Quick calculation tables

Quick calculation tables

2–1. Convert flow rate

$Q_{m^3/h}=0.06Q_{L/min}$

$Q_{m^3/h}=3.6Q_{L/s}$

For simultaneous nozzles:

$Q_{total}=\sum q_i$

The flow rate of the display circuit and the flow rate of the purification circuit must be defined separately.

2–2. Pressure and head conversion

For water with approximate density $1000kg/m^3$:

$H=\frac{p}{\rho g}$

$H_m\approx 10.20p_{bar}$

The pump head should not be taken directly to the height of the fountain. The required head includes the nozzle exit conditions and path drops. Head definition in KSB technical reference ^[440]

2–3. Initial estimate of the inner diameter of the pipe

$Q=vA$

$D_{mm}=18.81\sqrt{\frac{Q_{m^3/h}}{v_{m/s}}}$

The following table shows the flow rate in cubic meters per hour.

Method of use: For a flow rate of 12 cubic meters per hour and an assumed velocity of 1.5 meters per second, the calculated inner diameter is approximately 53.2 mm.

The inner diameter is not the same as the nominal diameter or outer diameter. The final selection should be controlled by product internal diameter, pressure drop, suction conditions, path length and transient performance. The speeds in the table are calculated scenarios and are not general allowable ranges for all circuits.

Frictional and local control relationship:

$h_f=f\frac{L}{D}\frac{v^2}{2g}+\sum K\frac{v^2}{2g}$

In this regard, $f$ is the Darcy friction coefficient and $K$ is the local drop coefficient of the connection or equipment.

2–4. Cable selection and voltage drop control

Cable selection depends on consumer current, conductor material, insulation, installation method, ambient temperature, cable proximity, path length, short circuit and protection. Guide for selecting the conductor cross-section from Schneider Electric ^[441]

For a two-wire DC circuit with parallel conductors:

$\Delta V=\frac{2\rho LI}{S}$

In this regard:

• $L$: one-way length of the route, meters;

• $I$ : current, ampere;

• $S$: cross-sectional area of each conductor, square millimeters;

• $\rho$: specific resistance of the conductor at the calculated temperature, ohm square millimeter/meter.

Educational example of voltage drop: 24 V DC circuit, constant current of 5 amps, one-way length of 20 meters and hypothetical special resistance of copper $0.0225$.

If the design limit of this example is 3%, the calculated cross section in terms of voltage drop is at least 6.25 mm2. Selecting a higher business class still requires other table controls.

This example does not authorize the use of 24 volts in any area of the fountain; Allowed voltage and type of protection must be determined separately.

2–5. Calculate the volume of the pond

All dimensions in the following relationships are in meters and volumes are in cubic meters.

A rectangular pond with a length of 12 meters, a width of 5 meters and a water depth of 0.4 meters:

$V=12\times 5\times 0.4=24m^3$

It means 24 thousand liters of water.

In the executive estimate, the volume occupied by the structure and equipment should be subtracted from the geometric volume. The free space above the water, usable storage volume and the capacity to receive return water when stopping are separate concepts.

2–6. Estimation of evaporation and compensatory water

If the evaporation rate of the free surface is $e$ in millimeters per day and the area is $A$ in square meters:

$V_{evap,L/day}=Ae$

The evaporation rate numbers in this table are hypothetical; The actual rate should be obtained from suitable climatic data or project measurements. This calculation does not cover additional evaporation from sprays and wet surfaces, droplet drift with wind, and filter washing losses.

For a period with approximately constant storage volume:

$V_{makeup}=V_{evap}+V_{drift}+V_{backwash}+V_{drain}+V_{leak}-V_{effectiverainfall}$

Registering the compensatory water meter is the main tool for controlling this estimate during operation.

2–7. Estimation of pump power

For water:

$P_h=0.002725QH$

$P_{shaft}=\frac{P_h}{{\eta }_p}$

$P_{electrical}=\frac{P_h}{{\eta }_p{\eta }_m{\eta }_d}$

In these relations, the power is in kilowatts, the flow is in cubic meters per hour, and the head is in meters. The efficiency of the pump, motor and drive are represented by ${\eta }_p,{\eta }_m,{\eta }_d$ respectively. KSB pump power and performance reference ^[442]

The following table is calculated with a hypothetical total efficiency of 60%:

For example, for a flow rate of 36, head of 15, and pump efficiencies of 0.65, motor 0.9, and drive 0.97, the input power is approximately 2.59 kW.

The nominal power of the motor must be controlled by the pump power curve within the permissible working range; Rounding the same number is not the only way to choose an engine.