You are designing a hydraulic power takeoff for a garden tractor. The hydraulic pump will be directly connected to the motor and supply hydraulic fluid at 250 psi for use by accessories. In order for the tractor to maintain normal operation, the maximum power the hydraulic system can use is limited to 11 hp. For what maximum hydraulic flow rate in gallons per minute (gpm) should you design

required flow rate is 75.44 gal/min  

Explanation:

Given the data in the question;

Power developed = 250 psi = 1.724 × 10⁶ Pa

hydraulic power W = 11 hp = 11 × 746 = 8206 Watt

now, Applying the formula for pump power

W = pgQμ

where p is density of fluid, Q is flow rate, μ is heat and W is power developed;

W = pgQμ

W = pgμ × Q  

W = P × Q let this be equ 1

so we substitute in our values;

8.2027 kW = 1.724 × 10⁶ Pa × Q

Q = 8206 / 1.724 × 10⁶

Q = 4.75986 × 10⁻³ m³/sec

We know that, 1 cubic meter per seconds = 15850.3 US liquid gallon per minute, so

Q = 4.75986 × 10⁻³  × 15850.3 gallon/min

Q = 75.44 gal/min    

Therefore, required flow rate is 75.44 gal/min    

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