Question

Carbon dioxide enters a compressor at \(100 \mathrm{kPa}\) and \(300 \mathrm{K}\) at a rate of \(0.2 \mathrm{kg} / \mathrm{s}\) and exits at \(600 \mathrm{kPa}\) and \(450 \mathrm{K}\) Determine the power input to the compressor if the process involved no irreversibilities. Assume the surroundings to be at \(25^{\circ} \mathrm{C}\).

Short answer

Answer: The power input to the compressor, if the process is reversible, is 601.51 kW.

Step-by-step solution

Determine the specific heats of carbon dioxide

We know the ideal gas relation between specific heat capacities: \(c_p - c_v = R\) where \(c_p\) is the specific heat at constant pressure, \(c_v\) is the specific heat at constant volume, and \(R\) is the specific gas constant for carbon dioxide. For carbon dioxide, \(R = 0.1889\ \mathrm{kJ/(kg \cdot K)}\). From the literature, we can find the specific heat values of CO2 as: \(c_p = 0.854\ \mathrm{kJ/(kg \cdot K)}\) and \(c_v = 0.665\ \mathrm{kJ/(kg \cdot K)}\)

Calculate the change in enthalpy and entropy

First, find the change in enthalpy using the specific heat at constant pressure (\(c_p\)): \(\Delta h = m c_p \Delta T\) where \(m\) is the mass flow rate and \(\Delta T\) is the change in temperature. \(m = 0.2\ \mathrm{kg/s}\) \(\Delta T = T_{out} - T_{in} = 450\ \mathrm{K} - 300\ \mathrm{K} = 150\ \mathrm{K}\) \(\Delta h = 0.2\ \mathrm{kg/s} \cdot 0.854\ \mathrm{kJ/(kg \cdot K)} \cdot 150\ \mathrm{K} = 25.62\ \mathrm{kJ/s}\) Next, find the change in entropy using the specific heat at constant pressure (\(c_p\)) and the given surrounding temperature: \(\Delta s = m c_p \ln{\frac{T_{out}}{T_{in}}}\) \(T_s = 25^\circ\mathrm{C} + 273.15\ \mathrm{K} = 298.15\ \mathrm{K}\) \(\Delta s = 0.2\ \mathrm{kg/s} \cdot 0.854\ \mathrm{kJ/(kg \cdot K)} \cdot \ln{\frac{450\ \mathrm{K}}{300\ \mathrm{K}}} = 1.9314\ \mathrm{kJ/(s \cdot K)}\)

Calculate the reversible work

To calculate the reversible work, we will use the following equation derived from the first law of thermodynamics for a reversible process: \(W_{rev} = \Delta h + T_s \Delta s\) \(W_{rev} = 25.62\ \mathrm{kJ/s} + 298.15\ \mathrm{K} \cdot 1.9314\ \mathrm{kJ/(s \cdot K)} = 25.62\ \mathrm{kJ/s} + 575.89\ \mathrm{kJ/s} = 601.51\ \mathrm{kJ/s}\).

Calculate the power input

Now that we have the reversible work, we can find the power input to the compressor by dividing the work by time: \(Power\ input = \frac{W_{rev}}{t}\) Since the mass flow rate, work and entropy are given per second, the power input has already been calculated in the previous step: \(Power\ input = 601.51\ \mathrm{kJ/s}\) or \(601.51\ \mathrm{kW}\). The power input to the compressor, if the process involved no irreversibilities, is \(601.51\ \mathrm{kW}\).