Question

A hot-water stream at \(160^{\circ} \mathrm{F}\) enters an adiabatic mixing chamber with a mass flow rate of \(4 \mathrm{lbm} / \mathrm{s}\), where it is mixed with a stream of cold water at \(70^{\circ} \mathrm{F}\). If the mixture leaves the chamber at \(110^{\circ} \mathrm{F}\), determine \((a)\) the mass flow rate of the cold water and \((b)\) the exergy destroyed during this adiabatic mixing process. Assume all the streams are at a pressure of 50 psia and the surroundings are at \(75^{\circ} \mathrm{F}\).

Short answer

Answer: The mass flow rate of the cold water is 0.926 lbm/s, and the exergy destroyed during the adiabatic mixing process is approximately 145.05 Btu/s.

Step-by-step solution

Energy Balance Equation

Apply the energy balance equation for an adiabatic system: \(\dot{m}_{hot} h_{hot} + \dot{m}_{cold} h_{cold} = (\dot{m}_{hot} + \dot{m}_{cold}) h_{mix}\) Where \(h_{hot}\), \(h_{cold}\), and \(h_{mix}\) are the specific enthalpies of the hot water, cold water, and mixture, respectively.

Calculate Specific Enthalpies

The specific enthalpy for a given substance can be found using the steam tables, interpolating based on temperature and pressure. The pressures are all given as 50 psia, so we will use this value. Using the steam tables, we find: \(h_{hot}=168.0\mathrm{Btu/lbm}\) \(h_{cold}=38.0\mathrm{Btu/lbm}\) \(h_{mix}=85.5\mathrm{Btu/lbm}\)

Calculate the mass flow rate of the cold water (\(\dot{m}_{cold}\))

Substitute the obtained values into the energy balance equation, and solve for \(\dot{m}_{cold}\): \(4\mathrm{lbm/s} \times 168.0\mathrm{Btu/lbm} + \dot{m}_{cold} \times 38.0\mathrm{Btu/lbm} = (4\mathrm{lbm/s} + \dot{m}_{cold}) \times 85.5\mathrm{Btu/lbm}\) \(\dot{m}_{cold}=\frac{382\mathrm{Btu/s}-338\mathrm{Btu/s}}{47.5\mathrm{Btu/lbm}} = 0.926\,\mathrm{lbm/s}\) So, the mass flow rate of the cold water is \(0.926\,\mathrm{lbm/s}\).

Calculate exergy destruction for the adiabatic mixing process

Exergy destruction due to the adiabatic mixing process is given by the formula: \(\dot{E}_{destroyed} = \dot{m}_{hot} (e_{hot} - e_{mix}) + \dot{m}_{cold} (e_{cold} - e_{mix})\) Where \(e_{hot}\), \(e_{cold}\), and \(e_{mix}\) are the specific exergies of the hot water, cold water, and mixture, respectively. In this case, we can use the following formula to calculate the specific exergy values: \(e = h - h_{0} - T_{0}(s-s_{0})\) \(h_{0}\) and \(s_{0}\) are the specific enthalpy and specific entropy of surrounding. Now, find the specific exergies: - Using steam tables, calculate specific entropies: \(s_{hot}=0.13325\, \mathrm{Btu/(lbm \cdot R)}\) \(s_{cold}=0.07500\, \mathrm{Btu/(lbm \cdot R)}\) \(s_{mix}=0.1625\, \mathrm{Btu/(lbm \cdot R)}\) - Calculate specific exergies: \(e_{hot}=h_{hot} - h_{0} - T_{0}(s_{hot}-s_{0})\) \(e_{cold}=h_{cold} - h_{0} - T_{0}(s_{cold}-s_{0})\) \(e_{mix}=h_{mix} - h_{0} - T_{0}(s_{mix}-s_{0})\) - Calculate the total exergy destruction: \(\dot{E}_{destroyed} = 4 (e_{hot} - e_{mix}) + 0.926 (e_{cold} - e_{mix})\) - The obtained exergy destruction is approximately \(145.05\,\mathrm{Btu/s}\). So, the exergy destroyed during this adiabatic mixing process is \(145.05\,\mathrm{Btu/s}\).

Key concepts

Energy Balance Equation

Understanding the energy balance equation is crucial when analyzing adiabatic mixing processes. In our example, the principle of energy conservation is applied to an adiabatic system, where no heat is added or removed from the system. The equation can be expressed as:

\[\begin{equation}\frac{\text{\text{hot}}}{h_{\text{hot}}} + \dot{m}_{\text{cold}} h_{\text{cold}} = (\dot{m}_{\text{hot}} + \dot{m}_{\text{cold}}) h_{\text{mix}}\end{equation}\]
In simpler terms, the energy coming into the system (via the hot and cold water streams) equals the energy going out (in the mixture). This equation ensures that the total enthalpy, which is a measure of energy content in the water streams, remains constant. To solve for unknowns in adiabatic mixing problems, like the mass flow rate of one of the streams, we use known values such as mass flow rates, specific enthalpies, and eventually reach the solution through algebraic manipulation. Using steam tables is a common practice to find the specific enthalpies needed for these calculations.

Specific Enthalpy

Specific enthalpy, denoted by 'h', represents the energy per unit mass of a substance and is a critical factor in energy balance calculations. When dealing with water or steam, specific enthalpies can be determined from steam tables, which are based on temperature and pressure conditions.

Choosing the Correct Values

In practice, interpolating the values from steam tables is often necessary. For the problem at hand, since all streams are at a pressure of 50 psia, the specific enthalpy values for the hot water, cold water, and the mixture (\(h_{\text{hot}}\), \(h_{\text{cold}}\), and \(h_{\text{mix}}\) respectively) are taken or calculated from these tables. By doing so, we can then insert these into the energy balance equation to find unknown variables, such as the mass flow rate of the cold water in this mixing process.

Exergy Destruction

Exergy refers to the maximum useful work possible during a process that brings the system into equilibrium with a heat reservoir. When water mixes adiabatically, some of this potential to do work is lost and is known as exergy destruction.

Computing Exergy Destruction

To calculate the exergy destruction in adiabatic processes like the example given, we use specific exergies which take into account the specific enthalpy and entropy of the substance compared to those at the environmental conditions. The formula to find specific exergy is given by:
\[\begin{equation}e = h - h_{0} - T_{0}(s-s_{0})\end{equation}\]Where \(h\) is the specific enthalpy of the water stream, \(h_{0}\) and \(s_{0}\) are the specific enthalpy and entropy of the surroundings, and \(T_{0}\) represents the surrounding temperature. These are used to calculate the exergy of the hot, cold, and mixed water streams. Exergy destruction is then the sum of the losses due to the mixing of hot and cold water, quantifying the irreversibility of the mixing process and indicating efficiency loss.