(Solved): Q4. For the open system shown below the density at point 1 and 2 is \( 850 \frac{\mathrm{kg}}{\math ...

Q4. For the open system shown below the density at point 1 and 2 is \( 850 \frac{\mathrm{kg}}{\mathrm{m}^{3}} \) and the density at point 4 is \( 750 \frac{\mathrm{kg}}{\mathrm{m}^{2}} \). The used venturi tube has \( k=0.048 \mathrm{~m}^{2} \) and its \( \beta=0.667 \). The reading at DPT is \( 25000 \mathrm{~Pa} \). The diameters for point 1,3 and 4 are the same. What will be the density at point 3 if you know that \( Q_{4}=0.2771 \frac{\mathrm{m}^{2}}{\mathrm{~m}} \) " EEnergy \( y_{4}-3861 \mathrm{~Pa}= \) EEnergy \( \mathrm{y}_{3} \). \[ \begin{array}{l} P_{1}=35000 P \alpha_{r} \\ P_{1}=P_{4} \text { and } \\ h=0.5 \mathrm{~m} \text { ? } \end{array} \] (12 marks)