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Anna71 [15]
3 years ago
12

At the inlet to the combustor of a supersonic combustion ramjet (or scramjet), the flow Mach number is supersonic. For a fuel-ai

r mass ratio of .03 and a combustor exit temperature of 4800 degr, calculate the inlet Mach number above which the flow will be unchoked. Assume one- dimensional frictionless flow with y = 1.4, with heat release per slug of fuel equal to 4.5 x 108 ft-lbf. (Anderson 3.11) The temperatures for the above combustion problems are pretty high. A more representative value of the heat capacity ratio for products of combustion at these temperatures is y = 1.28 The value of the specific gas constant R may still be approximated as equal to that of air. b) For problem #2, calculate the inlet Mach number above which the flow will be unchoked, based on y = 1.28

Engineering
1 answer:
Daniel [21]3 years ago
8 0

Answer:

The inlet match number is 3.42.

Explanation:

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3 years ago
A construction crew lifts approximately 400 lb. of material several times during a day from a flatbed truck to a 25 ft. rooftop.
Irina18 [472]

Answer:

2ib

Explanation:

if you divide 10 divided by 2 it gives you 5 and then subtract it by 2.2 = 2.8

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5 0
1 year ago
5. (5 points) Select ALL statements that are TRUE A. For flows over a flat plate, in the laminar region, the heat transfer coeff
finlep [7]

Answer:

The following statements are true:

A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction

C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface

E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.

Select ALL statements that are TRUE

B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant

D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed

Explanation:

7 0
3 years ago
A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest
sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

y = ax^{2} +bx+c\\dy/dx=2ax+b\\

At highest point of the curve dy/dx=o

2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275

-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\

Station PVT

Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)

3 0
3 years ago
A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De
andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0
2 years ago
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