Engineering Script

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Description	<div style="color: rgb(68, 68, 68); font-family: Consolas; font-size: 15px;"><div>A pump is a kind of equipment that consumes a considerable amount of energy in the industrial process. It is mainly driven by a motor, and in many cases, a kW meter or ampere meter is installed so that the power consumption can be calculated, but in many cases, this is not possible.</div><div><br>If know the pumping flow rate and pressure difference, user can calculate the amount of work required by the pump. Of course, the power usage of the motor requires assumptions.</div><div><br>The hydraulic power that drives the pump depends on mass flow rate, liquid density and height difference.</div><div><br></div><h2 style="margin: 0px; position: relative; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-alternates: normal; font-kerning: auto; font-optical-sizing: auto; font-feature-settings: normal; font-variation-settings: normal; font-variant-position: normal; font-weight: bold; font-stretch: normal; font-size: 14px; line-height: normal; font-family: "Trebuchet MS", Trebuchet, sans-serif; color: rgb(0, 0, 0);"><span style="font-family: Consolas;">Formula of pump power consumption</span></h2><div><br>Pump power equation is as follows.</div><div><br><b>W = ρ * Q * h * g / η</b></div><div><b><br></b>where<br>W = power (kg•m2/s3, ft-lbf/s)<br>ρ : density of fluid (kg/m3, lbm/ft3)<br>Q = volumetric flow (m3/s, gpm)<br>h = head (m or ft) the fluid has to be lifted<br>g : acceleration of gravity (9.8 m/s2, 32.174 ft/s2)<br>η = total efficiency (%, η of pump, η of motor)<br>1. In case of MKS (m/kg/sec/°C/bar/W) units<br>W = ρQhg/η = (1000 kg/m3 1000 m3/s * 25 m * 9.8 m/s2) / (3600 * 1000 * 0.8) = 85.1 kW<br><br></div><div>where,<br>W : power (W)<br>ρ : density of fluid (kg/m3)<br>Q : flow (m3/h)<br>h : differential head (m)<br>g : acceleration of gravity (9.8 m/s2)<br>η : friver efficiency (%)<br>2. In case of FLS (ft/lb/sec/°F/psi/bt) units<br>W = ρQh/η = (62.4 lbm/ft3 * 1000 gpm * 106 ft) / (7.481 gal/ft3 * 33,000 ft-lb/min-hp * 70%) = 38.3 hp<br><br></div><div>Where,<br>W : power (hp)<br>ρ : density of fluid (kg/m3 or lbm/ft3)<br>Q : flow (gpm)<br>h : differential head (ft)<br>g : acceleration of gravity (32.174 ft/s2)<br>η : friver efficiency (%)<br><br></div><div><h2 style="margin: 0px; position: relative; font-variant-numeric: normal; font-variant-east-asian: normal; font-variant-alternates: normal; font-kerning: auto; font-optical-sizing: auto; font-feature-settings: normal; font-variation-settings: normal; font-variant-position: normal; font-weight: bold; font-stretch: normal; font-size: 14px; line-height: normal; font-family: "Trebuchet MS", Trebuchet, sans-serif; color: rgb(0, 0, 0);"><span style="font-family: Consolas;">Python code of pump power consumption</span></h2></div><div><br></div><div>The following is a Python example that calculates the pump power.</div><div><br><span style="color: rgb(43, 0, 254);">def pumpower(unit, d, q, h, n):<br></span><span style="color: rgb(43, 0, 254);">    if unit == "MKS" and n > 0:<br></span><span style="color: rgb(43, 0, 254);">        W = (d * q * h * 9.8) / (3600 * 1000 </span><span style="color: rgb(43, 0, 254);">* n</span><span style="color: rgb(43, 0, 254);">) # kW<br></span><span style="color: rgb(43, 0, 254);">    elif unit == "FLS" and n > 0:<br></span><span style="color: rgb(43, 0, 254);">        W = (d * q * h) / (7.481 * 33000 * n) # hp<br></span><span style="color: rgb(43, 0, 254);">    else:<br></span><span style="color: rgb(43, 0, 254);">        pass<br></span><span style="color: rgb(43, 0, 254);">    return W<br></span><span style="color: rgb(43, 0, 254);"><br></span></div><div><span style="color: rgb(43, 0, 254);">pumppowerMKS = pumpower("MKS", 1000, 1000, 25, 0.8) # 1000 kg/m3, 1000 m3/hr, 25 m, 80%<br></span><span style="color: rgb(43, 0, 254);">pumppowerFLS = pumpower("FLS", 62.4, 1000, 106, 0.7) # 62.4 lbm/ft3, 1000 gpm, 106 ft, 70%<br></span><span style="color: rgb(43, 0, 254);"><br></span></div><div><span style="color: rgb(43, 0, 254);">print("Pump power = ", pumppowerMKS, " kW")<br></span><span style="color: rgb(43, 0, 254);">print("Pump power = ", pumppowerFLS, " hp")</span><br><span style="color: rgb(43, 0, 254);"><br></span></div><div>When run the code, you get the results below.</div><div><br></div><div><span style="color: rgb(43, 0, 254);">Pump power =  85.069 kW</span><br><span style="color: rgb(43, 0, 254);">Pump power =  38.275 hp</span></div><div><span style="color: rgb(43, 0, 254);"><br></span></div></div>