Question

Liquid water enters an adiabatic piping system at \(15^{\circ} \mathrm{C}\) at a rate of \(8 \mathrm{kg} / \mathrm{s} .\) If the water temperature rises by \(0.2^{\circ} \mathrm{C}\) during flow due to friction, the rate of entropy generation in the pipe is \((a) 23 \mathrm{W} / \mathrm{K}\) \((b) 55 \mathrm{W} / \mathrm{K}\) \((c) 68 \mathrm{W} / \mathrm{K}\) \((d) 220 \mathrm{W} / \mathrm{K}\) \((e) 443 \mathrm{W} / \mathrm{K}\)

Short answer

The rate of entropy generation in the pipe is approximately 55.3 W/K, or close to 55 W/K in this context.

Step-by-step solution

Calculate initial and final specific enthalpy

Using the given temperatures, we need to find the initial and final specific enthalpy of the water. We have: Initial temperature: \(T_1 = 15^{\circ}\mathrm{C} = 288.15\ \mathrm{K}\) Final temperature: \(T_2 = 15.2^{\circ}\mathrm{C} = 288.35\ \mathrm{K}\) Assume water as an incompressible substance; thus, from the enthalpy-temperature equation for an incompressible substance, we have \(h_2 - h_1 = c(T_2 - T_1)\), where \(c\) (specific heat capacity) is taken as a constant value of \(4.18\ \mathrm{kJ/kg\cdot K}\) for liquid water.

Determine the heat transfer rate

Given the mass flow rate \(m = 8\ \mathrm{kg/s}\), we can now calculate the heat transfer rate, \(Q_{in}\). Since the process is adiabatic (no heat loss to surroundings), we know that \(Q_{in} = m\Delta h = m(h_2 - h_1)\). Substituting the enthalpy difference from step 1, we get: \(Q_{in} = m\cdot c\cdot(T_2 - T_1) = 8\ \mathrm{kg/s} \cdot 4.18\ \mathrm{kJ/kg\cdot K} \cdot (0.2\ \mathrm{K}) = 6.688\ \mathrm{kW}\)

Calculate the rate of entropy generation

Now that we have the heat transfer rate, we can calculate the rate of entropy generation. For an incompressible substance, the rate of entropy generation is given by \(\dot{S}_{gen} = m\cdot c\cdot\ln\frac{T_2}{T_1}\). Substituting the values, we get: \(\dot{S}_{gen} = 8\ \mathrm{kg/s} \cdot 4.18\mathrm{kJ/kg\cdot K} \cdot\ln\frac{288.35\ \mathrm{K}}{288.15\ \mathrm{K}} \approx 0.0553\ \mathrm{kW/K}\) Therefore, the rate of entropy generation in the pipe is \(\dot{S}_{gen} = 55.3\ \mathrm{W/K}\). So, the correct answer is (b) \(55\ \mathrm{W/K}\).

Key concepts

Adiabatic Process

In an adiabatic process, no heat is exchanged between the system and its surroundings. This means that the process occurs within an isolated system, where the total heat remains constant. A key feature of adiabatic processes is that any change in energy within the system is purely through work done, either by or on the system.
For thermodynamic calculations, such as the one in our exercise, we assume that the system (in this case, the piping system) is perfectly insulated. Thus, any temperature change in the fluid is due to internal factors, like friction, and not due to external heat transfer.
The implications of this process are significant when considering entropy, where certain adiabatic processes can result in entropy generation due to irreversible changes like friction or viscous dissipation.

Specific Enthalpy

Specific enthalpy is a property of a substance that combines its internal energy with the product of its pressure and volume per unit mass. It is a vital concept in thermodynamics, especially in processes involving heat and work exchange.
Calculating specific enthalpy changes allows us to understand how much energy can be transferred in these forms. In the exercise, the change in specific enthalpy (\(h_2 - h_1\)>) occurs due to a change in the fluid's temperature.

• The equation for an incompressible substance: \(h_2 - h_1 = c(T_2 - T_1)\)
• Where \(c\) is the specific heat capacity, often assumed constant over small temperature ranges.

This specific enthalpy difference is critical to find, as it helps calculate the heat transfer rate, even in adiabatic processes.

Incompressible Substance

An incompressible substance is one that does not change its volume significantly when pressure changes. In practical terms, this means the density remains constant.
Liquid water is one of the common examples of such substances, as it only compresses slightly, even under high pressures. This assumption simplifies many calculations in fluid mechanics and thermodynamics, because changes in temperature alter properties without changing density.
For the exercise, assuming water as incompressible allows the use of simple equations for specific enthalpy and entropy without considering volume changes. Even minor temperature increases, such as 0.2°C, would not affect water’s density, making the analysis straightforward.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to change a unit mass of a substance's temperature by one degree Celsius (or one Kelvin). Different substances have unique heat capacities, which influence how they store and transfer energy.
For water, the specific heat capacity is taken as 4.18 kJ/kg·K, a convenient constant due to water's ability to absorb large amounts of heat compared to other substances.
In the described problem, specific heat capacity is crucial for calculating the enthalpy change due to the temperature rise. The formula: \(h_2 - h_1 = c(T_2 - T_1)\) relies on \(c\) for accurate calculation. This factor is fundamental to assess energy changes in both heating and cooling processes, providing insight into thermal dynamics during fluid flow.