Below figure shows a part of the production line of raspberry juice in F&B company. Initially,...
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Below figure shows a part of the production line of raspberry juice in F&B company. Initially, the production begins with the screw pressing of raspberies and topping of water to form the diluted raspberry juice (p = 1030 kg/m). The diluted raspberry juice is then split into three different streams (stream 2, 3, 4) to be stored in 3 different storage tanks. It is known that the gauge pressure inside the storage tank for raspberry juice is 63.78 kPa. Case 1 O P. =? V6 =? Z6 = ? gauge = 63. 78 kPa V = ? V = 11.39 m/s Di = 1.33 m V2 = 0.31 V, V2 = ? D2 = 0.78 m O P, =? Case 2 3.89 m Pg =? Ve = ? Ze =? Pa = ? O P, =?261 m Vs = ? 25 = ? D3 = 0.31D, + 0.57D, (3) V3 = ? Vg = ? V = 0.29 V, V = 8.39 m/s D, = ? Z9 =? V3 = ? (a) Compute the volumetric flow rates (V,, Vz, V3, and V,), diameters (Dz and D4), and velocity (v2 and v3). [8 marks] (b) Calculate the hydrostatic pressure (gauge) at points 5, 7, and 8. [4 marks] (c) If the pipe of the stream 3 rupture at the point 5, determine the maximum height that the raspberry juice can shoot up to. Make assumptions/analysis to simplify the calculation. [Hint: Use point 5 and 6 as the system] [4 marks] (d) If the old storage tank has a rupture point (with very large diameter hole) at the point 8, compute the discharge velocity of the raspberry juice as it leaves from point 8. Utilize some assumptions/analysis to ease the calculation. [Hint: Use point 8 and 9 as the system] [4 marks) Below figure shows a part of the production line of raspberry juice in F&B company. Initially, the production begins with the screw pressing of raspberies and topping of water to form the diluted raspberry juice (p = 1030 kg/m). The diluted raspberry juice is then split into three different streams (stream 2, 3, 4) to be stored in 3 different storage tanks. It is known that the gauge pressure inside the storage tank for raspberry juice is 63.78 kPa. Case 1 O P. =? V6 =? Z6 = ? gauge = 63. 78 kPa V = ? V = 11.39 m/s Di = 1.33 m V2 = 0.31 V, V2 = ? D2 = 0.78 m O P, =? Case 2 3.89 m Pg =? Ve = ? Ze =? Pa = ? O P, =?261 m Vs = ? 25 = ? D3 = 0.31D, + 0.57D, (3) V3 = ? Vg = ? V = 0.29 V, V = 8.39 m/s D, = ? Z9 =? V3 = ? (a) Compute the volumetric flow rates (V,, Vz, V3, and V,), diameters (Dz and D4), and velocity (v2 and v3). [8 marks] (b) Calculate the hydrostatic pressure (gauge) at points 5, 7, and 8. [4 marks] (c) If the pipe of the stream 3 rupture at the point 5, determine the maximum height that the raspberry juice can shoot up to. Make assumptions/analysis to simplify the calculation. [Hint: Use point 5 and 6 as the system] [4 marks] (d) If the old storage tank has a rupture point (with very large diameter hole) at the point 8, compute the discharge velocity of the raspberry juice as it leaves from point 8. Utilize some assumptions/analysis to ease the calculation. [Hint: Use point 8 and 9 as the system] [4 marks)
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