VIPSolutions## **Step 6: Maximum Production Rate** - **Cylinder volume per stroke:** \[ V_{stroke} = A \times \text{stroke length} = 0.001104\, m^2 \times 0.25\, m = 2.76 \times 10^{-4}\, m^3 \] - **Time per approach stroke:** \[ t_{approach} = \frac{V_{stroke}}{\text{Flow rate of P1}} = \frac{2.76 \times 10^{-4}}{0.000417} \approx 0.66\, s \] - **Assuming negligible punching time,** total cycle time ≈ 0.66 s. - **Parts per minute:** \[ \frac{60}{0.66} \approx 90\, \text{parts/min} \] write a short explanation in text form in 4 to 5 lines
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## **Step 6: Maximum Production Rate** - **Cylinder volume per stroke:** \[ V_{stroke} = A \times \text{stroke length} = 0.001104\, m^2 \times 0.25\, m = 2.76 \times 10^{-4}\, m^3 \] - **Time per approach stroke:** \[ t_{approach} = \frac{V_{stroke}}{\text{Flow rate of P1}} = \frac{2.76 \times 10^{-4}}{0.000417} \approx 0.66\, s \] - **Assuming negligible punching time,** total cycle time ≈ 0.66 s. - **Parts per minute:** \[ \frac{60}{0.66} \approx 90\, \text{parts/min} \] write a short explanation in text form in 4 to 5 lines
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Asked by: Kumar
Created at: 2025-07-26 20:54:47
Asked by: Kumar
Created at: 2025-07-26 20:54:47
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To determine the maximum production rate, we first calculated the cylinder volume per stroke and then divided it by the pump’s flow rate to find the time required for one approach stroke. Assuming the punching time is negligible, the total cycle time per part is approximately 0.66 seconds. This results in a maximum production rate of about 90 parts per minute. Thus, under ideal conditions, the system can produce up to 90 parts each minute.
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