# DC Machine Motor Speed and Field Current Analysis Given: - **DC machine operates as a motor** - **Terminal voltage \( V_t = 250\,\text{V} \) (constant)** - **Magnetization curve slope**: \( 150\,\text{V/A (of field current)} \) - **Ignore saturation** - **Armature reaction, losses, and other secondary effects are not specified—neglect unless otherwise stated** - **Field current for part (a): \( I_f = 1.67\,\text{A} \) (constant)** - **Shaft power varies from \( \) to \( 35\,\text{kW} \)** - **For (b): speed is maintained at \( 150\,\text{rpm} \) by adjusting \( I_f \)** --- ## Key Equations - **EMF equation**: \[ E = k \phi N \] Or, for a separately-excited DC machine (linear, unsaturated): \[ E = S \cdot I_f \cdot N \] where \( S \) is the slope relating field current to generated voltage per unit speed. - **Terminal voltage**: \[ V_t = E + I_a R_a \] If armature resistance \( R_a \) is not given, assume \( R_a \approx \) for simplicity unless specified. - **Mechanical power output**: \[ P_{\text{shaft}} = T \omega \] \[ T = \frac{P_{\text{shaft}}}{\omega} \] Where \( \omega = 2\pi N / 60 \), \( N \) in rpm. --- ## Step 1: Magnetization Curve Equation From the slope: \[ \left.\frac{dE}{dI_f}\right|_{N=150\,\text{rpm}} = 150\,\text{V/A} \] So, at any speed \( N \) (in rpm): \[ E = \left( \frac{150}{150} N \right) I_f = .1 N I_f \] --- ## Step 2: Armature Current and Shaft Power If armature resistance is neglected: \[ V_t = E \implies E = 250\,\text{V} \] But with load, there is armature current, which creates a voltage drop across \( R_a \): \[ I_a = \frac{P_{\text{shaft}}}{E} \] Assume \( R_a \) is negligible **unless specified**. --- ## Part (a): Constant Field Current (\( I_f = 1.67\,\text{A} \)), Vary Shaft Power ### 1. Express \( E \) as a function of speed: \[ E = .1 N I_f = .1 N (1.67) = .167 N \] So, \[ N = \frac{E}{.167} \] ### 2. Assume negligible \( R_a \), so \( E \approx V_t = 250\,\text{V} \) at no load. - **No load**: \[ N_ = \frac{250}{.167} \approx 1497\,\text{rpm} \] - **With load**: If \( R_a \) is neglected, \( E \) remains at 250 V regardless of load, so **speed is nearly constant**. - **If \( R_a \) is **not** negligible** (let's include it for realism): Let \( R_a \) be small, say \( .2\,\Omega \). \[ V_t = E + I_a R_a \] \[ I_a = \frac{P_{\text{shaft}}}{E} \] \[ E = V_t - I_a R_a = V_t - \left(\frac{P_{\text{shaft}}}{E}\right) R_a \] This leads to: \[ E^2 - V_t E + P_{\text{shaft}} R_a = \] \[ E = \frac{V_t \pm \sqrt{V_t^2 - 4 P_{\text{shaft}} R_a}}{2} \] Take the physically meaningful root (\( < V_t \)). For small \( R_a \), \( E \approx V_t - I_a R_a \). #### **Summary Table for Plotting** | \( P_{\text{shaft}}\,(\text{kW}) \) | \( I_a\,(\text{A}) \) | \( E\,(\text{V}) \) | \( N\,(\text{rpm}) \) | |:---:|:---:|:---:|:---:| | | | 250 | \( 1497 \) | | 5 | 20 | 246 | \( 1473 \) | | 10 | 40 | 242 | \( 1449 \) | | 15 | 60 | 238 | \( 1425 \) | | 20 | 80 | 234 | \( 1401 \) | | 25 | 100 | 230 | \( 1377 \) | | 30 | 120 | 226 | \( 1353 \) | | 35 | 140 | 222 | \( 1329 \) | _Calculation for each:_ \[ E = V_t - I_a R_a \] \[ N = \frac{E}{.167} \] \[ I_a = \frac{P_{\text{shaft}}}{E} \] #### **Sample Calculation for \( P_{\text{shaft}} = 10\,\text{kW} \):** \[ I_a = \frac{10,000}{250} = 40\,\text{A} \] \[ E = 250 - 40 \cdot .2 = 242\,\text{V} \] \[ N = \frac{242}{.167} \approx 1449\,\text{rpm} \] ### **Plot Description** - **X-axis**: Shaft Power (\( \) to \( 35\,\text{kW} \)) - **Y-axis**: Speed (\( N \), rpm) - **Shape**: Slightly decreasing from ~1497 rpm at no load to ~133 rpm at full load (35 kW). --- ## Part (b): Constant Speed (\( N = 150\,\text{rpm} \)), Vary Shaft Power Maintain \( N = 150\,\text{rpm} \) by varying \( I_f \). 1. At \( N = 150 \), \[ E = .1 \cdot 150 \cdot I_f = 150 I_f \] \[ V_t = E + I_a R_a \] \[ I_a = \frac{P_{\text{shaft}}}{E} \] \[ V_t = E + \left( \frac{P_{\text{shaft}}}{E} \right) R_a \] Rearranged: \[ V_t = E + \frac{P_{\text{shaft}} R_a}{E} \] \[ V_t E = E^2 + P_{\text{shaft}} R_a \] \[ E^2 - V_t E + P_{\text{shaft}} R_a = \] Solve for \( E \): \[ E = \frac{V_t + \sqrt{V_t^2 - 4 P_{\text{shaft}} R_a}}{2} \] (Use smaller root for correct physical sense.) 2. Then, \[ I_f = \frac{E}{.1 \cdot 150} = \frac{E}{150} \] ### **Summary Table for Plotting** | \( P_{\text{shaft}}\,(\text{kW}) \) | \( E\,(\text{V}) \) | \( I_f\,(\text{A}) \) | |:---:|:---:|:---:| | | 250 | 1.67 | | 5 | 246 | 1.64 | | 10 | 242 | 1.61 | | 15 | 238 | 1.59 | | 20 | 234 | 1.56 | | 25 | 230 | 1.53 | | 30 | 226 | 1.51 | | 35 | 222 | 1.48 | _Calculated as above; \( E = V_t - I_a R_a \), \( I_a = P_{\text{shaft}}/E \)._ ### **Plot Description** - **X-axis**: Shaft Power (\( \) to \( 35\,\text{kW} \)) - **Y-axis**: Field Current (\( I_f \), A) - **Shape**: Slightly decreasing from 1.67 A at no load to about 1.48 A at 35 kW. --- ## Summary ### (a) **Speed vs. Shaft Power** (Constant \( I_f \)) - Speed drops slightly as load increases. - At no load: ~1497 rpm - At 35 kW: ~133 rpm ### (b) **Field Current vs. Shaft Power** (Constant Speed) - Field current must decrease slightly as load increases to maintain speed. - From 1.67 A at no load to about 1.48 A at 35 kW. --- ## Example Plot Sketches **Speed vs. Shaft Power**  *Alt text: Line plot showing motor speed (rpm) slightly decreasing as shaft power (kW) increases from to 35 kW, for fixed field current.* **Field Current vs. Shaft Power**  *Alt text: Line plot showing field current (A) slightly decreasing as shaft power (kW) increases, with motor speed held constant.* --- ## References - DC Machine Equations (see any standard Electric Machines textbook) - Assumed \( R_a = .2\,\Omega \) for illustration; use actual value if provided. --- **If you need code (e.g., Python/Matplotlib) for plotting or want to see the calculations for a different \( R_a \), specify accordingly.**