A.2 An open tube is attached to a tank as shown in Fig. A.2....

A.2 An open tube is attached to a tank as shown in Fig. A.2. If the water rises to a height of 900 mm in the tube, determine the pressures of the air pA and pB above the water. Neglect the capillary effects in the tube. pA A 200mm water pB B 900mm 400mm Fig. A.2 A.3 A culvert is covered by a 1 x 1 m² gate. The gate is pivoted at the bottom as shown in Fig. A.3. Determine the depth of water h above the top of the culvert so that the gate will start to open clockwise to overcome the torque of 50 kNm at the pivot. water h square water gate Pivot Open to air Fig. A.3

Transcript

00:01 In this question we have given an open tube that is attached to a tank it is given that if the water rises to a height of 900 mm in the tube then we have to determine the pressure of the air pa and p b above the water so we have given that this is the water level in this tank this is the water and we have given that the water is rises up to a height of 900 mm so we have given that this height is 900 mm we have given that this height is 400 mm and this is over in this this is the level of the water so we have given that at this point the pressure is a and this is the point a we have given this height is 600 mm and this point the point is p b and at this point is b and this is a a dash and this point is b -b -b -dash and we can say that at the top at this point there is a pressure acting that is atmospheric pressure balancing pressure across section a a dash we can say that the pressure at a is equals to this is atmospheric pressure plus pressure due to the column so we can say that this is atmospheric pressure plus thousand into g into this height is 900 minus 600 so this total height minus this height so this is the pressure at a and we can say that the atmospheric pressure is 1 .01 into 10 to the power 5 into sorry this is plus into thousand into this is 9 .8 into this is 0 .3 and we can say that this will come out 1 .04 into 10 to the power 5 pascal so this is the pressure at a similarly we can find the pressure at b by balancing the pressure at this section because the pressure on the horizontal plane remains same at every point so you can say that balancing pressure across section b b dash so we can say that the pressure at b is equal to this is atmospheric pressure plus rog into h dash or you can say that this is atmospheric pressure plus this is 1 ,000 into 9 .8 into the height difference is this is 900 minus 400 into 10 to the power minus 3 in meter.

03:20 So this is 1 .01 into 10 to the power 5 plus 5 ,000 and we get this is 1 .06 into 10 to the power 5 pascal...

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