Extracted text: Fluid motor system problem: Water at 10°C is flowing at a rate of 115 L/min through the fluid motor shown in figure below. The pressure at A is 700 kPa and the pressure at B is 125 kPa. It is estimated that due to friction in the tubing there is an energy loss of 4.0 N.m/N of water flowing. At A the tubing entering the fluid motor is a standard steel hydraulic tube having an OD = 25 mm and a wall thickness of 2.0 mm. At B, the tube leaving the motor has OD = 80 mm and wall thickness of 2.8 mm. (a) Calculate the power delivered to the fluid motor by the water. (b) If the mechanical efficiency of the fluid motor is 85 percent, calculate the power output. (For water at 10°C, y= 9.81 kN/m³) %3D %3D HINT: USE THE GENERAL ENERGY EQUATION BETWEEN POINTS A & B TO FIND hR 25-mm OD PA/y + ZA + VA²/2g + h- hg- h = Pg/y+ ZB + vVg/2g x 2.0-mm wall Flow %3D BUT power delivered by a fluid to a motor is given by the following equation PR = hR x weight flow rate = Fluid motor hR y Q 1.8 m oDED ANS Efficiency (eM) is given by the following equation em = Po/PR ECK CORRECT. EXPLAIN W %3D 80-mm OD x 2.8-mm wall CHECK BELOW ANSWER: Po=0.9256 Kw B. WHAT IS THIS IN HORSEPOWER? 6.