PRESSURE DROP, POWER, AND COST Pressure Drop Consider flow through a device as shown below. The pressure drop across the device, ∆P, is the difference in static pressures at the outlet and the inlet, ∆P = h1 – h2 . h1Device h2 V1 V2 The total energy going into the device equals the total energy coming from the device, plus energy lost in the device. Total Energy In = Total Energy Out + Lost Energy We would like to develop an expression for ∆P that reflects what we know about energy in the system. Total energy is the sum of kinetic energy of the flowing fluid, KE, plus the potential energy, PE, due to the static pressure of the fluid. Energy is expressed per unit volume of fluid. Total Energy = KE + PE . Here, KE = 2gV21ρ and PE = ρL g h , where ρg = gas density, V = gas velocity ρL = density of fluid in manometer g = acceleration of gravity h = height of fluid in manometerNow let the energy lost in the device be expressed as a constant, ∆H, times the kinetic energy of the incoming gas: Energy lost in device = ∆H 21gV21ρ Why should this expression for lost energy be reasonable? Pressure drop can be thought of as the force/area necessary to push fluid past the surfaces and obstacles in its way as it flows through the device. Applying Newton’s Law for drag force in turbulent flow, ∆P = 2gD2gDV21CAreaV21AreaCAreaForceρ=ρ= . Thus, ∆H can be thought of as a sort of “drag coefficient” related to flow past or through the device. From this analysis, which applies for turbulent flow conditions, lost energy can be seen to be proportional to V2. If flow is laminar, then lost energy would be proportional to V because for laminar conditions CD ∝ 1/V; however, in most cases that involve industrial flows, turbulent conditions exist. After substitution, we find that 21g2L22g1L21gV21HhgV21hgV21ρ∆+ρ+ρ=ρ+ρ If the velocity of the fluid flowing into and out of the device is the same, V1 = V2. Because pressure drop, ∆P is given by h2 – h1, substitution yields Hg2VPLg21∆ρρ=∆. Values of ∆H can be determined experimentally by measuring ∆P with the other terms in the above equation known. These values have been tabulated for devices such as elbows, cyclones, and other objects that cause pressure drop in a stream of flowing fluid. Alternatively, for some shapes, ∆H can be calculated from theory. Power Recall that Power = timecetandisforcetimework×= ThusQPtimeareacetandisareaforcePower ∆=××= where Q is volumetric flow. If we use mks units, then power is given in watts if pressure drop is in Pascals (N/m2) and flow is in m3/s. If we choose to use English units, so that pressure drop is expressed in inches of water and flow is expressed in cubic feet per minute (cfm), then Powerwatts = 0.118 ∆Pinches of water Qcfm . Cost The power given in the equation above represents the power of the flowing gas stream. To convert this quantity to the electrical power necessary to run a fan that provides this power, we must divide by the mechanical efficiencies of the fan and the electrical motor, MEfan and MEmotor. With this modification, and expressing power in kilowatts instead of watts, gives Powerkw electrical = motorfancfm waterof inches-4MEME QP10 x 1.18 ∆ From this equation, we can determine the cost of running the fan if we know the cost of electricity, typically about $0.07/kwh, and the number of hours the fan operates. In addition, we can determine the horsepower for the electrical motor required, using the conversion that kw × 1.34 =