Question

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

Answer 1

The maximum hydraulic flow rate should be 0.0192 gpm.

Step-by-step explanation:

To determine the maximum hydraulic flow rate in gallons per minute, we need to convert the power limit from horsepower (hp) to watts and then use the formula: Flow rate (in gpm) = Power (in watts) / (Pressure (in psi) x 1714).

First, we need to convert 11 hp to watts. 1 horsepower is equal to 745.7 watts, so 11 hp is equal to 11 x 745.7 = 8202.7 watts.

Now, we can calculate the maximum hydraulic flow rate: Flow rate = 8202.7 watts / (250 psi x 1714) = 0.0192 gpm (rounded to four decimal places).

Answer 2

required flow rate is 75.44 gal/min

Step-by-step 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