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Lyrx [107]
2 years ago
15

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

Engineering
1 answer:
Marysya12 [62]2 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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I don’t understand this
blondinia [14]

Answer:

Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of 160cm^{2} by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.

Have an amazing day and I hope this can somewhat help :)

7 0
3 years ago
Marco is a franchisee with Daggies, a chain of sandwich shops. His business was doing well until several Daggies franchisees got
Oliga [24]

Answer:

The coattail effect

Explanation:

Clearly, the Daggies franchise were a huge part of Marco's success and attracted many customers for the business. This is why Marco deteriorated as soon as many Daggies franchisees closed.

The coattail effect is the phenomenon where an influencing member in a party (franchisee in this case)  contributes largely to the success of another, which is the case with Marco and Daggies

3 0
3 years ago
For the data points (1,1),(2,1/2),(3,1/3),(4,1/4), finde the natural cubic spline.
kow [346]

Answer:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

Explanation:

A cubic has the form:

y = ax³ + bx² + cx + d

Given four points, we can write a system of equations:

1 = a + b + c + d

1/2 = 8a + 4b + 2c + d

1/3 = 27a + 9b + 3c + d

1/4 = 64a + 16b + 4c + d

Solving this algebraically would be time-consuming, but we can use matrices to make it easy.

\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right]\left[\begin{array}{cccc}a\\b\\c\\d\end{array}\right]=\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right]

First, we find the inverse of the coefficient matrix.  This is messy to do by hand, so let's use a calculator:

\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right] ^{-1} =-\frac{1}{12}\left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]

Now we multiply by the solution matrix (again using a calculator):

-\frac{1}{12} \left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right] =\left[\begin{array}{cccc}-1/24\\5/12\\-35/24\\25/12\end{array}\right]

So the cubic is:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

7 0
3 years ago
The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate
Sonja [21]

Answer:

a) Mechanical efficiency (\varepsilon)=63.15%  b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, Wmec=0.95*6kW=5.7 kW.

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump \Delta P. So P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W.

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is \varepsilon=3.6kW/5.7kW=0.6315=63.15\%

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and \rho the density of the water:

[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T[/tex]

Finally, the temperature rise is:

\Delta T=2100/75348 \ºC=0.028 \ºC

7 0
3 years ago
A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where the
Archy [21]

Answer:

Work transfer is - 97.02 KJ. It means that work is given to the system.

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.

Explanation:

Given that

m= 3 kg

P₁=2 bar

T=T₁=T₂=30 °C

T=303 K

P₂=2.5 bar

PV=  Constant

This is the isothermal process .

We know that work for isothermal process  given as

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=mRT\ln \dfrac{P_1}{P_2}

For air

R= 0.287 KJ/Kg.K

Now by putting the values

W=mRT\ln \dfrac{P_1}{P_2}

W=5\times 0.287\times 303\ln \dfrac{2}{2.5}

W= - 97.02 KJ

So the work transfer is - 97.02 KJ. It means that work is given to the system.

We know that for ideal gas internal energy is the only function of temperature.The change in internal energy ΔU

ΔU = m Cv ΔT

Here ΔT= 0

So

ΔU =0

From first law of thermodynamics

Q= ΔU +W

ΔU = 0

Q= W

Q= - 97.02 KJ

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.

8 0
3 years ago
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