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3 years ago

What is the condition for maximum efficiency in a DC motor?

Engineering

1 answer:

Marysya12 [62]3 years ago

3 0

Answer:

The efficiency of a DC generator is maximum when those losses proportional to the square of the load current are equal to the constant losses of the DC generator. This relation applies equally well to all rotating machines, regardless of the type of machine.

Explanation:

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A spherical gas container made of steel has a(n) 17-ft outer diameter and a wall thickness of 0.375 in. Knowing that the interna

Arte-miy333 [17]

Answer:

Maximum Normal Stress σ = 8.16 Ksi

Maximum Shearing Stress τ = 4.08 Ksi

Explanation:

Outer diameter of spherical container D = 17 ft

Convert feet to inches D = 17 x 12 in = 204 inches

Wall thickness t = 0.375 in

Internal Pressure P = 60 Psi

Maximum Normal Stress σ = PD / 4t

σ = PD / 4t

σ = (60 psi x 204 in) / (4 x 0.375 in)

σ = 12,240 / 1.5

σ = 8,160 P/in

σ = 8.16 Ksi

Maximum Shearing Stress τ = PD / 8t

τ = PD / 8t

τ = (60 psi x 204 in) / (8 x 0.375 in)

τ = 12,240 / 3

τ = 4,080 P/in

τ = 4.08 Ksi

7 0

3 years ago

A closed system consists of 0.3 kmol of octane occupying a volume of 5 m³. Determine (a) the weight of the system, in N, and (b)

Leni [432]

Answer:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

Explanation:

To calculate the mass of the octane(m):

Number of mole of octane (n) =0.3kmol(given)

Molarmass of octane (M) =114.23kg/kmol

m=n*M

m=(0.3kmol)*(114.23kg/kmol)

m=34.269kg

To calculate for the weight of octane(W):

W=g*m

W=(9.81m/s^2)*(34.269kg)

W=336.18N

b) For specific volumes of Vn and Vm:

Given volume of octane (V) =5m^3

Vm=V/m

Vm=5m^3/34.269kg

Vm=0.1459m^3/kg

And Vn will be :

Vn=V/m=5m^3/0.3kmol

Vn=16.67m/Kmol

Therefore, the answers are:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

7 0

3 years ago

can anyone give me tips on adding HP to my 2014 dodge charger SE? it uses the 3.6L220 CI 24 valve v6 vvt (variable valve timing)

marissa [1.9K]

<h2>ANSWER</h2><h2></h2>

I had a couple of answers for this, but when I checked nothing

was right, so im not sure.

5 0

3 years ago

Calculate the magnitude of the velocity and the θ angular direction of the block and the bullet together when the 50 g bullet mo

almond37 [142]

Answer:

Magnitude of the velocity = 16.82 m/s

Angular direction, θ = 52.41°

Explanation:

As given ,

mass of bullet, m₁= 50g = 0.05 kg

speed of bullet , u₁ = 600 m/s

mass of the block , m₂ = 4 kg

speed of the block before collision , u₂ = 12 m/s

direction , θ = 30°

Now,

Assume that the combined velocity of bullet and block after collision = v

and the direction = θ

Now, from the conservation of momentum in x - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) vₓ

where v = final velocity after collision

u₁ = initial velocity of bullet before collision = 0

m₁ = mass of the bullet before collision = 0.05 kg

u₂ = velocity of block before collision = 12 cos(30° )

m₂ = mass of block before collision

m₁ + m₂ = combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (0) + 4(12 cos(30° ) ) = ( 0.05 + 4 ) vₓ

⇒ 0 + 4(6√3) = 4.05 vₓ

⇒24√3 = 4.05 vₓ

⇒vₓ = 10.26 m/s

Now, from the conservation of momentum in y - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) $v_{y}$

where v = final velocity after collision

u₁= initial velocity of bullet before collision = 600

m₁ = mass of the bullet before collision = 0.05 kg

u₂ = velocity of block before collision = 12 sin(30° )

m₂= mass of block before collision

m₁+ m₂= combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (600) + 4(12 sin(30° ) ) = ( 0.05 + 4 ) $v_{y}$

⇒ 30 + 4(6) = 4.05 $v_{y}$

⇒30 +24 = 4.05 $v_{y}$

⇒54 = 4.05 $v_{y}$

⇒ $v_{y}$ = 13.33 m/s

Now, the magnitude of the velocity = √vₓ² + $v_{y}$ ² = √(10.26)² + (13.33)²

= √105.26 + 177.68

= √282.95 = 16.82

The angular direction, θ = $tan^{-1}$ ( $\frac{v_{y} }{v_{x} }$ ) = $tan^{-1}$ ( $\frac{13.33}{10.26}$ ) = $tan^{-1}$ (1.299) = 52.41°

8 0

3 years ago

An article that discusses pharaohs and gives examples and explains how they look. Which text structure is that?

fgiga [73]

The correct answer to this open question is the following.

The text structure of an article that discusses pharaohs and gives examples and explains how they look is a description.

The text structure called description allows the reader to fully know the characteristics of the people it is referring to, including some important details. That is why the author of a description text adds words like "such as" and "for example."

When describing something, the write is giving structure to the text and sequence. What comes first., what is followed, and so on.

That is why The text structure of an article that discusses pharaohs and gives examples and explains how they look is a description. It includes cause and effect sentences, and some comparisons in order to contrast an idea.

3 0

3 years ago