4 for the fuel-air mixture, calculate M2, p2, and T2 at the exit of the combustor.
Engineering
1 answer:
snow_lady [41]3 years ago
5 0

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

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Multiple Choice
mote1985 [20]
I think it’s manufacturing
7 0
3 years ago
Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f
Papessa [141]

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit =  40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

4 0
3 years ago
Which of the following is a possible unit of ultimate tensile strength?
levacccp [35]

Answer:

Newton per square meter (N/m2)

Explanation:

Required

Unit of ultimate tensile strength

Ultimate tensile strength (U) is calculated using:

U = \frac{Ultimate\ Force}{Area}

The units of force is N (Newton) and the unit of Area is m^2

So, we have:

U = \frac{N}{m^2}

or

U = N/m^2

<em>Hence: (c) is correct</em>

4 0
2 years ago
A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground
7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

T ≈ 17199 lbf

4 0
3 years ago
Describe, in a general form, the equation, in time domain, that tells the voltage across a inductor, L, as a function of time wh
love history [14]

Answer:

a) V(t) = Ldi(t)/dt

b) If current is constant, V = 0

Explanation:

a) The voltage, V(t), across an inductor is proportional to the rate of change of the current flowing across it with time.

If  V represents the Voltage across the inductor

and i(t) represents the current across the inductor in time, t.

V(t) ∝ di(t)/dt

Introducing a proportionality constant,L, which is the inductance of the inductor

The general equation describing the voltage across the inductor of inductance, L, as a function of time when a current flows through it is shown below.

V(t) = Ldi(t)/dt ..................................................(1)

b) If the current flowing through the inductor is constant i.e. does not vary with time

di(t)/dt = 0   and hence the general equation (1) above becomes

V(t) = 0

4 0
3 years ago
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