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Mkey [24]
3 years ago
12

Two standard spur gears have a diametrical pitch of 10, a center distance 3.5 inches and a velocity ratio of 2.5. How many teeth

are on each gear?
Engineering
1 answer:
lubasha [3.4K]3 years ago
6 0

Answer:50 , 20

Explanation:

Given

Diametrical Pitch\left ( P_D\right )=\frac{T}{D}

where T= no of teeths

D=diameter

module(m) of gears must be same

m=\frac{D}{T}=\frac{1}{P_D}=0.1

Let T_1 & T_2be the gears on two gears

Therefore Center distance is given by

m\frac{\left ( T_1+T_2\right )}{2}=3.5

thus

0.1\frac{\left ( T_1+T_2\right )}{2}=3.5

T_1+T_2=70----1

and Velocity ratio is given by

VR=\frac{No\ of\ teeths\ on\ Driver\ gear}{No.\ of\ teeths\ on\ Driven\ gear}

2.5=\frac{T_1}{T_2}----2

From 1 & 2 we get

T_1=50, T_2=20

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At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t
nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

PGP = \rho *\frac{\pi r^2}{2} *V*e  \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0
3 years ago
A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3200
coldgirl [10]

Answer:

Speed of aircraft ; (V_1) = 83.9 m/s

Explanation:

The height at which aircraft is flying = 3000 m

The differential pressure = 3200 N/m²

From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3

Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.

Thus, let's apply the Bernoulli equation :

P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2

Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.

We'll obtain ;

P1/ρg + (V_1)²/2g = P2/ρg

Let's make V_1 the subject;

(V_1)² = 2(P1 - P2)/ρ

(V_1) = √(2(P1 - P2)/ρ)

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(V_1) = √(2 x 3200)/0.909)

(V_1) = 83.9 m/s

4 0
3 years ago
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otez555 [7]

Answer:

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Explanation:

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Answer:

Hope it helps...

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