3 QUESTION THREE (a) Define pressure. (2 marks) (b) U-tube...

3 QUESTION THREE (a) Define pressure. (2 marks) (b) U-tube manometer containing mercury was used to find the negative pressure in the pipe, containing water. The right limb was open to the atmosphere. Determine the vacuum pressure in the pipe, if the difference of mercury level in the two limbs was 100 mm and height of water in the left limb from the centre of the pipe was found to be 40 mm below. (8 marks) (c) The figure Q3 (c) below shows an inverted differential manometer having an oil of Specific gravity 0.8 connected to two (2) different pipes carrying water under pressure. Determine the pressure in the pipe B. The pressure in pipe A is 2.0 metres of water.

Transcript

00:01 Hello students for the first part of the question we need to sketch the flow configuration so first let's find out t that is s a will be equal to using kinematic equation it is u a plus u a into t plus half into a t square so we know that here in this question u a is zero therefore it can be written as 30 is equal to 0 plus 1 by 2 into 32 .17 is the acceleration into t square now let's sketch the flow configuration so this is a sketch and let's mark the distance that is here the distance will be 1 meter and here the angle making will be 45 degree so this is a mercury and the distance between this will be y and the distance from here to this point will be let's mark it as z and from here the distance to here it will be z2 and now the distance from here to here will be z1 and let's mark this as region 1 and this as region 2 so this is a sketch of flow configuration now let's move on to the second part of the question is to find out the velocity so from the continuity equation that is a1 v1 is equal to a2 v2 from this v2 can be equal to a1 divided by a2 into v1 this can be rewritten as v2 is equal to a1 is pi by 4 d1 square divided by pi by 4 d2 square into v1 here pi by 4 and pi by 4 get cancels therefore we can write and substitute which is v2 is equal to 80 square divided by 40 square into 3 which will be equal to 120 meter per second and now from bernoulli's equation between the section 1 and 2 we can write the equation as p1 by rho g plus v1 square by 2g plus z1 is equal to p2 by rho g plus v2 square divided by 2g plus z2 let's mark this as equation 1 and now here we know that z2 minus z1 will be equal to z and here z is equal to 1 into sin of theta that is 45 degree which is equal to z gives the value 0 .7071 and now we can write the equation as p1 minus p2 divided by rho g is equal to v2 square minus v1 square divided by 2g plus z2 minus z1 from this we can write p1 minus p2 will be equal to rho g into v2 minus v1 square divided by 2g plus z2 minus z1 and now the equation can be written as p1 minus p2 is equal to on substitution it gives 1000 into 9 .81 into 12 square minus 3 square divided by 2 into 9 .81 plus 0 .7071 here we'll get the answer as 74436 .651 now moving on to the third part of the question we need to find out the height difference can be written as p1 minus p2 divided by rho g will be equal to y into the height of the mercury by that of water minus 1 let's substitute the values into this and we'll get 74436 .651 divided by 9810 equal to y into 13 .56 divided by 1 minus 1 here we'll get the value of y as y is equal to 0 .6041 meter sorry this is the fourth part of the question and here the third part of the question is this is the answer for the pressure difference that is p1 minus p2 this is the pressure difference in between the section 1 and 2...

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