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 \( 71 \% \).

**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 \(R\,15.17\).