To achieve a desired nozzle pressure (DNP), a few factors must be considered. First, you must note the head loss (HL) or head gain (HG). Water head is the height of the water column (lift) due to imposing pressure. The head pressure is positive (gain) if the hose lay is downhill because the force of gravity is helping push the water down, consequently increasing the pressure. The head pressure is negative (loss) if the hose lay is uphill, since the force of gravity is pulling the water down, when it needs to be pumped up. Table 3.1 indicates that 1 foot of water head or lift produces 0.5 pounds per square inch of pressure. On that same note, 1 pound per square inch can produce 2 feet of water head). For every foot uphill or downhill, there is a change of 0.5 pounds per square inch of pressure. Note that this measurement represents the height of the hose (elevation) and not the length of the hose.
Water Pressure vs Height. The drawings correspond to the pressure on a square inch cross section caused by the height of water above it. Note that as the column's height doubles, so does the pressure. Both exact and rounded field application values are given.
The second consideration for pump pressure calculations involves friction loss (FL). As a field rule, the pressure in a line is reduced by 5 pounds per square inch for each appliance added to the line. For example, a hose lay with five wye valves will result in a 25 pounds per square inch pressure loss due to the friction introduced by these fittings. This approximation is used to simplify calculations and is not precisely what occurs in the field. See Table 3.3 for friction loss in forestry hose.
CALCULATING DESIRED NOZZLE AND PUMP PRESSURES
Engine and nozzle pressures are calculated as follows:
DNP = Desired Nozzle Pressure
EP = Engine Pressure
HG = Head Gain
HL = Head Loss
FL = Friction Loss
Desired Nozzle Pressure equals:
Engine (Pump) Pressure ± (Head Gain or Head Loss) - Friction Loss
DNP = EP ± (HG or HL) - FL
When calculating desired nozzle pressure in a downhill hose lay, add the head pressure. In uphill hose lays, subtract the head pressure. The calculations vary to account for the work of gravity.
The head pressure is expressed in terms of loss or gain. Because the pump and the nozzle are at opposite ends of the hose, head pressure that is positive at the pump will be negative at the nozzle and vice versa. It is crucial that the sign of the head pressure be correct. If the hose lay is uphill, the head pressure is negative, and if it is downhill, the head pressure is positive. Careful attention must be paid to the sign of the head gain or head loss term, and whether the gain or loss should be added or subtracted.
Head gain and head loss depend on the nozzle's position relative to the pump.
ESTIMATING IN THE FIELD
As mentioned earlier, it is often necessary to round numbers either up or down to make calculations easier. When precise calculation is not possible because there is no paper, pen, or calculator, rounded estimations are helpful. In the field, rounding does not greatly affect the results. It can helpful to round numbers to take outside disturbances of any kind into account. For example, rounding the pressure caused by water head up from 0.434 to 0.5 takes into account any additional friction that might be caused by the hose itself.
For a 100-foot vertical height of water in the hose, using the exact value of 0.434 pounds per square inch per foot would give a 43 pounds per square inch (psi) head loss. This loss is 7 pounds per square inch less than what was calculated earlier. By rounding up to 0.5 from 0.434, friction and head losses due to the hose itself are taken into account, and no additional calculation is needed. In the field, the 0.5 pounds per square inch head loss is therefore used. This approximation not only eases calculations, but is more realistic to use in the field.
FRICTION DUE TO HOSE LENGTH
For hose lengths longer than 100 feet, friction loss in the hose should be considered. Friction loss of a 100-foot, 1-inch hose, all synthetic, with volume rate of 15 gallons per minute, is typically 4 to 9 pounds per square inch. Friction loss of a 100-foot, 1-inch hose, cotton-synthetic, at 15 gallons per minute, is typically 3 to 6 pounds per square inch. Friction loss for a 1.5-inch hose at 15 gallons per minute is typically 1 pound per square inch for 100 feet. See Table 3.3.
Example 1 - A progressive hose lay has six gated wye valves along the length of the trunk line. The nozzle outlet is 200 feet below the engine. The desired nozzle pressure of the trunk line is 100 pounds per square inch. At what pressure does the engine need to perform?
Step 1. Find the appropriate conversion/estimation in Table 3.1 for the pressure caused by 1 foot of water head. 1 ft = 0.5 psi
Step 2. Set up the cancellation table so all units will cancel, except the desired unit, psi. There is a head gain due to the hose lay being downhill.
Step 3. Set up the cancellation table so all units will cancel, except the desired unit, psi, to calculate the friction loss due to the fittings. Guidelines indicate a 5 pounds per square inch loss per fitting.
Step 4. Use the equation for the engine pressure. EP = DNP ± HG (or HL) +FL
Step 5. Identify the DNP, the HG, and the FL. DNP = 100 psi, HG = 100 psi, FL = 30 psi
Step 6. Set up the problem and solve. EP = 100 psi - 100 psi + 30 psi
The engine pressure needs to be 30 pounds per square inch for a desired nozzle pressure of 100 pounds per square inch in this hose lay.
Kevin is fighting a fire and needs the nozzle pressure to be 100 pounds per square inch. He is 100 feet above the engine. What pump pressure does he need? Proceed through the questions below, which correspond to the steps necessary to solve the problem. After each step, click the next question button until you have reached the end of the exercise and have a value, in psi, for the engine pump pressure needed.
SIZE AND SHAPE VERSUS PRESSURE
The width or diameter of the tank does not affect the pressure. A column of water 100 feet high creates the same amount of pressure in a 2-foot diameter tank as it does in a 20-foot diameter tank. Think of people swimming in the ocean. They are not crushed by the pressure of such a large body of water because the pressure is the same if the height is the same, no matter how wide or what shape the container.
Different shaped tanks exhibit the same pressure.
Knowing that 1 pound per square inch of pressure can lift water vertically 2 feet, the water level of certain volumes of water (a cistern or tank) can also be calculated.
Example 2 - An engine's compound gauge is connected at the base of a 100-foot tall reservoir tank, and the gauge reads 35 pounds per square inch. How high is the water level in the reservoir?
Step 1. Find the appropriate conversion/estimation in table 3.1 for the height of water that creates 1 pound per square inch of pressure. 1 psi = 2 ft of water head
Step 2. Set up the cancellation table so all units will cancel, except the desired unit, feet, to calculate the height of water that creates the 35 pounds per square inch of pressure above the gauge.
The water level is 70 feet above the gauge.
Harvey has parked his engine 30 feet below the base of a nearby water tank. He connects his engine's compound gauge to the water line coming from the tank and obtains a reading of 40 pounds per square inch. How high is the water level in the tank?
In this problem, the pressure is read not at the base of the tank, but at 30 feet below.
Hint. Find the appropriate conversion/estimation in Table 3.1 for the height of water that creates 1 pound per square inch of pressure. 1 psi = 2 feet