6.48 programming project 1: encode/decode tic -tac-toe

What is Back EMF? How does it limits the speed of a permanent magnet DC?

ss7ja [257]

Answer and Explanation:

The DC motor has coils inside it which produces magnetic field inside the coil and due to thus magnetic field an emf is induced ,this induced emf is known as back emf. The back emf always acts against the applied voltage. It is represented by $E_b$

The back emf of the DC motor is given by $E_b=\frac{NP\Phi }{60A}$

Here N is speed of the motor ,P signifies the number of poles ,Z signifies the the total number of conductor and A is number of parallel paths

As from the relation we can see that back emf and speed ar dependent on each other it means back emf limits the speed of DC motor

8 0

3 years ago

Refer to Figure 9-18. A #_____ electrode lead and workpiece lead should be used to carry 150 amperes of electricity 75 feet to t

Anni [7]

Answer:

AC or

Explanation:

BC

8 0

2 years ago

Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?

zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

Th

To know more about Rankine cycle, visit: brainly.com/question/13040242

#SPJ4

4 0

1 year ago

Steam enters a turbine operating at steady state with a mass flow of 10 kg/min, a specific enthalpy of 3100 kJ/kg, and a velocit

Natali [406]

Answer:

$\dot W_{out} = 133.327\,kW$

Explanation:

The model for the turbine can be derived by means of the First Law of Thermodynamics:

$-\dot Q_{out}-\dot W_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right] =0$

The work produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right]$

The mass flow and heat transfer rates are, respectively:

$\dot m = (10\frac{kg}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot m = 0.167\,\frac{kg}{s}$

$\dot Q_{out} = (0.167\,\frac{kg}{s} )\cdot (1.1\times 10^{3}\,\frac{J}{kg} )$

$\dot Q_{out} = 183.7\,W$

Finally:

$\dot W_{out} = -183.7\,W + (0.167\,\frac{kg}{s} )\cdot \left(8\times 10^{5}\,\frac{J}{kg} -562,5\,\frac{J}{kg} +29.43\,\frac{J}{kg} \right)$

$\dot W_{out} = 133.327\,kW$

3 0

3 years ago

In wet mill ethanol plants, a total energy of 74,488 Btu (British thermal units, a common energy unit) is used to produce 1 gall

V125BC [204]

Answer:

The energy yield for one gallon of ethanol is 2.473 %.

Explanation:

The net energy yield ( $\% e$ ), expressed in percentage for one gallon of ethanol is the percentage of the ratio of the difference of the provided energy ( $E_{g}$ ), measured in Btu, and the energy needed to produce the ethanol ( $E_{p}$ ), measured in Btu, divided by the energy needed to produce the ethanol. That is: