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 thickness of 2.0 mm. At B, the tube leaving the motor has OD 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) 25 mm and a wall 80 mm and wall thickness of 2.8 INIH 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+ Zg + V³/2g x 2.0-mm wall Flow BUT power delivered by a fluid to a motor is given by the following equation PR = hg x weight flow rate = Fluid motor hR y Q 1.8 m %3D HonDED AN Efficiency (em) is given by the following equation HECKI CORRECT EXPLAIN W 80-mm OD x 2.8-mm wall CHECK BELOW ANSWER: Po 0.9256 Kw %3D B. WHAT IS THIS IN HORSEPOWER? 6.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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**Fluid Motor System Problem:**

Water at 100°C is flowing at a rate of 115 L/min through the fluid motor due to friction. The pressure at A is stated as 4 bar, and the pressure at B is stated as 3.6 bar, caused by the flow of water. Flow is through a standard steel hydraulic tube having an OD (outer diameter) of 25 mm and a wall thickness of 2.0 mm. The OD at point B is 80 mm with a wall thickness of 2.8 mm. The height from A to B is 1.8 meters. The mechanical efficiency of the fluid motor is 85 percent.

**Tasks:**

(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.

_Use the general energy equation between points A and B to find hR:_

\[ p_A/\gamma + Z_A + v_A^2/2g + h_R = p_B/\gamma + Z_B + v_B^2/2g \]

**Power Delivered Equation:**

BT power delivered by a fluid to a motor is given by the following equation:

\[ P_R = h_R \times \text{weight flow rate} = h_R \gamma Q \]

**Efficiency Equation:**

Efficiency (\( e_M \)) is given by the following equation:

\[ e_M = P_O / P_R \]

**Hint:**

For water at 100°C, \(\gamma = 9.81 \text{kN/m}^3\).

**Check Below Answer:**

Power Delivered: 0.29256 kW

**Conversion Question:**

What is this in horsepower?

**Diagram Explanation:**

The diagram depicts a fluid motor system, showing the flow of water from point A to point B through a steel tube. The dimensions are provided for two sections of the tube, with an outer diameter of 25 mm and a wall thickness of 2.0 mm at the inlet (A), and an outer diameter of 80 mm with a wall thickness of 2.8 mm at the outlet (B). The system's height difference between these points is 1.8 meters.
Transcribed Image Text:**Fluid Motor System Problem:** Water at 100°C is flowing at a rate of 115 L/min through the fluid motor due to friction. The pressure at A is stated as 4 bar, and the pressure at B is stated as 3.6 bar, caused by the flow of water. Flow is through a standard steel hydraulic tube having an OD (outer diameter) of 25 mm and a wall thickness of 2.0 mm. The OD at point B is 80 mm with a wall thickness of 2.8 mm. The height from A to B is 1.8 meters. The mechanical efficiency of the fluid motor is 85 percent. **Tasks:** (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. _Use the general energy equation between points A and B to find hR:_ \[ p_A/\gamma + Z_A + v_A^2/2g + h_R = p_B/\gamma + Z_B + v_B^2/2g \] **Power Delivered Equation:** BT power delivered by a fluid to a motor is given by the following equation: \[ P_R = h_R \times \text{weight flow rate} = h_R \gamma Q \] **Efficiency Equation:** Efficiency (\( e_M \)) is given by the following equation: \[ e_M = P_O / P_R \] **Hint:** For water at 100°C, \(\gamma = 9.81 \text{kN/m}^3\). **Check Below Answer:** Power Delivered: 0.29256 kW **Conversion Question:** What is this in horsepower? **Diagram Explanation:** The diagram depicts a fluid motor system, showing the flow of water from point A to point B through a steel tube. The dimensions are provided for two sections of the tube, with an outer diameter of 25 mm and a wall thickness of 2.0 mm at the inlet (A), and an outer diameter of 80 mm with a wall thickness of 2.8 mm at the outlet (B). The system's height difference between these points is 1.8 meters.
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