Question 3.6: A control volume of a nozzle section has surface pressures o...

• System sketch: The control volume is the outside of the nozzle, plus the cut sections (1) and (2). There would also be stresses in the cut nozzle wall at section 1, which we are neglecting here. The pressures acting on the control volume are shown in Fig. E3.6a. Figure E3.6b shows the pressures after 15 lbf/in^2 has been subtracted from all sides. Here we compute the net pressure force only.

• Assumptions: Known pressures, as shown, on all surfaces of the control volume.
• Approach: Since three surfaces have p = 15 lbf/in^2, subtract this amount everywhere so that these three sides reduce to zero “gage pressure” for convenience. This is allowable because of Eq. (3.42).

F_{UP} = \int{p_a(-n)  dA} = -p_a \int{n  dA} ≡ 0                           (3.42)

• Solution steps: For the modified pressure distribution, Fig. E3.6b, only section 1 is needed:

F_{press} = p_{gage,1} (-n)_1 A_1 = \left(25 \frac{lbf}{in^2}\right) \left[-(-i)\right] \left[\frac{\pi}{4}(3  in)^2\right] = 177i  lbf                        Ans.

• Comments: This “uniform subtraction” artifice, which is entirely legal, has greatly simplified the calculation of pressure force. Note: We were a bit too informal when multiplying pressure in lbf/in^2 times area in square inches. We achieved lbf correctly, but it would be better practice to convert all data to standard BG units. Further note: In addition to F_{press}, there are other forces involved in this flow, due to tension stresses in the cut nozzle wall and the fluid weight inside the control volume.