2.) The figure shows the schematic arrangement of a • differential level route from BM1 TP; to BM2. TP1 a.) Arrange the diagram into tabular form 10875 7035 $821 b.) Graph the profile of the ground c.) Determine the difference in elevation between TP1 & TP2. BM, EL 33.971 TP3 d.) Determine the difference in BM, elevation between the highest 7977 9736 3 560 7.186 and lowest point. e.) Determine the difference in elevation between BM1 and BM2. 5.741

A vertical summit curve has tangents of +4% and -2%. The allowable rate of change of grade is 0.3% per meter station. Stationing and elevation of P.T. ia at 10 + 020 and 143.63 m. respectively. Calculate the following: a. Length of the curve b. distance of the highest point of the curve from PC c. elevation of the highest point of the curve

(read image)

Sanitary Engineering

5. Water at 15°C (p = 999.1 kg/m³ and µ = 1.138 × 10¯³ kg/m-s) is flowing steadily in a 30-m-long and 6-cm-diameter horizontal pipe made of stainless steel at a rate of 10 L/s. Determine (a) the pressure drop, (b) the head loss, and (c) the pumping power requirement to overcome this pressure drop. Assume the entrance effects are negligible, and thus the flow is fully developed. Take the roughness of stainless steel as 0.002 mm. 10 L/s 30 m 6 cm

T=10.38 kNm

1. Find the diff. in elevation between TP1and TP3 2. Find the elevation of BM2 3. What is the difference in elevation between BM, and BM2? STA. BM BM, TPL TPH TP L ТРН TP L TPa-H BM2 BM B.S. ELEV. 749.06 F.S. 9.08 9.08 3.73 1.60 2.21 1.08 12.24 10.10 11.04 9.92 1.75 0.55 9.84 8.62 11.27 11.27

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A float to be used as a level indicator is being designed to float in oil, which has a specific gravity of 0.90. It is to be a cube 100 mm on a side, and is to have 75 mm submerged in the oil. Calculate the required specific weight of the float material.

A.) CREATE A SHEAR DIAGRAM BY AREA MOMENT AND BY PARTS B.) CREATE A M/EI DIAGRAM BY AREA MOMENT AND BY PARTS ITEM 1.0 180 kN 15 kN/m A В C- D -5 m- -5 m- -4 m- El = constant E = 70 GPa %3D I = 2,340 (10°) mm4 %3D