Thermodynamics Answer provided Please answer the questions with solution if you get the correct answers. Thank you Water (p 998 kg/m3 and u 0.001002 Pa s flows through the pipe system shown in the diagram on the next page. At Point 1, the water is already flowing at a velocity of 2.5 m/s through a 0.5 inch inner diameter smooth pipe. The pressure of the water at this point is 120 kPa. A pump with an efficiency of 80% is used to increase the pressure of the water. Note that after the pump there is a sudden change in diameter to 1 inch (smooth pipe). Referring to the diagram, the pressure at Point 2 is 275 kPa. The water then flows through an unknown length of pipe until a sharp exit open to the atmosphere. You may neglect Fioss at the sudden change in diameter from 05 to 1 inch a Determine the pump power (in kW) required to give a pressure of 275 kPa at Point 2 (161 12 W) b) Determine the length of pipe after Point 2 (Lin m). (824 140 m) Solution Volumetric flow rate is calculated first from inlet, by continuity of mas this must ramin const. = 2.5m/s * Area = 3.167 * 10-4m3/s Total head = given height plus friction losses estimate l/d for galve = approx 450, f = .03 for laminar flow head loss due to gate valve is .03*450 * V2/2g = 2.8 m add to head = 12.8 m pressure head = (274-120)*1000/( 998*9.81) =15.83 friction loss at 4 elbows: (l/d) =30*4 =120, head loss approx 1.1 m add to head , 14 m. Total head 30 m approx Power rhoQgH = 998* 3.167* 9.81* H = 116 w approx after factoring 80% efficiency 161 may be type o or some other friction factor used.