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Dvinal [7]
3 years ago
12

Air enters the combustor of a jet engine at Pressure=10 atm, T=1000R and M=0.2. Fuel is injected and burned with fuel-air ratio

(by mass) of 0.06. The heat released during the combustion is 4.50000000 ft-lb per slug of fuel. Assuming one-dimensional frictionless flow with r=1.4 for the fuel-air mixture, calculate M2, p2 and T2 at the exit of combustor?
Engineering
1 answer:
kkurt [141]3 years ago
5 0
Air enters the combustor of a jet engine at p1 = 10 atm, T1 = 1000 °R and M1 = 0.2. The
heat released during combustion is 4.5x108 ft-lb per slug of fuel. Assuming onedimensional
frictionless flow with γ = 1.4 for the fuel air mixture, calculate M2, p2, T2 at
the exit of the combustor. Hope this helps! Can I get brainly please? Plus safety first if you see any links don’t klick them please don’t.
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-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63
Fudgin [204]

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0
3 years ago
) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p
andrey2020 [161]

Answer:

number of pulses produced =  162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = \frac{linear\ displacement}{circumference} × No of pulses per revolution .................2

number of pulses produced = \frac{200}{314.16} × 256

number of pulses produced =  162 pulses

5 0
3 years ago
For an isotropic material, E and ν are often chosen as the two independent engineering constants. There are other elastic consta
pav-90 [236]

Answer:

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Explanation:

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8 0
3 years ago
Design a VHDL module
Aloiza [94]

Answer:

how we gonna do dat

Explanation:

5 0
2 years ago
When a retaining structure moves towards the soil backfill, the stress condition is called:__________.
Alecsey [184]

Answer:

(C) passive state.

Explanation:

The earth pressure is the pressure exerted by the soil on the shoring system. They are three types of earth pressure which are:

a) Rest state: In this state, the retaining wall is stationary, this makes the lateral stress to be zero.

b) Active state: In this state, the wall moves away from the back fill, this leads to an internal resistance. Hence the active earth pressure is less than earth pressure at rest

c) Passive state: In this state the wall is pushed towards the back fill, this leads to shearing resistance. Hence, the passive earth pressure is greater than earth pressure at rest

6 0
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