2.2. Pressure at Suction Side of the Pump

Part B will be solved by energy equation between reservoir A and pump B. So let Ps be the pressure at suction side of the pump, then we put the Bernoullis equation 95 + 0 + 0 will be equal to, the pump is at z = 100 and if the pressure Ps/gamma + V1 square/2g + losses in suction pipe.

In this chunk, we start by setting up the energy equation to analyze the pressure at the pump's suction side. We designate the pressure at the suction side as Ps. We apply the Bernoulli's equation between two points in the hydraulic system: reservoir A and pump B. In the equation, we consider the height of the reservoir (95 meters), the potential energy due to the height at the pump (z), and contributions from various energy components such as velocity head, pressure head, and energy losses in the suction pipe.

Imagine a water slide. When you stand at the top of the slide (similar to our reservoir), you have potential energy due to your height. However, as you slide down, you lose some of that energy to friction and other factors (which represent our energy losses). To find out how fast you’ll be moving at the bottom (like finding the pressure in our pump), we have to calculate the energy transformations that happen on the way down.

This will be 100 + Ps/gamma and V1 square/2g is going to be 0.065, because V1 we have already found out, plus losses in the suction pipe is 0.174 major and the minor losses 0.033.

Here, we continue the calculation by expressing the total head at the pump as a combination of the static head (height of the pump), the pressure head (Ps/gamma), the kinetic energy due to velocity (V1 square/2g), and the head loss in the suction pipe due to both major and minor losses. We substitute values to compute the pressure at the suction side of the pump more accurately.

Think of a water pump drawing water from a pool. The water from different heights (static head) and the speed at which it comes through the pipe (velocity head) combine to determine how hard the pump has to work to maintain a good flow. Any obstacles (losses) that slow the water down will mean the pump needs to exert more energy to compensate.

So Ps/gamma can be found out to be – 5.272 meter. This is gauge. So the real pressure is going to be 10 – 5.272, because we have already been told the absolute is 10, so it is going to be Ps = 4.728 multiplied by 9.79 to get it into form of kilopascal, so 46.29 kilopascal absolute.

By rearranging our earlier equation and solving for Ps/gamma, we find that the pressure head is negative, indicating a gauge pressure. We convert this gauge pressure into real pressure by adding atmospheric pressure (10 meters of water specifically given). Finally, by multiplying by the specific weight (9.79 kN/m³), we convert units from meters to kilopascals for practical use in engineering contexts.

This last calculation is like determining the actual water pressure you feel when you dive into a pool. Above you, the atmosphere presses down on you and the water pushes back. To find out how much 'push' you actually feel (your real pressure), you consider both the atmosphere above the pool and how deep you are in the water.

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