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Vsevolod [243]
3 years ago
6

The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of the air

craft is _____.
Engineering
1 answer:
navik [9.2K]3 years ago
8 0

Answer:

Ma=\frac{260}{330} \\Ma=0.7878

So, Ma < 1  Flow is Subsonic

Explanation:

Mach Number:

Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)

Mach < 1       Subsonic

Mach > 1       Supersonic

Ma= Speed of the object/Speed of the sound

Ma=\frac{260}{330} \\Ma=0.7878

So, Ma < 1 Flow is Subsonic

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Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo
AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

     = ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

           

7 0
3 years ago
In an ideal gas, specific enthalpy is a function of i. Entropy ii. Temperature iii, Pressure iv. Mass
Mice21 [21]

Answer:

Temperature

Explanation:

In an ideal gas the specific enthalpy  is exclusively a function of Temperature only this can be also written as h = h(T)  

A gas is said be ideal gas if obeys PV= nRT law

And in a ideal gas both internal energy and specific enthalpy are a function of Temperature only. Therefore the constant volume and constant pressure specific heats Cv and Cp are also function of temperature only.

5 0
3 years ago
Basic Question please help
Dvinal [7]
1(A)
2(B)
3(E)
4(C)
5(D)
6(B)
7(C)
7 0
3 years ago
A cantilever timber beam with a span of L = 4.25 m supports a linearly distributed load with maximum intensity of w0 = 5.5 kN/m.
9966 [12]

Answer:

the minimum width is b= 0.1414m = 141mm

Explanation:

]given,

L= 4.25

w₀ = 5.5kN/m,

allowable bending stress = 7MPa

allowable shear stress = 875kPa

h/b = 0.67

b = ?

for a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m,

the maximum moment, M exerted by the timber is =  \frac{w₀ L²}{9√3}[/texM = w₀ L²}/{9√3 = 99.34/15.6 =6.367kNmfor a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m, the shear force, V =  [tex]\frac{w₀ L}{2}[/texV = {w₀ L}/{3} = 7.79kNfor maximum bending stress of a rectangular timber, B, = [tex]\frac{6M}{bh²}

given h/b = 0.67, i.e h=0.67b

allowable bending stress = \frac{6M}{bh²} = 7000kPa

7000  = (6*6.37)/ (b *(0.67b)² ) = 38.21/0.449b³

3080b³=38.21

b³ = 38.21/3080 = 0.0124

b = 0.232m

h=0.67b = 0.67* 0.232 = 0.155m

for allowable  shear stress = (3V)/(2bh)

875 = (3V)/(2bh) = (3x 7.79)/(2xbx0.67b)

875 = 23.375/1.34b²

1172.5 b²= 23.375

b² =0.0199

b= 0.1414m

h=0.67b = 0.67* 0.1414 = 0.095m

the minimum width is b= 0.1414m = 141mm

4 0
3 years ago
The concentration of carbon monoxide (co) in an exhaust gas is 1x10^4 ppmv. what is the concentration in mg/m^3 at 25 and 1 atm
Mrrafil [7]

Answer:

1.1451 x 104 (11451.13)mg/m3

Explanation:

1 ppmv is defined as one volume of a contaminant or solid(CO)(mL) in 1 x 106 volume of solvent/water.

1ppmv = 1mL/m3

Concentration in mg/m3 = volume in ppm x molecular weight x pressure(kPa)/( gas constant x temperature(K)

Molecular weight of CO = 12 + 16

= 28g/mol

Temperature = 273.15 + 25

= 298.15K

Pressure = 1 x 101.325kPa

= 101.325kPa

Ppmv = 1 x 10-4ppmv

Gas constant, R = 8.3144 L.kPa/mol.K

Concentration in mg/m3 = (1 x 104 * 28 * 101.325)/(8.3144 * 298)

= 1.1451 x 104mg/m3

= 11451.13 mg/m3

3 0
3 years ago
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