# Calculate hydraulic torque from motor HP, hydraulic pump and gearbox

I am curious about the design considerations for a hydraulic system.

Let's say you have a 3HP 1200RPM motor you also have a hydraulic pump that's rated 4LPM and pressure of 40MPA rated for 40RPM MAX

You can put many different types of gear box on the 3HP motor to get it to operate in the N-40RPM range

Let's forget about the power loss through heat when using the gearbox.

I am trying to understand the relationship between HP and RPM of the electric motor, Pressure of the system and output Torque.

Let's say I am working at 200Bar, electric motor doing 40RPM and the hydraulic motor running at 4LPM.

How would that compare to say going to 20RPM and keeping 200Bar and 4LPM.

I do understand that raising the pressure would increase the heat and possibly require cooling.

Any suggestions on how to calculate these relationships?

I think you need an equation to compare the hydraulic work with the mechanical work. Additionally you can add the speed to get the power, but it's not necessary...

$$W_{hydr}=p*V_{geom}=M*2*\pi=W_{mech}$$

and now with speed

$$P_{hydr}=p*V_{geom}*n=p*Q=M*2*\pi*n=P_{mech}$$

Now the units:

• $$p$$ $$[bar]$$
• $$V_{geom}$$ $$[cm^3/U]$$
• $$n$$ $$[1/s]$$
• $$Q$$ $$[l/min]$$
• $$M$$ $$[Nm]$$

To summarize your problem description, you have ...

• 3 HP, 1200 RPM electric motor
• 40 MPa (400 bar) pump rated 4 LPM @ 40 RPM
• Gearbox ratio TBD between motor and pump
• Desired operation of 20MPa (200bar) at 4 LPM

First, we can make a few observations from the problem data.

1. The full-load torque of your electric motor. Note that this number is a nominal rating. Typical general-purpose induction motors can run at 10-15% higher torque than the full-load rating without damage, as long as cooling air flow is not obstructed. $$P_{motor} = 3 \ hp * {1 \ kW \over 1.341 \ hp} = 2.24 \ kW$$

$$M_{motor} = {P_{motor}\over2*\pi*{n/60}} = 17.8 \ Nm$$

1. Where did you find a pump with those specs? It's possible in theory but I've never seen anything close. A typical 4 LPM positive displacement pump would have a max speed over 1800 rpm. This avoids the need for a gearbox. Rexroth, Parker, Hydac, Eaton... many options available for pumps. Gear-type (200 bar) or radial piston (400 bar) pumps are typically the most cost-effective at your flow rate. Here is one catalog for example - https://dc-us.resource.bosch.com/media/us/products_13/product_groups_1/industrial_hydraulics_5/pdfs_4/re11263.pdf

2. In case you really do have a pump along those lines... here is the power transmitted through the hydraulic fluid: $$P_{flow} = p*Q*{100000 \ Pa \over 1 \ bar}*{1 \ m^3/s \over 60000 \ lpm} = 1.33 \ kW$$

3. Choose a typical pump efficiency of 85% and you have the power required to spin the pump: $$P_{pump} = {P_{flow} \over 0.85} = 1.57 kW$$ This is safely within the nominal motor power rating of 2.24 kW, including whatever gearbox efficiency you want to assume.

4. Lastly we can verify that motor torque is also sufficient to drive the pump at 40 rpm. $$M_{pump} = {P_{pump}\over2*\pi*{n/60}} = 374.8 \ Nm$$ So your gearbox would need to amplify torque by $${M_{pump} \over M_{motor}} = {374.8 \ Nm \over 17.8 \ Nm} = 21.1 \ min \ gear \ ratio$$ Which is accomplished with plenty of overhead thanks to the speed ratio: $${n_{motor} \over n_{pump}} ={1200 \ rpm \over 40 \ rpm} = 30.0 \ actual \ gear \ ratio$$

Units: $$P[kW], \ \ p[bar], \ \ n[rpm], \ \ Q[lpm], \ \ M[Nm]$$