130 x 1000 x 9.81 x 85 = 108,400,500 Kg.m2/h = 108,400,500 N-m/h. 141 0 obj <>/Filter/FlateDecode/ID[<2AC97D55E6451EDEE63209A70F0584E1><5E8D5153CF2EE04A8E3F0D84755798AD>]/Index[126 24]/Info 125 0 R/Length 76/Prev 182129/Root 127 0 R/Size 150/Type/XRef/W[1 2 1]>>stream %%EOF endobj – Medium pressure supply lines should be sized so the fluid velocity is between 15 and 20 feet per second. A = π.r² = π x (14 / 12)2 = 4.28 ft² m = ρ x A x v = 9200 lb/s v = 9200 lb/s = 9200 lb/s…… = A. ρ 4.28 ft² x 49 lb/ft³ Enter information into any 3 of the following 4 fields and press calculate. ������J@��� �w� %PDF-1.4 %���� If there’s no power on site, you need to spec a generator: • Most submersible pumps with 5HP or less are 230V/single phase. ���Ph>�h��YYY܆Iyd�\�Y��_��N�Mڲ��+�r5�7� sgﾈ4��(�z���5���C h��Ymo�6�+�����N���4�ץ �lm'胓h��\$lh��-��c� ��A�"���G>�� ��֋��tJ/��,��! How to select a centrifugal pump Examples of total head calculations - sizing a pump for a home owner application 14. 3 0 obj 0 231 x GPM ÷ RPM = 231 x 5 ÷ 1500 = 0.77 cubic inches per revolution. Hydraulic power, pump shaft power and electrical input power Hydraulic power P h = Q (m3/s) x Total head, h d - h s (m) x ρ (kg/m3) x g (m/s2) / 1000 Where h d – discharge head, h allow 0.1 Open isolation valve 0.4 Open control valve 10.8 Tee (flow from side branch) 1.2 Fluid Pressure in PSI Required to Lift Load (in PSI): Example: What pressure is needed to develop 50,000 pounds of push force from a 6″ diameter cylinder? 14,7) = 0,00836 m³/s. Power is consumed by a pump, fan or compressor in order to move and increase the pressure of a fluid. Target hydraulics assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information. Examples of common residential water systems 15. RPM x Pump Displacement ÷ 231 = 1200 x 2.5 ÷ 231 = 12.99 gpm. The power requirement of the pump depends on a number of factors including the pump and motor efficiency, the differential pressure and the fluid density, viscosity and flow rate. 104 0 obj <>stream – High pressure supply lines should be sized so the fluid velocity is below 30 feet per second. endobj Chapter 3: Review of Basic Vacuum Calculations Before we go any further, some time should be spent on some of the vocabulary specific to vacuum technology. Ukraine ��% 6��{=圁���2�C�AC��(O� ��j0�Y�C�hG��⢿�>�~�rb���(FG�|3\1�D^���%��/�r`��R��ҁF��l!,�^E��s���k����H'���E:/ݨQ¢Ǫ 7ɺx��}�x���̘-�M�6�pc �BC�v�����ѧ��'�&�=Ѥi� GPM of Flow Needed for Fluid Motor Speed: Example: How many GPM are needed to drive a 3.75 cubic inch motor at 1500 rpm? Pump Input Power = P. Formula – 1. stream [��g��(��m77���V�lv��%S���e��yi���.�%�KS_���j�3��Gx���SR���6 V�! You’ll need 1.5kW per motor HP So…a 5HP pump needs a minimum of 7.5kW You also need to spec the right power cable • PVC flat-jacketed wire is the most common. Cylinder Blind End Area =28.26 square inches, Cylinder Rod End Area = 21.19 square inches, Blind End Area ÷ Rod End Area x GPM In = 28.26 ÷ 21.19 x 15 = 20 gpm. H�|TMo�6��W쩐��ᷨ��-P��'�R��Ȕ�B QN��%e� �S� �vggv��*{�*�&3�4��� What pressure is needed to develop 50,000 pounds of pull force from a 6″ diameter cylinder which has a 3″ diameter rod? ?i��t?t��p�9�C(��p��/F���t8/��#^�!��I#�O�< Fluid Motor Torque from Horsepower and RPM: Example: How much torque is developed by a motor at 12 horsepower and 1750 rpm? v�ԁ�H��"I��b,W,��Wl1U�z�����=��>��{�[9�#my�m>ښy�Y��æmj�6�֛+γ�G�t�s�7��Cp㎎