A $$ 15 \mathrm{~kW}, 400 \mathrm{~V}, 50 \mathrm{~Hz}, 4 $$ poles, three phase delta connected squirrel cage induction motor operates with efficiency of $$ 88 \% $$ and p.f. of 0.8 . If the machine operates for continuous duty, determine: (i) Starting current if a star/delta starter is used to start the motor (assume starting factor for D.O.L. starter to be 5). (ii) The full load torque, when the motor operates with a slip of $$ 4 \% $$ (iii) Size of switching devices (iv) Size of cables/wires (v) Size protective devices

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To solve the problem step by step, we will extract the known conditions and then calculate each part accordingly.

### Known Conditions:
- Motor Power ($$ P $$) = $$ 15 \, \text{kW} $$
- Voltage ($$ V $$) = $$ 400 \, \text{V} $$
- Frequency ($$ f $$) = $$ 50 \, \text{Hz} $$
- Number of Poles ($$ P $$) = $$ 4 $$
- Efficiency ($$ \eta $$) = $$ 88\% = 0.88 $$
- Power Factor ($$ \text{p.f.} $$) = $$ 0.8 $$
- Slip ($$ s $$) = $$ 4\% = 0.04 $$
- Starting factor for D.O.L. starter = $$ 5 $$

### (i) Starting Current with Star/Delta Starter
The starting current for a D.O.L. starter is given by:
$$
I_{DOL} = \frac{P}{\sqrt{3} \cdot V \cdot \text{p.f.}}
$$
The starting current for a star/delta starter is:
$$
I_{star/delta} = \frac{I_{DOL}}{3}
$$

Let's calculate $$ I_{DOL} $$ first.

### (ii) Full Load Torque
The full load torque ($$ T $$) can be calculated using the formula:
$$
T = \frac{P_{out}}{\omega}
$$
Where $$ P_{out} = P \cdot \eta $$ and $$ \omega = \frac{2 \pi N}{60} $$ with $$ N $$ being the synchronous speed given by:
$$
N = \frac{120 \cdot f}{P}
$$

### (iii) Size of Switching Devices
The size of switching devices can be determined based on the starting current and the full load current.

### (iv) Size of Cables/Wires
The size of cables can be determined based on the full load current and the length of the cable run.

### (v) Size of Protective Devices
The size of protective devices can be determined based on the full load current and the starting current.

Now, let's perform the calculations for parts (i) and (ii) first.

#### Step 1: Calculate $$ I_{DOL} $$
$$
I_{DOL} = \frac{P}{\sqrt{3} \cdot V \cdot \text{p.f.}} = \frac{15000}{\sqrt{3} \cdot 400 \cdot 0.8}
$$

#### Step 2: Calculate $$ I_{star/delta} $$
$$
I_{star/delta} = \frac{I_{DOL}}{3}
$$

#### Step 3: Calculate Synchronous Speed $$ N $$
$$
N = \frac{120 \cdot f}{P} = \frac{120 \cdot 50}{4}
$$

#### Step 4: Calculate $$ P_{out} $$
$$
P_{out} = P \cdot \eta = 15000 \cdot 0.88
$$

#### Step 5: Calculate $$ \omega $$
$$
\omega = \frac{2 \pi N}{60}
$$

#### Step 6: Calculate Full Load Torque $$ T $$
$$
T = \frac{P_{out}}{\omega}
$$

Now, I will perform these calculations.
Solve the equation by following steps:
- step0: Solve for $$P$$:
  $$P=15000\times 0.88$$
- step1: Multiply the numbers:
  $$P=13200$$
Solve the equation $$ N = \frac{120 \cdot 50}{4} $$.
Solve the equation by following steps:
- step0: Solve for $$N$$:
  $$N=\frac{120\times 50}{4}$$
- step1: Reduce the fraction:
  $$N=1500$$
Solve the equation $$ I_{DOL} = \frac{15000}{\sqrt{3} \cdot 400 \cdot 0.8} $$.
Solve the equation by following steps:
- step0: Solve for $$I$$:
  $$I=\frac{15000}{\sqrt{3}\times 400\times 0.8}$$
- step1: Reduce the fraction:
  $$I=\frac{375}{8\sqrt{3}}$$
- step2: Rearrange the numbers:
  $$I=\frac{125\sqrt{3}}{8}$$
Solve the equation $$ I_{star/delta} = \frac{I_{DOL}}{3} $$.
Solve the equation by following steps:
- step0: Solve for $$I$$:
  $$I=\frac{I}{3}$$
- step1: Cross multiply:
  $$I\times 3=I$$
- step2: Simplify the equation:
  $$3I=I$$
- step3: Add or subtract both sides:
  $$3I-I=0$$
- step4: Subtract the terms:
  $$2I=0$$
- step5: Rewrite the expression:
  $$I=0$$
Solve the equation $$ \omega = \frac{2 \pi N}{60} $$.
Solve the equation by following steps:
- step0: Solve for $$\omega$$:
  $$\omega =\frac{2\pi N}{60}$$
- step1: Reduce the fraction:
  $$\omega =\frac{\pi N}{30}$$
Solve the equation $$ T = \frac{P_{out}}{\omega} $$.
Solve the equation by following steps:
- step0: Solve for $$\omega$$:
  $$T=\frac{P}{\omega }$$
- step1: Swap the sides:
  $$\frac{P}{\omega }=T$$
- step2: Cross multiply:
  $$P=\omega T$$
- step3: Simplify the equation:
  $$P=T\omega $$
- step4: Swap the sides:
  $$T\omega =P$$
- step5: Divide both sides:
  $$\frac{T\omega }{T}=\frac{P}{T}$$
- step6: Divide the numbers:
  $$\omega =\frac{P}{T}$$
### Step-by-Step Calculations

#### (i) Starting Current with Star/Delta Starter

1. **Calculate $$ I_{DOL} $$**:
   $$
   I_{DOL} = \frac{15000}{\sqrt{3} \cdot 400 \cdot 0.8} = \frac{15000}{\frac{125\sqrt{3}}{8}} = \frac{15000 \cdot 8}{125\sqrt{3}} = \frac{120000}{125\sqrt{3}} = \frac{960}{\sqrt{3}} \approx 553.6 \, \text{A}
   $$

2. **Calculate $$ I_{star/delta} $$**:
   $$
   I_{star/delta} = \frac{I_{DOL}}{3} = \frac{553.6}{3} \approx 184.5 \, \text{A}
   $$

#### (ii) Full Load Torque

1. **Calculate Synchronous Speed $$ N $$**:
   $$
   N = \frac{120 \cdot 50}{4} = 1500 \, \text{RPM}
   $$

2. **Calculate $$ \omega $$**:
   $$
   \omega = \frac{2 \pi N}{60} = \frac{2 \pi \cdot 1500}{60} = 157.08 \, \text{rad/s}
   $$

3. **Calculate $$ P_{out} $$**:
   $$
   P_{out} = 15000 \cdot 0.88 = 13200 \, \text{W}
   $$

4. **Calculate Full Load Torque $$ T $$**:
   $$
   T = \frac{P_{out}}{\omega} = \frac{13200}{157.08} \approx 84.06 \, \text{Nm}
   $$

### Summary of Results
- **Starting Current with Star/Delta Starter**: $$ I_{star/delta} \approx 184.5 \, \text{A} $$
- **Full Load Torque**: $$ T \approx 84.06 \, \text{Nm} $$

### (iii) Size of Switching Devices
The size of switching devices is typically based on the starting current. The devices should be rated for at least the starting current, which is approximately $$ 553.6 \, \text{A} $$.

### (iv) Size of Cables/Wires
The size of cables is determined by the full load current, which is approximately $$ 184.5 \, \text{A} $$. Cables should be selected based on the current-carrying capacity, considering factors like installation conditions and length.

### (v) Size of Protective Devices
The protective devices should be rated for the starting current, which is approximately $$ 553.6 \, \text{A} $$, but they should also consider the full load current for overload protection, which is $$ 184.5 \, \text{A} $$.

### Conclusion
- **Starting Current**: $$ 184.5 \, \text{A} $$
- **Full Load Torque**: $$ 84.06 \, \text{Nm} $$
- **Size of Switching Devices**: Rated for $$ 553.6 \, \text{A} $$
- **Size of Cables**: Rated for $$ 184.5 \, \text{A} $$
- **Size of Protective Devices**: Rated for $$ 553.6 \, \text{A} $$ (for starting) and $$ 184.5 \, \text{A} $$ (for overload)
- **Starting Current**: Approximately 184.5 A - **Full Load Torque**: Approximately 84.06 Nm - **Switching Devices**: Rated for 553.6 A - **Cables**: Rated for 184.5 A - **Protective Devices**: Rated for 553.6 A (starting) and 184.5 A (overload)

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