 Sludge Pumping Calculations Distance Learning Math Index Websites for Those Who Need Help Homepage Pressure Problems   1. Unit Volumes and Unit Weights: 1 cubic foot of water weighs 62.4 pounds 1 cubic foot of water equals 7.5 gallons 1 gallon of water weighs 8.34 pounds 2. Unit Pressure: Pressures are usually expressed in pounds per square foot (psf) or pounds per square inch (psi).  This is known as unit pressure. A container which has dimensions of 1'  x  1'  x  1'(the same as 12"  x  12"  x  12") and is filled with water, weighs 62.4 pounds.  The total  pressure on the bottom is therefore 62.4 lbs., and the unit pressure is 62.4 psf. Also, the unit pressure is: From this illustration we see that a column of water 1' high causes a pressure of 0.43 psi.   How high must a column of water stand to cause a pressure of 1 psi? Example: Find the pressure at the bottom of a standpipe in which water stands 60 feet deep. In the following diagram the unit pressure at each valve will be the same. The unit pressure at the valve in each case is 26 psi.  In the case of liquid at rest, the unit pressure, p, is determined solely by the height of the liquid surface, h, and the unit weight of the liquid, w (lbs./cu.ft.) or p  =  wh Care must be taken to use (w and h) in the proper units:  if (w) is in lbs./cu.ft., (h) must be in feet. Pressures are sometimes measured in terms of feet of head which means the same thing  as  height to the water level.  For example, the head on the pipes from either of the tanks is 60 feet.     3. Gage and Absolute Pressure: The dials of gages are designed so that the readings indicate pressures caused by the head of water. Air has weight and therefore exerts a pressure.  This pressure is usually taken to be 14.7 psi (atmospheric pressures).  The absolute pressure at the bottom of the tank is actually: Absolute pressure  =  gage  +  atmospheric                              =  26 psi  +  14.7 psi                              =  40.7 psi   4. Transmission of Pressure: The principle of transmission of pressure illustrated above is that the unit pressure at any point in a fluid at rest is transmitted equally in all directions. If  P1 = 100 lbs. and  A1 = 10 sq. inches Let (p) = pressure in pounds per square inch (psi) So an exerted pressure, P1, of 100 lbs. at A1 results in a pressure, P2, of 1,000 lbs. All unit pressures are 10 psi.  The pressure at any point in the vessel can be figured in the same manner. The hydraulic gage valve operates on the principle outlined above.   Water in Motion Pressure and head in relation to bodies of water at rest have been considered.  This is called static pressure and static head.  When water is in motion, such as from the mechanical force exerted by a pump in producing motion, the pressure created is called dynamic pressure.  There are factors other than elevation which affect pressures when water is in motion. Suppose a pump at a source of supply will pump water to a storage tank with an overflow elevation 100 feet above the centerline of the pump.  A pressure gauge placed at the outlet of the pump would show a pressure reading of 43 psi when the tank was full.  This would represent a back-pressure against the pump.  To pump water, the pump would have to exert a force greater than the pressure exerted by this 00 foot column of  water.  We would say this pump would operate against a static head of 100 feet. If the source of water for this pump came from a reservoir ten feet above the centerline of the pump, there would be a positive pressure on the suction side of the pump of ten feet (or 4.3 psi).  This is called suction head and would serve to relieve some of the work of the pump, and it would only have to exert a force against a net static head of 90 feet.(100 ft.  -  10 ft.). If the source of water came from a well where the ground water stood at an elevation  ten feet below the centerline of the pump, this would require an increased force to be exerted as a result of the need to lift the water ten feet to the pump.  This is called  suction lift and the head against which the pump must operate would be increased to 100  feet. The maximum lift which can be accomplished by suction is 33.9 feet.  This is the theoretical maximum, determined and limited by atmospheric pressure, and is not attainable in field operations.  It takes just as much energy for a pump to lift water 10 feet  by suction as it does to force water 10 feet in elevation by pressure. Fluid Flow Calculations The amount of water passing a given section of a pipe line in a unit time is called the discharge, Q.      If:  A  =  cross sectional area of pipe (square feet)           V  =  velocity of water in pipe (feet per second) Then:  Q  =  AV  =  cubic feet of water per second   Find the quantity of water passing through a 6" diameter pipe if the velocity is 5 feet per second: The cross-sectional area, A, of a pipe is In water works practice it is customary to express discharge in gallons per minute (gpm) or in million gallons per day (mgd).  450 gpm  x  60 min.  =  27,000 gal./hr. 2,700 gph  x  24 hrs.  =  648,000 gal./day                                   =  0.65 mgd The effect of reducing a pipe diameter is to increase the velocity.  For example, if  the 6" pipe just mentioned discharges into a 4" pipe, what will be the velocity in the 4" pipe? Since the discharge through each pipe is the same, (too high for water works practice).     Loss of Pressure If only static head was involved in pumping operations, any force exerted by the pump which was greater than the static head (or back pressure), would result in  water being pumped.  But as soon as water is in motion, there are other factors to  be considered.  Most important of these factors is friction loss. There is always friction in any moving body of water.  In pressure pipe lines, friction is caused by the water rubbing on the inside surface of the pipe. Friction  results in loss of pressure.  Friction loss is usually measured in "feet per 1,000 feet of pipe" and may be easily converted to pressure loss in pounds per square inch. Pumps are designed to operate under specific head conditions.  In addition to the  static head, all friction losses and minor losses should be computed in order to  determine the total head against which the pump will operate.  The pump will then be specifically designed or selected from a standard design to provide the desired capacity for the conditions under which it will operate. The total pressure provided at the discharge side of the pump represents the discharge pressure of the discharge head. When water flows through a pipe, the water molecules rub against each other and against the wall of the pipe.  This is called friction and it requires pressure to overcome it.  The amount of pressure required (or feet of head) to overcome the  friction cannot be used for any other purpose and it is therefore called lost pressure  or lost head.   Factors Influencing Head Loss in Pipes Friction loss in a pipe depends on the length of the line since the head loss varies directly to the pipe length.  The rate of flow and the condition of the interior surface  of the pipe also influence head loss. Type of pipe, pipe coating (if any), age of pipe, and the smoothness or roughness of the interior surface of the pipe affects the friction loss.  A smooth interior surface would result in very little friction loss at a reasonable velocity, whereas a very rough  surface would develop a considerable loss through friction, at the same velocity. The rate of flow is referred to as velocity, which is the speed of the water. Velocity is measured in feet per second.  Friction loss increases as the velocity through the pipe increases.  If the flow through a pipe is doubled, the friction loss in the pipe will be increased by almost four times. The diameter of the pipe also affects the head loss, as the diameter determines the area of wall in contact with flowing water.  Also, for a given discharge the diameter determines the velocity of the water. Increased friction may also be due to any of the following factors: Sedimentation:  mud, silt or sand. Obstruction of the pipe due to debris:  sticks, boards, stones, tools and other things that may have gotten into the pipe during construction. Partly closed valves. Accumulation of air at summits. Mineral deposits and slime growths on walls of  pipe. Tuberculation. There are other minor losses such as losses resulting from changes in size of pipe or direction of flow, meter losses, losses through check valves or undersize gate valves, pump friction, etc. The amount of friction developed is the criterion by which the size of pipe and the amount of power required for pumping are determined.  When a given amount of water is to be transported, the total amount of friction developed depends on the diameter and length of pipe and the condition of its interior. In determining the proper size of a new water pipe to be installed in any water distribution system, it is necessary that the main be large enough to deliver the expected demand at adequate pressure.  The probable maximum demand must be estimated and should be based on use in comparable areas.  Fire demands may be exceeded by lawn sprinkling demands in residential areas.  Allowance should be made for probable future increases in demand.  It may be more economical to anticipate future demand rather than to replace the main with a larger one at some future date. Mains which serve fire hydrants should not be less than 6 inches in diameter. Mains of smaller size will not be recognized by fire insurance underwriters. Many large cities do not install any mains smaller than 8 inches in size.  An 8 inch main costs little more than a 6 inch main to install, since trenching and other costs remain the same.  Yet an 8 inch main has more than twice the capacity of a 6 inch main, under the same head.  It is much more desirable to oversize, rather than to undersize, when installing new mains. When mains are not of sufficient size to meet peak demands in an area, pressures drop rapidly and inadequate service results.  Conditions of this nature have been rather prevalent in many cities during the past few years because of increasing demands resulting from improved standards of living, rapid population growth, new appliances, air conditioning and other factors. The American Water Works Association recommends a normal static pressure of 60 to 75 psi through the system.  A minimum pressure of 45 psi is desirable in a residential area and 75 psi would be desirable in a commercial or industrial area.   Calculation of Head Loss There are several formulas for computing the carrying capacities of water mains.  The one most commonly used in the water works industry is the Hazen-Williams Formula.  A factor called "C" represents the coefficient of  friction.  It is a measure of the roughness of the interior surface of pipe.  A "C" factor of 140 would indicate a very smooth pipe, whereas a "C" factor of 70  would indicate a very rough pipe.  The smoother the surface of the pipe wall, the larger the value of "C" and, consequently, the greater the carrying capacity.  To save time, these formulas can be converted into chart form from which head loss values can be read directly. A good example to show differences in head loss is a flow of 200 gallons per minute in the following: 2 inch pipe - 690 feet per 1000 feet 3 inch pipe - 96 feet per 1000 feet 6 inch pipe - 3.31 feet per 1000 feet 8 inch pipe - .82 feet per 1000 feet 10 inch pipe - .28 feet per 1000 feet 12 inch pipe - .11 feet per 1000 feet   Compare the head loss due to friction between the 2 inch and 12 inch pipe.     Problem: A ground storage tank is to be constructed on a hill with an overflow elevation of 586 feet.  An 8" line is to be used to serve the town approximately 20,000 feet away from the tank.  The elevation of the town is approximately 380 feet.  Find: What fire flow can be expected in the center of town 20,000 feet away while maintaining a 20 psi pressure residual? With a fire flow of 700 gpm what will be the head loss in the 20,000 feet of pipe? With a fire flow of 700 gpm what will the pressure residual be at the hydrant in the center of town 10,000 feet from the tank?   Solution: a.  First find the difference in elevation 586'  -  380'  =  206'. Determine the feet of head necessary to provide a 20 psi pressure residual.  (2.31' of water is equal to 1 pound of pressure, therefore, 20  x  2.31  =  46.2')  Therefore, 46.2' of head is necessary. Subtract this figure from the elevation difference as this head must be constant and maintained at all times.  206'  -  46.2'  =  159.8'.  This 159.8' of head may be used up in friction losses or 7.92' per 1000' of pipe may be used. (Using 159' at Head). Referring to the friction loss chart, in an 8" pipe with a head loss of 7.92' per 1000' of pipe length, slightly less than 700 gpm fire flow can be expected. (Because of fitting loss and changes in the pipe with time). From the solution in  b), 158.4' of head is lost due to friction.  To find the available head and resulting pressure remaining, subtract 158.4' from the total elevation difference (head) as found in  a) above or 206'.  Therefore, 206'  -  158.4'  =  47.6'.  Since 2.31' of head equals 1 psi, divide 47.6' by 2.31' to result in a residual pressure of 20.6 psi. b.  Using the table with a 700 gpm fire flow, the friction loss per 1000' of pipe is 7.92/1000'. The correct multiplier for  b.1) is 20,000 of pipe per 1000' of pipe, or 20. Therefore, 7.92'  x  20 results in 158.4' of frictional head loss in 20,000 feet of pipe. c.                   At 10,000' the pressure drop is 79.2'.  The residual pressure is 206' - 79.2' = 126.8' Diagram for Lost Pressure or Lost Head. The pressure lines shown in the above sketch are called hydraulic gradients.  The elevation of any point on the hydraulic gradient represents the height to which water would rise due to pressure in the line at a point immediately below the given point on the gradient. The discharge pressure of a high service pump equals the elevation through which the water is pumped plus pipe friction. 