Question

Air is compressed at steady state from 1 bar, \(300 \mathrm{~K}\), to 6 bar with a mass flow rate of \(4 \mathrm{~kg} / \mathrm{s}\). Each unit of mass passing from inlet to exit undergoes a process described by \(p v^{1.27}=\) constant. Heat transfer occurs at a rate of \(46.95 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing to cooling water circulating in a water jacket enclosing the compressor. If kinetic and potential energy changes of the air from inlet to exit are negligible, calculate the compressor power, in \(\mathrm{kW}\).

Short answer

The compressor power is approximately 187.68 kW.

Step-by-step solution

- Identify given data

Extract the given information: - Initial pressure, \( p_1 = 1 \text{ bar} \) - Initial temperature, \( T_1 = 300 \text{ K} \) - Final pressure, \( p_2 = 6 \text{ bar} \) - Mass flow rate, \( \frac{dm}{dt} = 4 \text{ kg/s} \) - Heat transfer per unit mass, \( q = 46.95 \text{ kJ/kg} \) - Process relationship, \( p v^{1.27} = \text{const} \)

- Use process equation

Use the given polytropic process relationship \( p v^{1.27} = \text{const} \). Apply this to find the relationship between temperatures: \[ T_2 = T_1 \times \bigg(\frac{p_2}{p_1}\bigg)^{(\frac{k-1}{k})} \] where \( k = 1.27 \), \( p_1 = 1 \text{ bar} \), and \( p_2 = 6 \text{ bar} \).

- Calculate final temperature

Calculate the final temperature using the relationship derived: \[ T_2 = 300 \text{ K} \times \bigg(\frac{6}{1}\bigg)^{(\frac{1.27-1}{1.27})} \] Simplify the exponent and calculate: \[ T_2 = 300 \text{ K} \times 6^{0.2126} \] \[ T_2 \text{ }\text{approximately} = 393.4 \text{ K} \]

- Apply first law of thermodynamics

Apply the first law of thermodynamics for a steady-flow process: \[ h_2 - h_1 = q + w_s \] where: - \( h_1 \) and \( h_2 \) are the specific enthalpies at the initial and final conditions, - \( q = 46.95 \text{ kJ/kg} \) is the heat transfer, - \( w_s \) is the specific work done by the compressor.

- Relate specific enthalpies

For air, approximating as ideal gas: \[ \frac{dh}{dT} = c_p \Rightarrow \text{ }h_2 - h_1 = c_p (T_2 - T_1) \] where \( c_p \text{ is approximately } 1.005 \text{ kJ/kg·K} \): \[ h_2 - h_1 = 1.005 \times (393.4 - 300) \] \[ h_2 - h_1 = 1.005 \times 93.4 = 93.87 \text{ kJ/kg} \]

- Solve for specific work

From the first law equation: \[ w_s = h_2 - h_1 - q \] Substitute the values: \[ w_s = 93.87 \text{ kJ/kg} - 46.95 \text{ kJ/kg} \] \[ w_s = 46.92 \text{ kJ/kg} \]

- Calculate compressor power

Power required is given by: \[ \text{Power} = \frac{dm}{dt} \times w_s \] Substitute mass flow rate and specific work: \[ \text{Power} = 4 \text{ kg/s} \times 46.92 \text{ kJ/kg} \] Convert to kW: \[ \text{Power} \text{ in } kW = 4 \times 46.92 \text{ kW} \] \[ \text{Power} = 187.68 \text{ kW} \]

Key concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with heat, work, and energy. It defines how energy is transferred within a system and its environment. There are four laws of thermodynamics that describe these interactions. In this exercise, we are focusing mainly on the first law of thermodynamics to analyze the energy changes in the air compression process.

• The first law, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transformed from one form to another.
  This is crucial to understand because it means that the energy added to the system as work and heat equals the change in internal energy of the system.
• We relate this to our problem by calculating the energy added to the air through compression and heat transfer, then determining the resulting work done by the compressor.

Polytropic Process

A polytropic process is one where the relationship between pressure and volume follows the equation: \[ p v^{n} = \text{constant} \] In this exercise, we have a polytropic process with an exponent of 1.27. This specific type of process is common in engineering when considering the compression and expansion of gases.
We're given the relationship \[ pv^{1.27} = \text{const} \] and we use it to find the relationship between temperatures at the inlet and outlet. Knowing that the process follows this equation, we can solve for the final temperature using< \br> \[T_2 = T_1 \times \bigg(\frac{p_2}{p_1}\bigg)^{\frac{n-1}{n}}. \] Forget about complex scenarios. For now, just remember that the polytropic process describes how pressure and volume change together while some heat transfer occurs.

Steady-Flow Process

In thermodynamics, a steady-flow process is one where a fluid (in this case, air) flows through a control volume steadily, meaning that its properties do not change with time at any specific point. Mass and energy through a control surface remain constant over time.

• For our steady-flow process, the analysis will involve examining the compressing air from the inlet to the exit of a compressor.
• This approach simplifies our calculations because we assume the properties of air and the flow rate are consistent.

This means, we don't have to worry about changes with time but only spatial changes from one point to another within the system—tracking air as it gets compressed.

First Law of Thermodynamics

The First Law of Thermodynamics, often written as: \[Q - W = \text{difference in internal energy}. \] for a control volume, but in specific terms for steady-flow processes, it's usually written as \[ \text{Heat Transfer} (q) + \text{Work done} (w_s) = \text{Change in Enthalpy} (\text{{h2 - h1}})\].This means that the change in energy (enthalpy) of the air is equal to the heat added to the air minus the work done by the compressor.
This helps us solve the problem as we apply this law to the system and find out that: \[ q = 46.95 \text{kJ/kg} \text{ and } h_2 - h_1 = c_p \times (T_2 - T_1). \] Given these, we equate and solve for work done and power for the compressor.