QUESTION 4 [6] A DC motor is driven from a 225 V supply for 180 minutes. Calculate how much current will be drawn; and also; what process costs if the ESKOM fee is R $$ 1,73 / \mathrm{kWh} $$. The motor delivers a torque of 18 Nm to the load with a rotation speed of 1100 rpm . The efficiency of the system is $$ 71 \% $$.

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**Step 1. Calculate the Mechanical Output Power**

The motor delivers a constant torque of  
$$
\tau = 18\ \text{Nm}
$$
at a rotational speed of  
$$
\text{RPM} = 1100.
$$  
First, convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s). Recall that one revolution is $$2\pi$$ radians and there are 60 seconds in a minute:
$$
\omega = \frac{1100}{60} \times 2\pi \quad \text{rad/s}.
$$
Calculating:
$$
\omega \approx 18.33 \times 6.2832 \approx 115.2\ \text{rad/s}.
$$

The output mechanical power is given by:
$$
P_{\text{out}} = \tau \times \omega.
$$
Thus:
$$
P_{\text{out}} = 18\ \text{Nm} \times 115.2\ \text{rad/s} \approx 2073.6\ \text{W}.
$$

---

**Step 2. Calculate the Electrical Input Power Required**

The overall system efficiency is given as $$71\%$$ or $$0.71$$. The relationship between the output power and the input power is:
$$
\eta = \frac{P_{\text{out}}}{P_{\text{in}}}.
$$
Solving for the input power:
$$
P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{2073.6}{0.71} \approx 2919\ \text{W}.
$$

---

**Step 3. Calculate the Current Drawn from the Supply**

The supply voltage is given as:
$$
V = 225\ \text{V}.
$$
The input power is related to the current drawn ($$I$$) by:
$$
P_{\text{in}} = V \times I.
$$
Thus:
$$
I = \frac{P_{\text{in}}}{V} = \frac{2919}{225} \approx 12.97\ \text{A}.
$$
Rounded as needed, the motor draws approximately:
$$
\sim 13\ \text{A}.
$$

---

**Step 4. Calculate the Energy Consumed Over 180 Minutes**

The duration of operation is:
$$
t = 180\ \text{minutes} = 3\ \text{hours}.
$$
The energy consumed (in kilowatt-hours) is:
$$
E = P_{\text{in (in kW)}} \times t.
$$
Converting $$P_{\text{in}}$$ to kilowatts:
$$
P_{\text{in}} \approx 2.919\ \text{kW}.
$$
Thus:
$$
E \approx 2.919\ \text{kW} \times 3\ \text{h} \approx 8.757\ \text{kWh}.
$$

---

**Step 5. Calculate the Process Cost**

The cost of energy, as provided, is:
$$
\text{Cost per kWh} = R\,1.73/\text{kWh}.
$$
Therefore, the total cost is:
$$
\text{Cost} = E \times \text{Cost per kWh} \approx 8.757 \times 1.73 \approx R\,15.17.
$$

---

**Final Answers:**

- **Current Drawn:** Approximately $$13\ \text{A}$$.
- **Process Cost:** Approximately $$R\,15.17$$.
The motor draws approximately 13 A and costs about R15.17 for 180 minutes of operation.

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