Question

Helium is to be compressed from 105 kPa and \(295 \mathrm{K}\) to \(700 \mathrm{kPa}\) and \(460 \mathrm{K}\). A heat loss of \(15 \mathrm{kJ} / \mathrm{kg}\) occurs during the compression process. Neglecting kinetic energy changes, determine the power input required for a mass flow rate of \(60 \mathrm{kg} / \mathrm{min}\).

Short answer

Answer: The power input required is 872 kW.

Step-by-step solution

Write the first law of thermodynamics for an open system

The first law of thermodynamics states, for an open system with steady-state flow and negligible kinetic energy changes: \(Q - W = \Delta H\) Where Q is the heat transfer, W is the work input, and ΔH is the change in enthalpy.

Determine the heat transfer, Q

We are told that there is a heat loss of 15 kJ/kg during the compression process. Therefore, Q = -15 kJ/kg, because heat loss is negative.

Find the change in enthalpy, ΔH

The change in enthalpy can be expressed as: \(\Delta H = m (\Delta h) = m (h_2 - h_1)\) Where m is the mass flow rate, and h_1 and h_2 are the specific enthalpies at the initial and final states, respectively. We are not given specific enthalpy values, but we can determine them using the ideal gas equation: \(h = c_p \cdot T\) Where c_p is the specific heat at constant pressure (for helium \(c_p = 5.193 \mathrm{kJ/(kg \cdot K)}\)), and T is the temperature in kelvin.

Determine h_1 and h_2 using the ideal gas equation and given temperatures

We are given the initial temperature \(T_1 = 295\mathrm{K}\) and the final temperature \(T_2 = 460\mathrm{K}\). Calculate the specific enthalpies: \(h_1 = c_p \cdot T_1 = 5.193 \mathrm{kJ/(kg \cdot K)} \times 295 \mathrm{K} \approx 1532 \mathrm{kJ/kg}\) \(h_2 = c_p \cdot T_2 = 5.193 \mathrm{kJ/(kg \cdot K)} \times 460 \mathrm{K} \approx 2389 \mathrm{kJ/kg}\)

Calculate the change in enthalpy, ΔH

Using the mass flow rate, \(m = 60\mathrm{kg/min}=1\mathrm{kg/s}\), and the specific enthalpies found in Step 4: \(\Delta H = m (\Delta h) = m (h_2 - h_1) = 1\mathrm{kg/s} \times (2389 \mathrm{kJ/kg} - 1532 \mathrm{kJ/kg}) = 857 \mathrm{kJ/s}\)

Calculate the work input, W, using the first law of thermodynamics equation

Now we can determine the work input using the first law of thermodynamics equation: \(Q - W = \Delta H\) \(-15 \mathrm{kJ/kg} - W = 857 \mathrm{kJ/s}\) \(W = -15 \mathrm{kJ/kg} - 857 \mathrm{kJ/s} \approx -872 \mathrm{kJ/s}\)

Convert the work input to power input

Since we have the work input value, we can convert it to power input by dividing it by the time taken: Power input = \(\dfrac{-872 \mathrm{kJ/s}}{1 \mathrm{s}} = 872 \mathrm{kW}\)

Express the final answer

The power input required for a mass flow rate of 60 kg/min is 872 kW.

Key concepts

Heat Transfer

Understanding heat transfer is crucial when analyzing thermodynamic processes. Heat transfer, denoted by the symbol Q, refers to the transfer of energy between systems or surroundings because of a temperature difference. It plays a fundamental role in the first law of thermodynamics, which is all about the conservation of energy.

During a thermodynamic process, when heat is added to a system, it can do work on its surroundings or change the internal energy of the system, often observed as a change in the temperature of the substance. In our helium compression example, the system loses heat, indicated by a negative value for Q, which stands for a heat loss of 15 kJ/kg. It's important to note that heat transfer can occur through various modes: conduction, convection, and radiation.

• Conduction is the transfer of heat through a solid material.
• Convection occurs in fluids and is aided by fluid motion.
• Radiation is the transfer of energy through electromagnetic waves.

Whether heat is being transferred into or out of the system, it impacts the energy balance and the work that can be done by or on the system.

Enthalpy Change

Enthalpy change, represented as \( \Delta H \), is the measure of the total heat content in a thermodynamic system during a constant pressure process. It's a central concept in thermodynamics as it helps determine how much energy is required for a process or is released during a process.

More formally, enthalpy change is the difference in the enthalpy of a system between the final state and the initial state and can be calculated as \( \Delta H = m(h_2 - h_1) \), where \(h_2\) and \(h_1\) are the specific enthalpies at the final and initial states respectively, and m is the mass flow rate.

In our scenario, the positive \( \Delta H \) implies that the internal energy of the helium increases, largely because work is done on it to compress it, even as it loses heat. This increase in enthalpy reflects the need for significant power input to achieve the desired compression. For an ideal gas, we calculate the specific enthalpy by multiplying the specific heat at constant pressure (\(c_p\)) with temperature (\(T\)). This is crucial as it simplifies the determination of enthalpy change without directly measuring heat content.

Compression Process

A compression process, such as the one experienced by the helium gas in our exercise, involves increasing the pressure and often the temperature of a gas by reducing its volume. This is achieved by doing work on the gas, usually in a compressor, and can be adiabatic, isothermal or polytropic, depending on how the process is carried out relative to heat transfer.

In adiabatic compression, no heat is exchanged with the surroundings, implying Q = 0. However, in our helium example, the process is not adiabatic as it involves heat loss. The work performed on the helium not only raises its pressure but also its temperature, which in an ideal gas, signifies an increase in average kinetic energy of the molecules.

Importance of Compression

Compression processes are fundamental in numerous mechanical and chemical engineering applications, such as refrigeration, internal combustion engines, and industrial gas production. They are designed to manipulate the gas pressure and temperature to suit a particular end-use or facilitate subsequent chemical reactions.

Understanding the power input during these processes is important for the design and operation of systems that undergo compression. In the helium example, a correct calculation of power input leads to effective energy consumption estimates and informs the compressor's power requirements.