Pipe Pressure Drop Detail Calculation due to friction, Valves, Fittings Real Practical Example скачать в хорошем качестве

Pipe Pressure Drop Detail Calculation due to friction, Valves, Fittings Real Practical Example

#Water #mechanical #process The video notes are given below for your quick reference - I tried to put the total note altogether, but due to space limitation, had to delete many lines. Please watch the video to get the detail and proven calculation. Water flows through pipe due to pressure difference. A pump discharges water at higher pressure to transfer the water to a higher elevation. When water flows through pipes, its pressure gets dropped due to friction. We will calculate the pressure drop across a piping system. Problem – Let’s assume there is a tank on the ground floor of a building filled with water. We have an empty tank on the rooftop of the building at 40m height. We need to transfer the water from the ground tank to the rooftop tank through MS pipe. Temperature of water is 30°C. We have a pump of flow capacity 100m3/hr. We have to find the minimum pump discharge pressure at which the water can be transferred to the rooftop tank easily. Solution – At first we need the pipe diameter. Here it is given as 150MM in the figure. (It can also be calculated easily. You may refer my earlier video Pipe Sizing for detail calculation )    • Basic Pipe Sizing Pump Suction Discharge L...   Minimum pump discharge pressure should be more than the cumulative value of the following – A. Pressure required to overcome the 40m height (static head) B. Pressure drop in the pipeline due to friction C. Pressure drop due to pipe fittings like bend, valve, tee, reducer etc For the first part (A), Static Head = difference in elevation + Tank Height = 40 + 2 = 42m So, pressure required to overcome the vertical lift = pressure of 42m water column = 411877 Pa Now the second part (B) – Pressure drop in the pipeline due to friction The pressure drop can be calculated using any of the following equations – Darcy-Weisbach Equation Hazen-Williams Equation Data we have – Q = 100m3/hr Q = 0.0278 m3/s L = (40 + 2) + 5 = 47m D = 0.1542m ρ =1000 kg/m3 Q = (πD2/4) x V V = 4Q / πD2 V = 1.49 m/s f = friction factor The friction factor is to be taken from the Moody diagram or using the Colebrook equation. It depends on the flow type and roughness of a pipe surface. For laminar flow, f = 64 / Re (Re = Reynolds number) But, in the case of water flow through pipe, it is always turbulent flow. We will also find the Reynolds Number (Re) which will clearly indicate the flow type. To get the Friction Factor value for turbulent flow, either we can use Colebrook equation or Moody Diagram. We will rather use Moody diagram to find the friction factor. The tentative roughness (ε) is already mentioned in the Moody diagram in a table format. In our case, for MS pipe (of course not rusted), ε = 0.025 mm Relative pipe roughness = ε / D = 0.025 / 154.2 (all units in mm) = ε / D = 1.621 x 10-4 Reynolds Number, Re = ρVD / μ ρ = 1000 kg/m3 (density of water) V = 1.49 m/s (water velocity in pipe we calculated) D = 0.1542 m (pipe internal diameter) μ = dynamic viscosity of the fluid [Pa-s or N-s/m2 or kg/m-s] Dynamic Viscosity of water varies with temperature as indicated in the table (experimental values) – In our case, the temperature is 30°C. μ = 0.79722 mPa-s = 7.9722 x 10-4 Pa-s Re = ρVD / μ = 2.9 x 105 So, using the data ε / D & Re , we can find the friction factor f from the Moody Diagram. f = 0.016 (approximately) Pressure drop ∆p = 0.016 x (47 / 0.1542) x (1000 x 1.492 / 2) ∆p = 5413.5 Pa Now the Part C - Pressure drop due to pipe fittings & valves Valves and fittings can cause pressure loss greater than those caused by the pipe alone. More the restricted passage, greater is the loss of head. It can be calculated using the following formula – ∆p = KρV2 / 2 Where – K = loss coefficient depending on geometry and size Another way to express these losses is in terms of an equivalent length of an unobstructed straight pipe in which an equal loss would occur for the same average flow velocity. That means - equivalent lengths (Le) of pipe fittings, bends etc. need to be taken into account while calculating pressure drop using Darcy-Weisbach equation. There, the length L will be pipe length (horizontal + vertical) + equivalent length of pipe for the fittings & valves. The equivalent length for a valve or fitting (Le) is calculated by multiplying internal diameter of pipe with (Le / D) ratio of that fitting. The (Le / D) ratio are experimental values. Some typical data is shown in the table below for a few frequently used fittings: In our problem, we will use the K factor formula ∆p = KρV2 / 2 Now, if we look at our drawing – we find the following valves and fittings – Check valve (NRV) – 1 Gate Valve (assume full open in general) – 1 90° elbow – 1 Total pressure drop to be overcome = (411877+5413.5+7261) = 424551.5 Pa = 4.33 kg / cm2 Taking a safety margin of 15%, total discharge pressure requirement is = 4.33 x 1.15 = 5 kg / cm2