**Liquid Flow Lohm Laws**

The Lohm Laws predict the actual performance of fluidic devices beyond the definition conditions of water at 25 psid and 80°F. In Liquid Flow several variables must be related, including:

**I** = Flow rate

**H** = Differential pressure

**V** = Viscosity correction factor. V factors compensate for the interaction of viscosity and device geometry and are unique to each class of device. See "V" factors graphs for typical Lee orifices. Use 1.0 for water @ 80°F

**S** = Specific gravity. Use 1.0 for water @ 80°F

**K** = A constant to take care of units of measure. Use 20 for psi and gpm. See table below for Units Constant K values.

*Swipe to the right for more table information*

The Lohm Law for Liquid Flow is: | |

When testing with water at 25 psid ( = 5), 80°F and flow rate in gallons per minute,

Notes:

1. V and S are equal to 1 for water at 80°F

*Swipe to the right for more table information*

d | = | orifice diameter (inches) |

C_{d} |
= | coefficient of discharge |

C_{v} |
= | flow coefficient |

For special flow requirements, The Lee Company can determine the required Lohm rating.

The following formulas are presented to extend the use of the Lohm laws to many different liquids, operating over a wide range

of pressure conditions.

**Nomenclature**

L = Lohms

H = Differential pressure

I = Liquid flow rate: Volumetric

S = Specific gravity*

V = Viscosity compensation factor**

w = Liquid flow rate: Gravimetric

K = Units Constant – Liquid

*S = 1.0 for water at 80°F.

**V = 1.0 for water at 80°F.

These formulas introduce compensation factors for liquid density and viscosity. They are applicable to any liquid of known

properties, with minimum restrictions on pressure levels or temperature.

The units constant (K) eliminates the need to convert pressure and flow parameters to special units.

*Swipe to the right for more table information*

VOLUMETRIC FLOW UNITS |
|||

Pressure Units | |||
---|---|---|---|

Flow Units | psi | bar | kPa |

GPM | 20 | 76.2 | 7.62 |

L/min | 75.7 | 288 | 28.8 |

ml/min | 75700 | 288000 | 28800 |

in^{3}/min |
4620 | 17600 | 1760 |

*Swipe to the right for more table information*

GRAVIMETRIC FLOW UNITS |
|||

Pressure Units | |||
---|---|---|---|

Flow Units | psi | bar | kPa |

PPH | 10000 | 38100 | 3810 |

gm/min | 75700 | 288000 | 28800 |

**Problem 1.** A restrictor is required to flow 0.1 GPM of 50/50 ethylene glycol/water blend (specific gravity = 1.07) at 45°F and 6 psid. How many Lohms are required?

**Solution:**

- Read kinematic viscosity; v = 5.0 cs from viscosity curve.
- Use v and ?P to determine viscosity correction factor, V = .87, from "V" Factor curve.
- Select unit constant K from table.
- Compute Lohms required.

**Problem 2.** What pressure drop will result from a flow of 57 mL/min. of 50/50 ethylene glycol/water mixture (specific gravity = 1.07) at 45°F, flowing through a 1000 Lohm restrictor?

**Solution:**

- Find viscosity. v = 5 cs
- Use knowledge of system to assume initial solution.

H = 4 psid - Use assumed H to determine V = 0.75 from "V" Factor Curve.
- Select units constant K from table.
- Compute trial ?P

- Make trials as required to find correct solution.

H = 2 psid V = .55

**Problem 3**. What restriction will permit a flow of 1 gallon of water per hour at 50 psi ?P?

**Problem 4.** A jet with a hole diameter of .012" flows 18 lb/hr of water at 100 psi ?P. How many Lohms?

**Problem 5. **What ?P will be required to flow 20 GPH of water through a 2000 Lohm Jet?

**Problem 6.** What water flow will result from a restriction of 500 Lohms and a ?P of 500 psi?

Note: For special flow requirements, The Lee Company can determine the required Lohm rating.

**Problem 7.** A restrictor is required to flow 0.15 GPM of MIL-H-83282 hydraulic fluid at 80°F and 100 psi ?P. What restriction is required?

**Solution:**

- Read specific gravity; S = 0.84 from chart.
- Read viscosity; v = 21cs. from chart.
- Use viscosity and ?P to determine viscosity compensation factor V = 0.95 from graph.
- Select units constant, K = 20 from table.

**Problem 8.** What pressure drop will result from a flow of 5 PPH of SAE #10 lubricating oil at 20°F, flowing through a 1000 Lohm restrictor?

**Solution:**

- Read specific gravity and viscosity.

S = 0.90, v = 600cs. - Use knowledge of system to assume solution.

H = 50 psid. - Use assumed ?P to determine V = 0.18
- Select units constant, K = 10,000 from table
- Compute trial ?P.

- Make trials as required to find correct solution.

H = 26 psid.

**Problem 9.** A Safety Screen is required to flow 775 lb/hr of JP4 @ 80°F with a maximum pressure drop of 5 psid. What is the maximum Lohm rate allowed for the Safety Screen?

**Solution:**