TITLESuitable pump for a new projectEXECUTIVE SUMMARYBefore the end of the semester 3 on 2014, our particular section have been given a mini project by our instructor Dr. Fairna as a final task for SKKK 2043 in which the due date is on 18th December 2014. The main objective of our project is to make a research in order to find and suggest a suitable pump as requested by the task given.In order to complete our task, we use internet and Fluid Mechanics text book authorized by Yunus A. Cengel and John M. Cimbala as our main sources. This project was prepared by a group of three members which are Tengku Muhammad Solihin Bin Tengku Razali, Muhamad Amirul Bin Azhar and Nur Syahidatul Ridzuan.The major finding of this project is we need to sketch a graph and make a calculation based on data and information provided in the task before came out with the solution. The task given quite difficult to solve, a lot of try and error of the calculations we had done in order to find the best answer as requested. There are several steps that we must take a look and will be discussing in this project. Based on the given data, we substitute the data into the Energy Equation to get head loss in pump values. Then, we applied the equation that stated in text book to give the values of rotational speed (N) and impeller diameter (D) at maximum efficiency, thus it lead us to get value of power input to the pump as wanted in the question given. The value that we get by steps above had been substitute into the available NPSH equation so that we can compare to the value of required NPSH. Graph of pump performance will plotted which is containing values of head loss (H), discharge (Q), power input (P) and efficiency (). Lastly, a lot of things that we had learnt and achieved while completing this task. The following pages summarize the key status indicators for the project as a whole. Thank you to our instructor for ongoing support in this project

INTRODUCTIONWith the given value of Q, H, and P, we must find the suitable pump which is has maximum efficiency value. Besides, we must suggest one pump that available commercially that could be from brochure or website and give reason why we choose that pump. Other than that, we should provide the brand and specification of pump from the supplier in our project.THEORY AND CALCULATIONProperties: The pump operates at room temperature and the density of water is 1000 kg/m3.Lets see what we have given. Data available (discharge, differential head and power input) with an impeller diameter of 0.15m and a rotational speed of 1500 rpm are shown below. However, efficiency needs to be calculated by using a formula given:Efficiency, = Discharge, Q (L/s)0510152025

Differential Head, H, (m)9.258.817.856.484.812.96

Power Input, P, (KW)00.961.031.191.261.45

Efficiency (%)0.000.450.750.800.750.50

Table 1.1These are plotted in figure 1-1, along with a curve for brake horsepower.

Next we need to convert the dimensional data of table 1-1 into non-dimensional pump similarity parameters. We can use the formula given below;

Efficiency0.000.450.750.800.750.50

CP x10000.3260520.3498260.4041680.4279430.492474

CQ00.009430.018860.0282910.0377210.047151

CH x101.634091.556361.3867681.1447460.8497270.522909

Table 1-2

Figure 1-2A vertical line at the best efficiency point (BEP) is sketched in figure 1-2. The curve-fitted data yield the following non-dimensional pump performance parameter at the BEP;CQ = 0.028 CH = 0.114 CP = 0.00404 Efficiency = 0.80 Now, we design the suitable pump (B) such that its best efficiency point is homologous with BEP of the available pump (A), but with different pump diameter and different rotational speed. Namely, are known but we still do not know the differential head for pump B. So, we use energy equation to find the answer as we assume the flow of water is at steady state and incompressible.

Q = AvA = 0.025, D = 0.178mUse Energy Equation = Assumption: z2- z1 = 10m, V1 and V2 is zero due to the large tank. Besides, we know that [=50kPa, =150kPa, =2.55m, =0.143m] from the calculation and also based on the question.Thus : = = 22.88 mFrom that, we will calculate for Impeller diameter (DB) by using Affinity Law. We know that [= 0.05, = 0.114, = 22.88, =0.028]

= 0.19 m

With the value DB known, we return to solve the unknown of WB, the design rotational for pump. We know that =0.028, = 0.19 m, = 0.05 /s

WB = 41 rps or 2460 rpm

Finally, the required brake horsepower for pump (B) is needed to be calculate. We know that [=0.00404 =1000kg/m3 =2460 rpm = 0.19m ]

bhpb= = 17.413x103 kW

DATA ANALYSISDischarge, Q (L/s)0510152025

Differential Head, H, (m)9.258.817.856.484.812.96

Power Input, P, (KW)00.961.031.191.261.45

Efficiency (%)0.000.450.750.800.750.50

Table 1-1

Figure 1-1

Efficiency0.000.450.750.800.750.50

CP x10000.3260520.3498260.4041680.4279430.492474

CQ00.009430.018860.0282910.0377210.047151

CH x101.634091.556361.3867681.1447460.8497270.522909

Table 1-2Figure 1-2

CONCLUSIONWe choose single stage electric power centrifugal water pump because the pump required power to generate the pump which is 17.413 kW, 27.88m of differential head, 0.19 m of impeller diameter and 41.0rps of rotational speed required for operation at maximum efficiency. The data below shows the specification of pump that we choose.Required chosen criteria:ModelFlowdifferential headRotation speedMotor powerDiameter(mm)Diameter(mm)

(m3/h)(m)(r/min)(kw)SuctionOutlet

IS125-100-25020080290075125100

Required criteria Flowdifferential headRotation speedMotor powerDiameter(mm)Diameter(mm)

(m3/h)(m)(r/min)(kw)SuctionOutlet

18027.882403.1217.14125100

We choose the pump because the calculated criteria fall into the range of the criteria of the pump have almost same value with the criteria of chosen pump as show in table above.REFERENCEYunus A.Cengel, John M. Cimbala (2014). Fluid Mechanics: Fundamental and Applications (3rd edition). Singapore: Mcgraw-Hill Educationhttp://www.alibaba.com/product-detail/High-Quality-Single-stage-Electric-Power_60084549955.html?s=p