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## **Step 5: Electric Motor Power Calculation** ### **Flow rates:** - **P1:** 25 lpm = \( 25/1000 \) m³/min = 0.000417 m³/sec - **P2:** 5 lpm = 0.000083 m³/sec ### **Hydraulic Power:** \[ P_{hyd} = P \times Q \] - For P1: \[ P_{P1} = 1 \times 10^6\, Pa \times 0.000417\, m^3/sec = 417\, W \] - For P2: \[ P_{P2} = 5.88 \times 10^6\, Pa \times 0.000083\, m^3/sec \approx 488\, W \] **Total hydraulic power:** \[ P_{total} = 417 + 488 = 905\, W \] ### **Considering efficiency (\( \eta = 75\% \)):** \[ P_{motor} = \frac{905}{0.75} \approx 1207\, W \] **Motor HP:** \[ \frac{1207}{746} \approx 1.62\, HP \] **Choose a standard motor rating:** **2 HP**. write a short explanation in text form in 4 to 5 lines

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## **Step 5: Electric Motor Power Calculation** ### **Flow rates:** - **P1:** 25 lpm = \( 25/1000 \) m³/min = 0.000417 m³/sec - **P2:** 5 lpm = 0.000083 m³/sec ### **Hydraulic Power:** \[ P_{hyd} = P \times Q \] - For P1: \[ P_{P1} = 1 \times 10^6\, Pa \times 0.000417\, m^3/sec = 417\, W \] - For P2: \[ P_{P2} = 5.88 \times 10^6\, Pa \times 0.000083\, m^3/sec \approx 488\, W \] **Total hydraulic power:** \[ P_{total} = 417 + 488 = 905\, W \] ### **Considering efficiency (\( \eta = 75\% \)):** \[ P_{motor} = \frac{905}{0.75} \approx 1207\, W \] **Motor HP:** \[ \frac{1207}{746} \approx 1.62\, HP \] **Choose a standard motor rating:** **2 HP**. write a short explanation in text form in 4 to 5 lines

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Asked by: Kumar
Created at: 2025-07-26 20:53:24

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To determine the required electric motor power, the hydraulic power for each pump was calculated based on their respective pressures and flow rates, totaling 905 W. Accounting for a system efficiency of 75%, the required motor power increases to approximately 1207 W, or about 1.62 HP. To ensure reliable operation and accommodate standard ratings, a 2 HP electric motor is recommended for the system. This provides adequate capacity for both pumps and accounts for potential load variations.

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