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:
$$
T = \frac{P \cdot 60}{2 \pi N}
$$
Where $$ N $$ is 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 Full Load Torque $$ T $$
$$
T = \frac{P \cdot 60}{2 \pi N}
$$

Now, I will perform these calculations.
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 $$ I_{star/delta} = \frac{I_{DOL}}{3} = \frac{\frac{125\sqrt{3}}{8}}{3} $$.
Solve the system of equations by following steps:
- step0: Solution:
  $$\left\{ \begin{array}{l}I=\frac{I}{3}\\frac{I}{3}=\frac{\frac{125\sqrt{3}}{8}}{3}\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}I=0\I=\frac{125\sqrt{3}}{8}\end{array}\right.$$
- step2: Find the intersection:
  $$I \in \varnothing $$
Solve the equation $$ T = \frac{15000 \cdot 60}{2 \pi N} = \frac{15000 \cdot 60}{2 \pi \cdot 1500} $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}T=\frac{15000\times 60}{2\pi N}\\frac{15000\times 60}{2\pi N}=\frac{15000\times 60}{2\pi \times 1500}\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}T=\frac{450000}{\pi N}\\frac{450000}{\pi N}=\frac{300}{\pi }\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}T=\frac{450000}{\pi N}\N=1500\end{array}\right.$$
- step3: Substitute the value of $$N:$$
  $$T=\frac{450000}{\pi \times 1500}$$
- step4: Simplify:
  $$T=\frac{300}{\pi }$$
- step5: Calculate:
  $$\left\{ \begin{array}{l}N=1500\T=\frac{300}{\pi }\end{array}\right.$$
- step6: Check the solution:
  $$\left\{ \begin{array}{l}N=1500\T=\frac{300}{\pi }\end{array}\right.$$
- step7: Rewrite:
  $$\left(N,T\right) = \left(1500,\frac{300}{\pi }\right)$$
### Results of Calculations

1. **Starting Current with Star/Delta Starter:**
   - The D.O.L. starting current is given by:
     $$
     I_{DOL} = \frac{125\sqrt{3}}{8} \, \text{A}
     $$
   - The starting current for a star/delta starter is:
     $$
     I_{star/delta} = \frac{I_{DOL}}{3} = \frac{125\sqrt{3}}{24} \, \text{A}
     $$

2. **Full Load Torque:**
   - The synchronous speed $$ N $$ is calculated as:
     $$
     N = 1500 \, \text{RPM}
     $$
   - The full load torque $$ T $$ is approximately:
     $$
     T \approx 95.49 \, \text{Nm}
     $$

### Next Steps
Now, we need to calculate the size of switching devices, cables/wires, and protective devices.

### (iii) Size of Switching Devices
The size of switching devices can be determined based on the starting current. The rating of the switching device should be higher than the starting current.

### (iv) Size of Cables/Wires
The size of cables can be determined based on the full load current. The full load current can be calculated as:
$$
I_{FL} = \frac{P}{\sqrt{3} \cdot V \cdot \text{p.f.}} = \frac{15000}{\sqrt{3} \cdot 400 \cdot 0.8}
$$

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

Let's calculate the full load current $$ I_{FL} $$ and then use it to determine the sizes of switching devices, cables, and protective devices.

#### Step 1: Calculate Full Load Current $$ I_{FL} $$
$$
I_{FL} = \frac{15000}{\sqrt{3} \cdot 400 \cdot 0.8}
$$

Now, I will perform this calculation.
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:
  $$I=\frac{125\sqrt{3}}{8}$$
### Full Load Current Calculation
The full load current ($$ I_{FL} $$) is given by:
$$
I_{FL} = \frac{125\sqrt{3}}{8} \, \text{A}
$$

### Summary of Results
1. **Starting Current with Star/Delta Starter:**
   $$
   I_{star/delta} = \frac{125\sqrt{3}}{24} \, \text{A}
   $$

2. **Full Load Torque:**
   $$
   T \approx 95.49 \, \text{Nm}
   $$

3. **Full Load Current:**
   $$
   I_{FL} = \frac{125\sqrt{3}}{8} \, \text{A}
   $$

### (iii) Size of Switching Devices
- The size of the switching device should be rated for at least the starting current $$ I_{star/delta} $$.

### (iv) Size of Cables/Wires
- The size of cables should be rated for at least the full load current $$ I_{FL} $$.

### (v) Size of Protective Devices
- The size of protective devices should be rated for the full load current and should also consider the starting current.

### Conclusion
To finalize the sizes of switching devices, cables, and protective devices, you would typically refer to standard electrical engineering tables or guidelines that provide the appropriate ratings based on the calculated currents.

If you need specific numerical values for the sizes of switching devices, cables, and protective devices, please provide the relevant standards or guidelines you would like to use for those calculations.
Starting current with star/delta starter: approximately 27.39 A Full load torque: about 95.49 Nm Full load current: approximately 27.39 A Size of switching devices: rated for at least 27.39 A Size of cables: rated for at least 27.39 A Size of protective devices: rated for at least 27.39 A

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