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bogdanovich [222]

3 years ago

12

A three-phase, 480 Volt, 120 horsepower, 50 Hertz four-pole induction motor delivers rated output power at a slip of 4%. Determi

ne the synchronous speed, the actual motor speed and the slip speed?

2 answers:

Bezzdna [24]3 years ago

8 0

Answer:

Explanation:

Given:

Number of poles, np = 4 poles

E = 480 V

Frequency, f = 50 Hz

Slip, s = 4%

A.

Speed, Vs = 120 x Frequency (Hz)/number of poles

= 120 × 50/4

= 1500 rpm

B.

Actual motor speed, V = Vs × (1 - s)

V = 1500 × (1 - 0.04)

= 1440 rpm

C.

Slip speed, s = Vs - V

= 1500 - 1440

= 60 rpm

kkurt [141]3 years ago

4 0

Answer:

a) $n_{s} = 750\,rpm$ , b) $n_{r} = 720\,rpm$ , c) $n_{p} = 30\,rpm$

Explanation:

a) Synchronous speed is:

$n_{s} = f\cdot \frac{60}{n_{poles}}$

$n_{s} = (50\,hz)\cdot \left(\frac{60}{4}\right)$

$n_{s} = 750\,rpm$

b) Actual motor speed is:

$S = \frac{n_{s}-n_{r}}{n_{s}} \times 100\%$

$n_{r} = \left(1-\frac{S}{100}\right)\cdot n_{s}$

$n_{r} = \left( 1 - \frac{4}{100} \right)\cdot 750\,rpm$

$n_{r} = 720\,rpm$

c) Slip speed is:

$n_{p} = n_{s} - n_{r}$

$n_{p} = 750\,rpm - 720\,rpm$

$n_{p} = 30\,rpm$

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An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$

From the question the, the mass flow rate at the inlet $m_{1}}$ is 2Kg/s $\\$

Since 5% flow is delivered into the feedwater heating, $\\$

$m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s$ $\\$

Also for the outlet 3 the remaining 95% will flow out. Hence

$m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s$ $\\$

Now, from $m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$ we substitute values

$W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}$

$W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)$

$\\$

$W_{out} = 2285.19 kW$ .

Hence the power produced is 2285kW

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Explanation:

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kompoz [17]

Answer:

Explanation:

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