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jarptica [38.1K]
3 years ago
9

The Stokes-Oseen formula for drag force F on a sphere of diameter D in a fluid stream of low velocity V, density p and viscosity

μ is
F=3πμDV+9π/16∗pV2d2
Is this formula dimensionally homogenous?
Engineering
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer:

\frac{ML}{T^2}=\frac{ML}{T^2}

Hence it is proved that Stokes-Oseen formula is dimensionally homogenous.

Explanation:

For equation to be dimensionally homogeneous both side of the equation must have same dimensions.

For given Equation:

F= Force, μ= viscosity, D = Diameter, V = velocity, ρ= Density

Dimensions:

F=\frac{ML}{T^2}

\mu=\frac{M}{LT}

D=L\\\\V=\frac{L}{T}\\ \\\rho=\frac{M}{L^3}

Constants= 1

Now According to equation:

\frac{ML}{T^2}=[\frac{M}{LT}][L] [\frac{L}{T}] + [\frac{M}{L^3}][\frac{L^2}{T^2}][L^2]

Simplifying above equation, we will get:

\frac{ML}{T^2}=2*\frac{ML}{T^2}

Ignore "2" as it is constant with no dimensions. Now:

\frac{ML}{T^2}=\frac{ML}{T^2}

Hence it is proved that Stokes-Oseen formula  is  dimensionally homogenous.

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I need help with this question please
solniwko [45]

Answer:

The resultant moment is 477.84 N·m

Explanation:

We note that the resultant moment is given by the moment about a given point

The length of the sides of the formed triangles are;

l = sin(40°) × 4/sin(110°) ≈ 2.736

Taking the moment about the lower left hand corner of the figure, with the convention that clockwise moments are positive, we have;

The resultant moment, ∑m, is given as follow;

∑M = 250 N × 4 m + 400 N × cos(40°) × 4 m - 400 N × cos(40°) × 2 m + 400 N × sin(40°) × 2 m × tan(40°) - 600 N × cos(40°) × 2 m - 600 N× sin(40°) × 2 m × tan(40°) = 477.837084 N·m

Therefore, the resultant moment, ∑m ≈ 477.84 N·m clockwise.

6 0
2 years ago
Assume the average fuel flow rate at the peak torque speed (1500 rpm) is 15kg/hr for a sixcylinder four-stroke diesel engine und
sveticcg [70]

Answer:

Q = 8.845 DEGREE

Explanation:

given data:

combine Mass for 6 cylinder (M) =15 Kg/hr

mass of  each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec

Engine speed (N)= 1500rpm

Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m

Discharge Coefficient (Cd) = 0.75

Pressure difference = 100 MPa

Density of fuel = 800 kg/m^3

velocity of fuel is v  = cd\sqrt{\frac{2*P}{p}}

v = 0.75 \sqrt{\frac{2\times 100\times 10^6}{800}} = 375 m/sec

injected fuel volume  (V) =Area of given  Orifices × Fuel velocity × time of single injection × no of injection/sec

we know that p = m/ V

SoV = \frac{0.000694}{800} =8.68\times10^{-7} m3/sec

putting these value in volume equation and solve for Discharge 8.68\times 10^{-7} = (\frac{(3.14}{4})\times 6\times( .0002\times .0002) \times  375 \times  \frac{(Q}{360}) \times \frac{30}{750} \times \frac{(750}{60)}

Q = 8.845 DEGREE

4 0
2 years ago
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are current
postnew [5]

Answer:

D=0.016m

Explanation:

From the question we are told that:

Discharge Rate F_r=0.5kgls

Pressure P=15Kpa

Temperature T=25=>298K

Ambient pressure is 1 atm.

Generally the equation for Density is mathematically given by

\rho=\frac{PM}{RT}

\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}

\rho=16.958kg/m^2

Generally the equation for Flow rate is mathematically given by

F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}

Where

Q=Heat coefficient\ ratio\ of\ Nitrogen

Q=1.4

\mu= Discharge\ coefficient

\mu=0.68

Therefore

0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}

A=2.129*10^{-4}

Where

A=\frac{\pi}{4}D^2

\frac{\pi}{4}D^2=2.129*10^{-4}

D=0.016m

8 0
2 years ago
1 po
Dmitry_Shevchenko [17]

Answer:

D. Brake

Explanation:

NJMVC is an acronym for New Jersey Motor Vehicle Commission and it is an agency of government that was established in 2003. NJMVC is saddled with the responsibility of inspecting, titling of number plates and registration of motor vehicles, as well as licensing of drivers in New Jersey, United States of America.

The NJMVC subject the applicants (drivers) to a series of test before they are issued a valid driver's license, some of these tests include;

1. Vision test.

2. Knowledge test.

3. Road test.

All of the above mentioned tests must be passed by a driver before he or she are issued a valid driver's license.

Some of the test drive requirements are;

- Valid registration documents.

- Valid sticker of inspection.

- No obstacle or center consoles should prevent the examiner from accessing both the foot brake and parking brake.

Hence, NJMVC will reject a vehicle from the road test if the examiner does not have access to a brake. Thus, the brakes are required to be in good working condition prior to the road test.

However, if an applicant (driver) passes the road test he or she would be issued an authorization for licensing by the examiner and then a digital driver's license by the motor vehicle commission.

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3 years ago
Assume all surface are smooth Draw the FBD for each body
victus00 [196]

Answer: I’ll send picture

Explanation:

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