Question

Steam at \(0.7 \mathrm{MPa}, 355^{\circ} \mathrm{C}\) enters an open feedwater heater operating at steady state. A separate stream of liquid water enters at \(0.7 \mathrm{MPa}, 35^{\circ} \mathrm{C}\). A single mixed stream exits as saturated liquid at pressure \(p\). Heat transfer with the surroundings and kinetic and potential energy effects can be ignored. (a) If \(p=0.7 \mathrm{MPa}\), determine the ratio of the mass flow rates of the incoming streams and the rate at which entropy is produced within the feedwater heater, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of liquid exiting. (b) Plot the quantities of part (a), each versus pressure \(p\) ranging from \(0.6\) to \(0.7 \mathrm{MPa}\).

Short answer

Find the enthalpies and entropies for all streams, then apply energy and entropy balances. Calculate the ratio \( \frac{m_1}{m_2} \) and entropy production rate. Plot these against pressure \( p \).

Step-by-step solution

Identify Given Parameters

For steam: Pressure, \( p = 0.7 \, \text{MPa}\) Temperature, \( T = 355 \, ^\circ\text{C} \)For liquid water: Pressure, \( p = 0.7 \, \text{MPa} \)Temperature, \( T = 35 \, ^\circ\text{C} \)Exit stream: Pressure, \( p = 0.7 \, \text{MPa} \)Saturated liquid condition.

Determine Properties of Inlet Streams

Using steam tables:For steam, \( h_1 \) is the specific enthalpy and \( s_1 \) is the specific entropy at \( 0.7 \, \text{MPa} \) and \( 355 \, ^\circ\text{C} \). For liquid water, \( h_2 \) is the specific enthalpy and \( s_2 \) is the specific entropy at \( 0.7 \, \text{MPa} \) and \( 35 \, ^\circ\text{C} \).

Determine Properties of Exit Stream

For the exit stream which is a saturated liquid at \( 0.7 \, \text{MPa} \):Using steam tables: \( h_e \) and \( s_e \) are the specific enthalpy and specific entropy of saturated liquid at \( 0.7 \, \text{MPa} \).

Apply Mass and Energy Balance

The feedwater heater is in steady state condition and heat transfer with surroundings is ignored.From energy balance:\[ m_1 h_1 + m_2 h_2 = (m_1 + m_2) h_e \]Solve for the ratio of mass flow rates \( \frac{m_1}{m_2} \):\[ \frac{m_1}{m_2} = \frac{h_e - h_2}{h_1 - h_e} \]

Apply Entropy Balance

Since kinetic and potential energy effects are negligible and heat transfer with surroundings is ignored, the entropy balance gives us:\[ m_1 s_1 + m_2 s_2 + S_{gen} = (m_1 + m_2) s_e \]\( S_{gen} \) is the rate of entropy production.Rearrange to solve for \( \frac{S_{gen}}{m_2} \):\[ \frac{S_{gen}}{m_2} = (m_1 + m_2) s_e - m_1 s_1 - m_2 s_2 \]

Calculate Numerical Values

Use the steam tables to find numerical values for \( h_1, h_2, h_e, s_1, s_2, \) and \( s_e \). Substitute these into the formula to find the ratio \( \frac{m_1}{m_2} \) and \( \frac{S_{gen}}{m_2} \) numerically.

Plot Results for Part (b)

Vary the pressure \(p\) from \(0.6 \, \text{MPa}\) to \(0.7 \, \text{MPa}\), repeating steps 2 through 6 for each value. Plot \( \frac{m_1}{m_2} \) and \( \frac{S_{gen}}{m_2} \) against \(p\).

Key concepts

Mass Flow Rate Ratio

In thermodynamic systems like an open feedwater heater, understanding the mass flow rate ratio is crucial. This ratio, often denoted as \( \frac{m_1}{m_2} \), represents the proportion of the mass flow rate of steam to the mass flow rate of liquid water entering the heater. The energy balance equation simplifies to: \[ \frac{m_1}{m_2} = \frac{h_e - h_2}{h_1 - h_e} \] where \(h_1, h_2, \) and \(h_e\) are the specific enthalpies of the steam, liquid water, and the mixed exit stream, respectively. Knowing this ratio helps in determining the performance and efficiency of the feedwater heater and is essential in designing and optimizing thermal systems.

Entropy Production

Entropy production is a measure of the irreversibilities or losses in a thermodynamic system. For an open feedwater heater, it is determined using the entropy balance equation: \[ m_1 s_1 + m_2 s_2 + S_{gen} = (m_1 + m_2) s_e \] Rearranging, we find, \[ \frac{S_{gen}}{m_2} = (\frac{m_1}{m_2} + 1)(s_e) - \frac{m_1}{m_2} s_1 - s_2 \] where \(s_1, s_2,\) and \(s_e\) are the specific entropies of the steam, liquid water, and the exit stream, respectively. \[ S_{gen}\big) \] quantifies the entropy generated within the system. Minimizing entropy production is key for achieving higher energy efficiency.

Steady State Energy Balance

Under steady-state conditions, the properties of the streams entering and leaving the feedwater heater remain constant over time. This is expressed through the steady-state energy balance equation: \[ m_1 h_1 + m_2 h_2 = (m_1 + m_2) h_e\] where \(h_1\) and \(h_2\) are the specific enthalpies of the incoming steam and liquid water, and \(h_e\) is the specific enthalpy of the exit stream. This balance ensures that the energy from the entering streams is accounted for in the exit stream, ensuring the system is in equilibrium and helps in accurately calculating the mass flow rate ratio.

Steam Tables

Steam tables are crucial tools for determining thermodynamic properties like enthalpy (\( h \)) and entropy (\( s \)). For the given problem, steam tables help find these properties at specific pressures and temperatures. For example, at \(0.7 \text{MPa}\) and \(355 \degree \text{C}\), steam tables provide the values of \( h_1 \) and \( s_1 \). For liquid water at \(35 \degree \text{C}\), they provide \( h_2 \) and \( s_2\). Accurate data from steam tables is vital for performing energy and entropy calculations, leading to precise and reliable results.

Thermodynamic Properties

Thermodynamic properties such as enthalpy (\( h \)), entropy (\( s \)), pressure (\( P \)), and temperature (\( T \)) are fundamental in analyzing thermal systems. Enthalpy and entropy are especially important in steam cycles. Enthalpy changes indicate heat added or removed, while entropy changes reveal the system's disorder and efficiency. By understanding these properties, one can predict how fluids behave under different conditions, optimize system performance, and minimize energy losses. Accurate knowledge of these properties aids in designing efficient and effective thermal systems.