Saturated water at 25C enters a tank through a 10-cm diameter pipe at a velocity of 3 m/s. At the bottom of the tank, water exits at 6 m/s through a 5.25- cm diameter pipe. The depth of the water in the tank is 1.5 m. The tank has a diameter of 1.25 m and a height of 2.5 m. Water in the tank is also magically kept at 25C, which is the same to the surrounding temperature. Assume that pressures at inlet, outlet, and the surrounding are roughly the same and pressure effect is negligible.

In this scenario, we can use the principle of conservation of mass to analyze the flow of water into and out of the tank. Since the system is at steady state (the depth of water doesn't change), the mass flow rate of water entering the tank must equal the mass flow rate of water exiting the tank.

The mass flow rate (\( \dot{m} \)) can be calculated using the formula:

\[ \dot{m} = \rho A V \]

Where: - \( \rho \) is the density of water (constant at 25°C), - \( A \) is the cross-sectional area of the pipe, and - \( V \) is the velocity of the water.

Let's calculate the mass flow rates for both the inlet and outlet:

For the 10-cm diameter inlet pipe: - \( A_{inlet} = \pi d_{inlet}^2 / 4 = \pi (0.1 \, \text{m})^2 / 4 \) - \( V_{inlet} = 3 \, \text{m/s} \)

For the 5.25-cm diameter outlet pipe: - \( A_{outlet} = \pi d_{outlet}^2 / 4 = \pi (0.0525 \, \text{m})^2 / 4 \) - \( V_{outlet} = 6 \, \text{m/s} \)

Plug in the values to find the respective mass flow rates. For water at 25°C, the density \( \rho \) is approximately \( 997 \, \text{kg/m}^3 \).

\[ \dot{m}_{inlet} = \rho A_{inlet} V_{inlet} = 997 \times (\pi (0.1)^2 / 4) \times 3 \] \[ \dot{m}_{outlet} = \rho A_{outlet} V_{outlet} = 997 \times (\pi (0.0525)^2 / 4) \times 6 \]

Now we can set the mass flow rates equal to each other because the mass entering and leaving the system per unit of time must be the same.

\[ \dot{m}_{inlet} = \dot{m}_{outlet} \] \[ 997 \times (\pi (0.1)^2 / 4) \times 3 = 997 \times (\pi (0.0525)^2 / 4) \times 6 \]

By solving this equation, you can confirm that the mass flow rate entering is equal to the mass flow rate exiting, or if not, identify the discrepancy. In this case, if the equation doesn't balance out, then the assumption that the system is at steady-state may be incorrect, or there might be a source of error or additional unaccounted for inflows or outflows.

Extra: Understanding fluid dynamics and the principle of the conservation of mass is crucial in many engineering applications, especially in fields such as civil, mechanical, and chemical engineering. This principle asserts that the mass of a closed system must remain constant over time. This means that the amount of mass that enters a system must either remain within the system or exit it; mass is neither created nor destroyed.

In the given problem, it was assumed that the water temperature remains constant, which suggests that the water's density remains constant as well. This simplification eliminates the need to account for changes in density due to temperature, which can occur in real-world scenarios. It also assumes that pressures at various points are roughly the same and hence that the pressure effect can be neglected, which would otherwise require application of the Bernoulli equation to include pressure, kinetic energy, and potential energy terms for incompressible fluid flow analysis. Additionally, neglecting pressure changes implies that we don't consider any energy changes due to work being done on or by the fluid, and we focus solely on the mass flow rates.