Question

Nitrogen \(\left(\mathrm{N}_{2}\right)\) enters a compressor operating at steady state at \(1.5 \mathrm{MPa}, 300 \mathrm{~K}\) and exits at \(8 \mathrm{MPa}, 500 \mathrm{~K}\). If the work input is \(240 \mathrm{~kJ}\) per \(\mathrm{kg}\) of nitrogen flowing, determine the heat transfer, in kJ per \(\mathrm{kg}\) of nitrogen flowing. Ignore kinetic and potential energy effects.

Short answer

Heat transfer is \( 447.8 \mathrm{~kJ/kg} \).

Step-by-step solution

Identify known values

Given: Initial pressure, \(P_1 = 1.5 \mathrm{~MPa}\)Initial temperature, \(T_1 = 300 \mathrm{~K}\)Final pressure, \(P_2 = 8 \mathrm{~MPa}\)Final temperature, \(T_2 = 500 \mathrm{~K}\)Work input, \(W = 240 \mathrm{kJ/kg}\)

Use the first law of thermodynamics for control volumes

For a control volume at steady state, the first law of thermodynamics states:\[ \frac{d}{dt}(\text{Energy in Control Volume}) = Q_{in} - Q_{out} + \frac{{dE_{cv}}}{dt} + W_{in} - W_{out} = 0\] Here, the changes in kinetic and potential energies are ignored, so the equation simplifies to: \[ Q - W = \frac{{dE}}{dt}\]

Apply the specific form of the energy balance

The specific energy balance can be written as:\[ Q - W = h_2 - h_1\]where:\( h_1 \) is the specific enthalpy at the inlet condition, and \( h_2 \) is the specific enthalpy at the exit condition.

Find enthalpies using tables or equation of state (ideal gas assumption)

Assuming perfect gas behavior for nitrogen, the specific enthalpy can be calculated as:\[ h = c_p T \]where \(c_p\) is the specific heat at constant pressure. For nitrogen \(c_p = 1.039 \frac{ kJ }{ kg \bullet K } \).So the enthalpies are:\[ h_1 = c_p T_1 = 1.039 \times 300 = 311.7 \mathrm{~kJ/kg} \]\[ h_2 = c_p T_2 = 1.039 \times 500 = 519.5 \mathrm{~kJ/kg} \]

Solve the energy balance equation

Substituting known values into the energy balance equation:\[ Q - 240 = 519.5 - 311.7 \]Solving for the heat transfer \(Q\):\[ Q = 519.5 - 311.7 + 240 \]\[ Q = 447.8 \mathrm{~kJ/kg} \]

Key concepts

First Law of Thermodynamics

The first law of thermodynamics is a fundamental concept for understanding energy conservation. It states that energy cannot be created or destroyed, only transferred or converted. When working with a control volume, it is represented as:

\[ \frac{d}{dt}(\text{Energy in Control Volume}) = Q_{in} - Q_{out} + \frac{dE_{cv}}{dt} + W_{in} - W_{out} = 0 \]

In simpler terms, this equation balances the heat transfer (\text{Q_{in}} and \text{Q_{out}}), work (\text{W_{in}} and \text{W_{out}}), and changes in energy (\text{dE_{cv}/dt}) of a system. For a steady-state process, the system's energy doesn't change over time, so \text{dE_{cv}/dt} = 0. This helps simplify calculations considerably.

Steady State Process

A steady-state process implies that all properties within the system do not change over time. This means that the mass, energy, and entropy within the control volume remain constant.

In our example problem, the compressor operating at steady state means that the input and output conditions are continuous and unchanging over time, allowing the balance of energy inputs and outputs efficiently.Understanding steady-state conditions helps simplify complex thermodynamic problems by removing time dependency from the equations, focusing only on the input and output values.

Enthalpy Calculations

Enthalpy (\text{h}) is the measure of energy in a thermodynamic system. It includes internal energy and the product of pressure and volume. For our purposes, it's easier to understand as the heat content per unit mass.

In this problem, enthalpy changes are used to determine heat transfer. Using the formula for specific enthalpy (\text{h}):

\[ h = c_p T \]

where \text{c_p} is the specific heat at constant pressure, and \text{T} is the temperature. For nitrogen, \text{c_p} is given as 1.039 kJ/kg·K.
Hence, we calculate:

• \text{h_1} = 1.039 × 300 = 311.7 kJ/kg
• \text{h_2} = 1.039 × 500 = 519.5 kJ/kg

The specific enthalpy change (\text{∆h}) helps in solving the overall energy balance equation.

Ideal Gas Behavior

In thermodynamics, ideal gas behavior is an assumption made to simplify calculations. An ideal gas follows the equation:

\[ PV = nRT \]
where \text{P} is pressure, \text{V} is volume, \text{n} is the number of moles of the gas, \text{R} is the universal gas constant, and \text{T} is the temperature.

In our exercise, assuming nitrogen behaves as an ideal gas allows us to use the simple relationship \text{h = c_p T} for calculating enthalpies, avoiding complex real-gas equations.

Ideal gas assumptions work well for many gases under a range of conditions, simplifying the process of thermodynamic calculations and making results more approachable.