Question

Carbon-dioxide gas at \(3 \quad \mathrm{MPa}\) and \(500 \quad \mathrm{K}\) flows steadily in a pipe at a rate of \(0.4 \mathrm{kmol} / \mathrm{s}\). Determine \((a)\) the volume and mass flow rates and the density of carbon dioxide at this state. If \(\mathrm{CO}_{2}\) is cooled at constant pressure as it flows in the pipe so that the temperature of \(\mathrm{CO}_{2}\) drops to \(450 \mathrm{K}\) at the exit of the pipe, determine ( \(b\) ) the volume flow rate at the exit of the pipe.

Short answer

(a) The volume flow rate V can be calculated using the ideal gas law equation: \(V = \frac{(0.4 \,\text{kmol/s})(8.314 \,\text{kJ/kmol*K})(500 \,\text{K})}{(3000 \,\text{kPa})} = 0.5549 \,\text{m}^3\text{/s}\) The mass flow rate \(\dot{m}\) can be found using the relation \(\dot{m} = n\dot{V}\frac{m}{M}\): \(\dot{m} = (0.4 \, \text{kmol/s}) \, \frac{44.01 \, \text{kg/kmol}}{1 \, \text{kmol/s}} = 17.604 \, \text{kg/s}\) The density \(\rho\) of CO2 at this state can be found using the following relation: \(\rho = \frac{m}{V}\): \(\rho = \frac{(0.4 \,\text{kmol/s})(44.01 \,\text{kg/kmol})}{0.5549 \,\text{m}^3\text{/s}} = 31.68 \, \text{kg/m}^3\) (b) The volume flow rate at the exit of the pipe when the temperature drops to 450 K at constant pressure is: \(V_2 = \frac{0.5549 \, \text{m}^3\text{/s}(450\,K)}{500\,K} = 0.4989 \, \text{m}^3\text{/s}\) Therefore, the volume flow rate at 3 MPa and 500 K is 0.5549 m³/s, the mass flow rate is 17.604 kg/s, the density is 31.68 kg/m³, and the volume flow rate at the exit of the pipe when the temperature drops to 450 K is 0.4989 m³/s.

Step-by-step solution

(a) Volume flow rate calculation

First, we need to calculate the volume flow rate of the CO2 gas at 3 MPa and 500 K using the ideal gas law equation. Rearranging the equation to solve for \(V\), we have \(V = \frac{nRT}{P}\). Given, n = 0.4 kmol/s, R = 8.314 kJ/kmol*K, P = 3 MPa = 3000 kPa, and T = 500 K. Substituting these values in the equation, we get: \(V = \frac{(0.4 \,\text{kmol/s})(8.314 \,\text{kJ/kmol*K})(500 \,\text{K})}{(3000 \,\text{kPa})}\) Calculate the volume flow rate, V.

(a) Mass flow rate calculation

Next, we need to calculate the mass flow rate, which can be found using the relation \(\dot{m} = n\dot{V}\frac{m}{M}\). The molar mass of CO2 is 44.01 kg/kmol. Thus, we have: \(\dot{m} = (0.4 \, \text{kmol/s}) \, \frac{44.01 \, \text{kg/kmol}}{1 \, \text{kmol/s}}\) Calculate the mass flow rate, \(\dot{m}\).

(a) Density calculation

Now we will calculate the density of CO2 at this state using the following relation: \(\rho = \frac{m}{V}\). We just found the volume and we have the molar mass of CO2, we can find the density. Given the molar mass, M = 44.01 kg/kmol and calculated volume V earlier. Density, \(\rho = \frac{(0.4 \,\text{kmol/s})(44.01 \,\text{kg/kmol})}{V}\) Calculate the density, \(\rho\).

(b) Volume flow rate at exit of the pipe

Finally, we need to find the volume flow rate at the exit of the pipe when the temperature drops to 450 K at constant pressure. Since the pressure is constant, we can use the relation \(V_1T_2 = V_2T_1\), where \(V_1\) and \(T_1\) are the initial volume flow rate and temperature, and \(V_2\) and \(T_2\) are the final volume flow rate and temperature. Rearranging for \(V_2\), we get: \(V_2 = \frac{V_1T_2}{T_1}\) Given \(T_2 = 450\,K\) and calculated \(V_1\) and \(T_1 = 500\,K\) earlier. Substitute the values and calculate the volume flow rate at the exit of the pipe, \(V_2\).