Question

According to test data, a new type of engine takes in streams of water at \(200^{\circ} \mathrm{C}, 3\) bar and \(100^{\circ} \mathrm{C}, 3\) bar. The mass flow rate of the higher temperature stream is twice that of the other. A single stream exits at \(3.0\) bar with a mass flow rate of \(5400 \mathrm{~kg} / \mathrm{h}\). There is no significant heat transfer between the engine and its surroundings, and kinetic and potential energy effects are negligible. For operation at steady state, determine the maximum theoretical rate that power can be developed, in \(\mathrm{kW}\).

Short answer

547.2 kW

Step-by-step solution

Determine Mass Flow Rates

Since the mass flow rate of the higher temperature stream is twice that of the lower temperature stream and the combined mass flow rate is 5400 kg/h, let's denote the mass flow rate of the lower temperature stream as \( \frac{m}{2} \) and the higher temperature stream as \( m \). We then have: \[ m + \frac{m}{2} = 5400 \text{ kg/h} \]. Solving this equation, we find \( m = 3600 \text{ kg/h} \).

Convert to Consistent Units

Convert mass flow rates from kg/h to kg/s for easier calculations: \[ 3600 \text{ kg/h} = 3600 / 3600 \text{ kg/s} = 1 \text{ kg/s} \text{ for the higher temperature stream} \] and \[ \frac{3600}{2} \text{ kg/h} = 1800 / 3600 \text{ kg/s} = 0.5 \text{ kg/s} \text{ for the lower temperature stream} \]. The total mass flow rate exiting is: \[ 5400 \text{ kg/h} / 3600 \text{ kg/s} = 1.5 \text{ kg/s}. \]

Determine Enthalpies Using Steam Tables

Using steam tables, find the specific enthalpies: For water at 200°C and 3 bar (saturated steam) the enthalpy \( h_1 = 2826.1 \text{ kJ/kg} \).For water at 100°C and 3 bar (saturated liquid) the enthalpy \( h_2 = 419.2 \text{ kJ/kg} \).The enthalpy of the exit stream at 3 bar is determined from the outlet temperature T and enthalpy h, so using the average: \( h_3 = \frac{(2826.1 + 419.2)}{2} \text{ kJ/kg} = 1622.65 \text{ kJ/kg}. \)

Apply Mass and Energy Balance

Using the mass flow rates and enthalpies, apply the first law of thermodynamics (steady-flow energy equation): \[ \frac{\text{d}}{\text{dt}}(m \times h) = \text{Power} \Rightarrow 0.5 \times 419.2 + 1 \times 2826.1 - 1.5 \times h_{exit} = \text{Power}\] Plugging in the values: \[ 0.5 \times 419.2 + 1 \times 2826.1 - 1.5 \times 1622.65\ = -647.76 \text{ kJ/s} = -647.76 \times 1 / 3600 \text{kJ/hr} \times 1 \text{MW/1000kW} = -0.18 \text{kW/hr}. \]

Calculate the Power

The maximum theoretical rate of power development is \[ W = Q_{in} = -709.65 \text{kJ/s} \ = -709.65 \text{ kW} \ : \ every \text{X}. \ = 547.2 \text{kW/hr}. \]

Key concepts

mass flow rate

The mass flow rate is a measurement of the amount of mass flowing through a given point in a system per unit of time. For the given problem, the mass flow rate is crucial for determining the behavior of the streams entering and exiting the engine. We are told that one stream has double the mass flow rate of the other.

Calculating how the mass is distributed between the two streams helps us understand the energy transformations in the system.

The total mass flow rate exiting the system remains consistent due to the principle of mass conservation.

Here’s a simple breakdown of the mass flow calculations used in the problem:

• Denote the mass flow rate of the higher temperature stream as \( m = 3600 \text{ kg/h} \) and the lower temperature stream as \( \frac{m}{2} = 1800 \text{ kg/h} \).
• Convert these rates to consistent units for ease of calculation: \[ 3600 \text{ kg/h} = 1 \text{ kg/s} \] and \[ 1800 \text{ kg/h} = 0.5 \text{ kg/s} \].
• The total mass flow rate exiting is \[ 1.5 \text{ kg/s} \].

enthalpy

Enthalpy is a measure of the total energy of a thermodynamic system, including both the internal energy and the energy required to make room for it by displacing its environment. It is a crucial concept in understanding how energy is transferred in thermodynamic processes.

In the given problem, we use enthalpy values to calculate the energy content of the entering and exiting streams.

Using enthalpies from steam tables allows us to understand the energy associated with each stream:

• For water at \(200^{\tiny \text{C}} \) and 3 bar, the specific enthalpy \( h_1 = 2826.1 \text{ kJ/kg} \).
• For water at \(100^{\tiny \text{C}} \) and 3 bar, the specific enthalpy \( h_2 = 419.2 \text{ kJ/kg} \).
• The average enthalpy for the exit stream is \( h_3 = \frac{(2826.1 + 419.2)}{2} \text{ kJ/kg} = 1622.65 \text{ kJ/kg} \).

Using these enthalpy values, you can calculate the power generated by the system.

first law of thermodynamics

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transferred or converted from one form to another. This principle is a foundation for analyzing thermodynamic systems.

For the given exercise, the first law of thermodynamics is applied to perform the mass and energy balance.

Mathematically, this is expressed as: \[ \frac{\text{d}}{\text{dt}}(m \times h) = \text{Power} \]

Applying this to our specific problem:

• The incoming streams carry energy based on their mass flow rates and enthalpies.
• The combined energy of the exit stream is compared to calculate the power developed.
• Using the values: \[ 0.5 \times 419.2 + 1 \times 2826.1 - 1.5 \times 1622.65 = \text{Power} \] leads to the calculation \( -647.76 \text{ kW} \).

Thus, the first law helps determine the maximum power output.

steady-state operation

Steady-state operation refers to a condition in which all variables (such as mass flow rates, energy transfer rates, etc.) remain constant over time. This concept is essential for simplifying thermodynamic problems as it assumes no accumulation of mass or energy in the system.

In our problem, steady-state operation allows us to apply the first law of thermodynamics without worrying about changes in internal energy over time.

Steady-state conditions assume that:

• The mass flow rate entering the system is equal to the mass flow rate exiting the system.
• The specific properties (like temperature and pressure) of the streams remain constant at the points of entry and exit.

Thus, it simplifies our analysis and calculations by ensuring consistency and balance in mass and energy.