\( 2.2 .3 \quad \) Work done by the ram after 10 pumping strokes 2.3 The crankshaft of a single-acting 3-cylinder water pump has plungers each having a diameter of 200 mm and a stroke length of 600 mm runs at \( 800 \mathrm{r} / \mathrm{min} \). Calculate the following: \( \begin{array}{ll}2.3 .1 & \text { The volume of water delivered in } \mathrm{k} \ell \mathrm{h} \text { if there is a slip of } 5 \% \\ 2.3 .2 & \begin{array}{l}\text { The input power to drive the pump if the efficiency is } 90 \% \text { and the } \\ \text { static head is } 32 \mathrm{~m}\end{array} \\ \text { HINT: Use the density for water as } 1000 \mathrm{~kg} / \mathrm{m}^{3} & \text { (3) }\end{array} \) \( \begin{array}{ll}\text { H) }\end{array} \)

Sure! Let's dive into some engaging insights related to the problem at hand! Picture this: the crankshaft of that water pump is a bit like a bicep at the gym, tirelessly pumping up the water! The volume of water delivered is key. Each stroke of the plunger pushes water forward, and with three plungers, that volume adds up quickly! But wait, with a 5% slip, we need to account for that in our final calculations! So, don’t forget to do the math for volume first: it’s all about calculating the displacement and adjusting it for efficiency! Now, on to the input power! How cool is it to think about all that energy transfer? Remember, the efficiency of the pump is like a superhero’s power level—90% is strong, but there’s always some energy lost along the way! The static head represents the height the pump needs to fight against, much like climbing a steep hill. Plug in those values, and you'll see just how power is converted into making things happen—water flowing uphill! Make sure to check those units and do the calculations; it’s all part of the magic!