Net positive suction head (NPSH) is a quantity important in suction pump design, which is a task that a chemical engineer might undertake. It has units of height (meters).
Consider the fact that the pressure on a point in a fluid depends only on one variable, the depth of the point, when the density is constant. In exact terms, this is:
p = ρgh
is the density of the fluid (kg/m3
is the gravity constant (N/kg). Express the same in this form:
p/ρg = h
Now we see that the pressure is directly proportional to a height, which is called "head" in chemical engineering
A suction pump sucks. That is, the pressure in its suction inlet is lowered. This causes a problem: if the pressure in the suction inlet drops below the vapor pressure of the liquid, vapor bubbles are free to form, because the pressure of the liquid is not enough to collapse them. This phenomenon is called cavitation. So, instead of pumping effectively, the energy is wasted on boiling the liquid. Not only the efficiency suffers: the bubbles are an abrasive and damage the pump. The abrasion generates heat, which only increases the cavitation. rootbeer277 informs me that the pump may heat up so badly that if its cover opened for maintenance, the liquid boils explosively and causes severe burns to the maintenance worker. Clearly, we don't want this to happen.
The only solution here is to ensure that the pressure in every point inside the pump is always higher that the vapor pressure of the fluid. This means that we have to keep a minimum pressure in the input flow. That is, we must have enough net positive suction head.
It is a mathematically difficult task to compute the required NPSH, usually denoted (NPSH)R. Instead, the pump manufacturer measures it experimentally and supplies the results with the pump. The engineer must then find out the available NPSH or (NPSH)A before installing the pump.
Let us set out to find out the available NPSH. I'll use the equations expressed in the form of energy per unit mass (J/kg) here. They describe the "potential energy of pressure" and can be easily converted to pressure (Pa) or head (m) equations.
The NP in NPSH stands for "net positive". That is, it isn't an absolute quantity, but the overpressure with respect to the vapor pressure. So, to calculate (NPSH)A for a piping system, first we'll set a "zero" point. We wish to find the pressure above the vapor pressure, so we'll have a negative term for the vapor pressure:
Then, we aren't interested in the velocity head. A flow is mass that is moving, so it has some kinetic energy per unit mass. The kinetic energy is
Ekin = mv2/2, so if we divide by the mass, we get the velocity energy per unit mass:
Ekin/m = v2/2. A pump doesn't change the velocity of the fluid, which means that this energy is irrelevant, so we'll add a negative term to cancel out the kinetic energy:
-v2/2. The vapor pressure and the velocity head are measured in the absolute, not net pressure.
Two things add pressure to the liquid. First is the weight of the fluid above the pump, which depends on the depth below the liquid surface. In energy terms, this becomes
z is the depth from the liquid surface to the pump centerline. The second is the pressure that is subjected to the liquid in the first place at the surface. (In an open-air tank, this would be the atmospheric pressure.) So, we'll add
p0 is the pressure above.
The pressure can be lowered, too. A piping system always wastes some energy to friction. Thus, we'll need to add a negative term,
-Σ(F), which is the sum of all energies lost in the pipe, pipe bends, valves, water separators, motionless mixers and other hindrances to flow. Remember that also the inlet has a resistance!
Summing up all this, the engineer is left with this equation:
p0/ρ + gz - Σ(F) - pvap/ρ - v2/2
Deciphered, it says: energy available as NPSH is: pressure at the surface by density (
p0/ρ), plus the pressure at the depth by density (
gz), minus losses to fluid friction (
Σ(F)), minus the vapor pressure by density (
pvap/ρ), minus the kinetic energy by mass (
Now the engineer has to calculate this, compare the value to the (NPSH)R found from the chart supplied with the pump and note that at least 0.5 - 1 m more NPSH is available as a safety margin. If it isn't, he must choose another setup: lift the tank higher (higher liquid surface), place the pump lower or install another pump to supply the NPSH.
Example: We install a centrifugal pump 3.0 m above a lake surface. The fluid velocity is 0.9 m/s. Flow friction losses are calculated to be 1.0 J/kg. (This value is realistic - see flow friction loss) The density of the water is 1000 kg/m3, Earth's gravity is the familiar 9.81 N/kg and the atmospheric pressure is about 100 kPa. The vapor pressure of water at this temperature is 872 Pa. What (NPSH)A is available?
Noting that the depth is negative and substituting the data:
(100000 Pa)/(1000 kg/m3) + (9.81 N/kg)(-3.0 m) - 1.0 J/kg - (872 Pa)/(1000 kg/m3) - (0.9 m/s)2/2
= 68.293 J/kg
Dividing this value of energy per unit mass with
, we get:
(NPSH)A = 6.96 m
The result is about 7 m higher head than that of the vapor pressure. That means that the pump needs to have a (NPSH)R
of 6 m or less. (Realistic values: 2 to 5 m) If it needs more NPSH and we'll decide to install it anyway, cavitation
results even with this cold water!
Thanks to rootbeer277 for assistance.
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