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Chapter 4.7 - Water Pumps

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Selected questions Chapter4.7: Water Pumps Short type questions Long type questions 1. a1 available for b1 measuring pump flow and write briefly about What are the types of field test methods any two methods available? Ans Some of the measurement methods for large water flow are as under: Tracer method BS5857 Ultrasonic flow measurement Tank filling method Installation of an on-line flowmeter a2 b2 Tracer Method Tracer methods are particularly suitable for cooling water flow measurement because of their a3 b3 sensitivity, accuracy and minimum permanent head loss. The method is based on injecting a tracer into the cooling water for a few minutes at an accurately measured constant rate. A series of samples is extracted from the system at a point where the tracer has become completely mixed with the cooling water. The mass flow rate is calculated from : qcw where, qcw = q1 C1/C2 = cooling water mass flow rate, kg/s q1 = mass flow rate of injected tracer, kg/s C1 = concentration of injected tracer, kg/kg C2= concentration of tracer at downstream position during the plateau period of constant concentration, kg/kg The tracer normally used is sodium chloride. Ultrasonic Flow meter Operating under Doppler effect principle these meters are non-invasive type of measurements which can be taken without disturbing the system. Scales and rust in the pipes are likely to impact the accuracy. For better accuracy, a section of the pipe can be replaced with new pipe for flow measurements. Tank filing method In open flow systems such as water getting pumped to an overhead tank or a sump, the flow can be measured by noting the difference in tank levels for a specified period during which the outlet flow is stopped. Installation of an on-line flow meter If the application to be measured is going to be critical and periodic then the best option would be to install an on-line flowmeter which can rid of the major problems encountered with other types. 2. Explain the Pump operating point with a diagram? Ans When a pump is installed in a system the effect can be illustrated graphically by superimposing pump and system curves. The operating point will always be where the two curves intersect. .7 - Pumps (table format) 40 Selected questions If the actual system curve is different in reality to that calculated, the pump will operate at a flow and head different to that expected. Numerical type questions 1. The following table gives the centrifugal water pump details: Rated flow : 90 m /h Rated head : 4.5 kg/cm (g) Motor Rating pump : 37 kW Considering 65% pump efficiency and 85% motor efficiency, (a) find out whether the sizing of the drive correct? If not what should be the size of motor? (b) If the above pump is drawing 18.5 kW and the required head is 30m, the rated flow rate is 90 m /h, what should be the size of the new pump? And what would be the savings considering 70% pump efficiency and 89% motor efficiency? (a) The liquid horse power of the pump is Hydraulic power Ph = Q (m3/s) x Total head, hd - hs (m) x (kg/m3) x g (m/s ) / 1000 = 90 x 4.5 x 10 x 9.81 3600 = 11.04 kW Considering 65% pump efficiency and 85% motor efficiency The required power = 11.04 0.65 x 0.85 = 19.98 kW Higher size motor has been chosen, which is incorrect. The reduced standard size motor for this pump would be 22 kW. (b) The measured parameters are: Flow = 90 m /h Head = 3.0 kg/cm Power = 18.5 kW The operating efficiency of the pump is (considering 85% motor efficiency) Pump output power .7 - Pumps (table format) = 90 x 3.0 x 10 x 9.81 3600 x 0.85 = 8.65 kW 41 Selected questions Pump efficiency = 8.65 18.5 = 46.8% The new sizing of the pump should be 90 m /h, 30m head. Considering a pump efficiency of 70% and the motor efficiency of 89%, the power consumption should be: = 90 x 30 x 9.81 3600 x 101.9 x 0.7 x 0.89 Existing power consumption = 18.5 kW Proposed power consumption = 11.8 kW Net savings = 6.7 kW .7 - Pumps (table format) = 11.8 kW 42 Selected questions 2. In a large paper plant, the following are the designed and measured parameters for a clear water pump. Particulars Design Operating 800 576 Head, m of WC 55 24 (after control valve) Power, kW 160 124 Speed, rpm 1485 1485 Flow, m /h The pump delivery has been throttled to about 30% (closed) manually to get the required flow rate. Normal required water flow rate is 500 m /h to 700 m /h. Calculate the present operating efficiency and in your opinion what should be the optimum solution to get the required flow rate variation? And what would be the savings if the pump is delivering the flow rate of 550 m /h. (Consider efficiency of motor as 93%). Present pump output Pump input power = QxHxg 3600 x p x m = Ans. 576 x 24 x 9.81 = 40.5 kW 3600 x 0.93 = 124 kW pump operating efficiency = 40.52 x 100 124 = 32.67% The pump is operating at a poor efficiency of 32.67% due to throttling of the flow. Since the pump discharge requirement varies from 500 m /h to 700 m /h, the ideal option would be to operate with a VSD. According to affinity laws: Q1 N = 1 Q2 N2 H1 N 1 = H2 N2 P1 N 1 = P2 N2 2 3 For a flow rate 550 m /h, the reduced speed of pump would be: = N1 N1 550 = 800 1485 = 1021 rpm With the reduction in speed the reduction in terms of head would be: 2 = .7 - Pumps (table format) 1021 x 5.5 1485 = 2.6kg / cm 43 Selected questions The reduction in power would be: 3 = ~ 124 40.3 = 3. 40.3 kW = the reduction in power 1021 x 124 1485 83.7 kW = 40.3 kW Analyze the following figure and answer the questions. 7.5 m NRV Dia = 6 P 2.5 Pump rated parameters Q H P P 30 lps 18 m 9.3 kW 65% a b Is the pump operating at its design efficiency Ans. a. In the present case, flow rate from the pump will be higher than rated flow rate. It is mainly due to lower operating head (around 10 m) as against rated head (18 m). b. 4. In normal operation what would be the flow rate from pump compared to rated value? Pump operating efficiency will be less than design efficiency. It is due to higher flow rate and lower operating head. In one of the chlor Alkali plant, analysis of one of the operating parameter of a titanus impeller pump for flow of brine were as follows: m3/hr Head kW Rated 310 45 90 Actual 210 40 67 On detailed examination of the flow/head requirement (maximum) was assessed to be 260 m3/h and 30 m. Though change of pump was one of the option, considering cost of special pumps impeller impeller cutting was one of the options suggested which involves Rs 3.0 lakh as cost. Calculate likely annual saving after impeller cutting with pump efficiency at 65% and motor efficiency at 85%, fluid density 1160 kg/m3 operating hours: 8000, unit rate: Rs 5/Ans. Hydraulic power Ph = Q(m3/s)xTotal head, hd-hs(m)x (kg/m3) x g (m/s ) / 1000 After trimming impeller power consumption = (260 / 3600) x 30 x 1160 x 9.81 = 44.6 kW 1000 x 0.65 0.85 Estimated power savings = 67 - 44.6 Annual savings (Rs) = 22.4 x 8000 x 5 = Rs 8.96 lakh Simple pay back period = .7 - Pumps (table format) = 22.4 kW 3 .0 : 4 months 8 .96 44

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