Liquid Flow: Lohm Rate vs. Orifice Diameter Lohm Laws Viscosity Compensation Factor for Single Orifice Viscosities of Typical Fluids vs. Temperature

Lohm Rate vs. Orifice Diameter Lohm Laws

Liquid Flow

The Lohm has been selected so that a 1 Lohm restriction will permit a flow of 100 gallons per minute of water with a pressure drop of 25 psi at a temperature of 80°F.

 I = Flow rate (gallons per minute). H = Differential pressure (psi). L = Lohms, a measure of resistance to liquid flow. It includes all density, viscosity, Reynolds number, coefficient of discharge & area units. When testing on water at 25 psi and the above formulas simplify as follows: Some useful relationships:

1. 1000 Lohms will permit a flow of 50 lb/hr water at 25 psi ∆P.
2.
 Flow Coefficient, 3. d = Orifice diameter (inches)
Cd = Coefficient of discharge
A = Orifice area (inches)2

4.
 For metric units where d = orifice diameter in millimeters

Liquid Flow – Examples

Problem 1. What restriction will permit a flow of 1 gallon of water per hour at 50 psi ∆P? Problem 2. An orifice with a hole diameter of .012" flows 18 lb/hr of water at 100 psi ∆P.
How many Lohms? Problem 3. What ∆P will be required to flow 20 GPH of water through a 2000 Lohm orifice? Problem 4. 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.

Liquid Flow – Two Forumlas for Combinations of Restrictors

PARALLEL FLOW, the total Lohm rating is: Please note that this relationship is identical to the electrical equation. SERIES FLOW, the total Lohm rating is: Please note that this relationship is not the same as in electrical problems. The difference is due to the non-linearity of  Please note that this relationship is not the same as in electrical problems. The difference is due to the non-linearity of

 When L1 = L2 = L3 , then For passageway size: DT = D/ N1/4

DT = Diameter of a single equivalent orifice, with a Lohm rate = LT

D = Diameter of the actual orifices, each with a Lohm rate = L1

Liquid Flow – Series Flow

One of the reasons for using two restrictors in series is to allow fine tuning of a total resistance value. If L1 is known and is more than 90% of LT, then L2 may vary by ±5% without altering the value of LT by more than ±1%, even though the value of L2 may be as high as 40% of LT . This effect becomes even more pronounced as L1 approaches LT.

To determine the intermediate pressure between two resistances in series, the following formulas may be used. Liquid Flow – Flow Formula

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* (click here) V = Viscosity compensation factor** (click here) w = Liquid flow rate: Gravimetric K = Units Constant – Liquid (click here) *S = 1.0 for water at 80°F. **V = 1.0 for water at 80°F.

Liquid Flow

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.

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.

Volumetric Flow Units Gravimetric Flow Units Liquid Flow – Units Constant K

Volumetric Flow Units VOLUMETRIC FLOW UNITS Flow Units Pressure Units psi bar kPa GPM 20 76.2 7.62 L/min 75.7 288 28.8 ml/min 75,700 288,000 28,800 in3/min 4,620 17,600 1,760

Gravimetric Flow Units GRAVIMETRIC FLOW UNITS Flow Units Pressure Units psi bar kPa PPH 10,000 38,100 3,800 gm/min 75,700 288,000 28,800

Liquid Flow Calculations – Examples

Problem 1. An orifice 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:

1. Read specific gravity; S = 0.84 from chart here.
2. Read viscosity; v = 21cs. From chart here.
3. Use viscosity and ∆P to determine viscosity compensation factor V = 0.95 from graph here.
4. Select units constant, K = 20 from table here. Problem 2. What pressure drop will result from a flow of 5 PPH of SAE #10 lubricating oil at 20°F, flowing through a 1000 Lohm orifice?

Solution:

1. Read specific gravity and viscosity.
S = 0.90, v = 600cs.
2. Use knowledge of system to assume solution.
H = 50 psid.
3. Use assumed ∆P to determine V = 0.18
4. Select units constant, K = 10,000 from table here.
5. Compute trial ∆P. 6. Make trials as required to find correct solution.
H = 26 psid.

Problem 3. A Safety Screen is required to flow 775 lb/hr of diesel fuel @ 80°F with a maximum pressure drop of 5 psid.
What is the maximum Lohm rate allowed for the Safety Screen?

Solution:

1. Find specific gravity; S = 0.87 from curve here.
2. Find viscosity; v = 3.1cs from curve here.
3. Use v and ∆P to determine viscosity compensation factor, V = 0.87 from curve here.
4. Select units constant, K = 10,000 from table here. Viscosity Compensation Factor For Single Orifice Viscosities of Typical Fluids vs. Temperature    The Lee Company, Industrial Microhydraulics Group • 82 Pequot Park Road, Westbrook, CT 06498-0424 • Tel: 860-399-6281 • Fax: 860-399-7037 • inquiry@TheLeeCo.com