Question

At steady state, a device receives a stream of saturated water vapor at \(210^{\circ} \mathrm{C}\) and discharges a condensate stream at \(20^{\circ} \mathrm{C}, 0.1 \mathrm{MPa}\) while delivering energy by heat transfer at \(300^{\circ} \mathrm{C}\). The only other energy transfer involves heat transfer at \(20^{\circ} \mathrm{C}\) to the surroundings. Kinetic and potential energy changes are negligible. What is the maximum theoretical amount of energy, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of steam entering, that could be delivered at \(300^{\circ} \mathrm{C} ?\)

Short answer

1335.32 \text{kJ/kg} is the maximum theoretical energy.

Step-by-step solution

Identify the given data

Temperature of steam entering: \( T_1 = 210^{\c} \). Temperature of condensate stream: \( T_2 = 20^{\circ} \). Temperature of heat delivery: \( T_H = 300^{\circ} \). Temperature of heat reject: \( T_L = 20^{\circ} \). Specific heat capacity values will be required for the calculations.

Define the energy equation

At steady state, the energy balance equation for the device can be written as: \[ Q_h - Q_c = m (h_1 - h_2) \] Here, \( Q_h \) is the heat delivered at \( T_H \, Q_c \) is the heat rejected, \( m \) is the mass of the steam, \( h_1 \) is the enthalpy of the steam entering, and \( h_2 \) is the enthalpy of the condensate stream.

Calculate specific enthalpy values

From steam tables, find the specific enthalpy of water vapor at \(210^{\circ} \text{C}\): \( h_1 \). And condensate at \(20^{\circ} \, 0.1 \text{MPa}\): \( h_2 \). \( h_1 \) for saturated steam at \(210^{\circ} \) is approximately 2820.4 \text{kJ/kg} \ \( h_2 \) for liquid at \(20^{\circ} \, 0.1 \text{MPa}\) is approximately 84.1 \text{kJ/kg} \

Determine the maximum theoretical energy delivered

To find the maximum energy delivered, assume that all the heat input is converted into useful energy without losses: \[ Q_h = m (h_1 - h_2) \] By substituting the values: \[ Q_h = 1 \, \text{kg} \times (2820.4 \, \text{kJ/kg} \ - 84.1 \, \text{kJ/kg}) \] So, \[ Q_h = 2736.3 \text{kJ/kg} \]

Compute the thermal efficiency

For maximum theoretical energy delivery at \( 300^{\circ} \text{C} \): We use the Carnot efficiency formula: \[ \eta = 1 - \frac{T_L}{T_H} \] Convert temperatures to Kelvin (\text{K}): \[ T_H = 300 + 273 = 573 \text{K} \] \[ T_L = 20 + 273 = 293 \text{K} \] Now, \[ \eta = 1 - \frac{293}{573} = 0.488 \]

Calculate the maximum theoretical energy delivered at \(300^{\circ} (C)\text{C} \)

Using Carnot efficiency, \ \ \[ W_{max} = \eta \times Q_h = 0.488 \times 2736.3 \] Simplifying: \[ W_{max} \approx 1335.32 \text{kJ/kg} \]

Key concepts

steady state

When a system is in a steady state, its properties do not change over time. In other words, the energy entering the system is equal to the energy leaving it. This balance is crucial when analyzing thermodynamic systems.

In our exercise, the device receives steam and discharges a condensate stream. The temperatures and properties of these streams remain constant, which means the system is in a steady state.

At steady state, the energy equation can be simplified as no accumulation occurs in the system. This means we can accurately use the energy balance equation: \[ Q_h - Q_c = m (h_1 - h_2) \] where all variables are defined consistently over time.

enthalpy calculations

Enthalpy is a measure of the total energy of a thermodynamic system, including internal energy and energy required to displace its surroundings. In our problem, we need to calculate the specific enthalpy values for both the incoming and outgoing streams.

From steam tables, the specific enthalpy of water vapor at 210°C (saturated steam) is found to be approximately 2820.4 \text{kJ/kg}. For the condensate at 20°C and 0.1 MPa, the specific enthalpy is approximately 84.1 \text{kJ/kg}.

Using these values in the energy equation: \[ Q_h = m (h_1 - h_2) \] we find: \[ Q_h = 1 \text{kg} \times (2820.4 \text{kJ/kg} - 84.1 \text{kJ/kg}) = 2736.3 \text{kJ/kg} \] This computation tells us the amount of energy delivered by heat transfer.

Carnot efficiency

Carnot efficiency represents the maximum theoretical efficiency that a heat engine can achieve, operating between two temperatures. It is given by the formula: \[ \text{Efficiency} = 1 - \frac{T_L}{T_H} \] where temperatures must be in Kelvin.

For our system, the high temperature (\text{T}_H) is 300°C, and the low temperature (\text{T}_L) is 20°C. Converting those to Kelvin: \[ T_H = 300 + 273 = 573 \text{K} \] \[ T_L = 20 + 273 = 293 \text{K} \] Using these values, we calculate the Carnot efficiency: \[ \text{Efficiency} = 1 - \frac{293}{573} \] Simplifying: \[ \text{Efficiency} = 0.488 \] This efficiency shows the fraction of heat energy that can theoretically be converted into work.

heat transfer

Heat transfer is the process by which thermal energy is exchanged between physical systems. In this exercise, there are two key heat transfers: heat delivered at 300°C and heat rejected at 20°C.

The energy delivered, represented by \(Q_h\), is calculated as approximately 2736.3 \text{kJ/kg} using the enthalpy values. This energy is the input for our system.

Using the calculated Carnot efficiency of 0.488, the maximum theoretical energy delivered, or work \(W_{max}\) at 300°C becomes: \[ W_{max} = \text{Efficiency} \times Q_h \] \[ W_{max} = 0.488 \times 2736.3 \text{kJ/kg} = 1335.32 \text{kJ/kg} \] This result shows the maximum amount of energy extractable through heat transfer at the given conditions.