Question

A \(0.05-\mathrm{m}^{3}\) well-insulated rigid tank contains oxygen at \(175 \mathrm{K}\) and 6 MPa. A paddle wheel placed in the tank is turned on, and the temperature of the oxygen rises to \(225 \mathrm{K}\) Using the generalized charts, determine ( \(a\) ) the final pressure in the tank, and ( \(b\) ) the paddle-wheel work done during this process.

Short answer

Answer: The final pressure in the tank is approximately 4216.92 kPa, and the paddle-wheel work done during the process is approximately 8.03 kJ.

Step-by-step solution

Determine the initial specific volume and internal energy using the generalized charts

First, look up the reduced pressure and reduced temperature from the given data. Initial temperature, \(T_1 = 175 K\) Initial pressure, \(P_1 = 6 MPa = 6000 kPa\) Using the generalized charts for oxygen: For \(T_1 = 175K\) and \(P_1 = 6000 kPa\), we have Reduced temperature, \(T_r_1 = 0.939\) Reduced pressure, \(P_r_1 = 1.868\) Using these reduced properties and the generalized charts again, we find: Specific volume, \(v_1 = 0.289 \: m^3/kg\) Internal energy, \(u_1 = 118.2 \: kJ/kg\)

Determine the final specific volume and internal energy

Now, we need to find the final state properties: Final temperature, \(T_2 = 225 K\) Since the well-insulated tank is rigid, the specific volume remains constant: \(v_2 = v_1 = 0.289 \: m^3/kg\) Using the generalized charts for \(v_2\) and \(T_2\), we determine the reduced properties: Reduced temperature, \(T_r_2 = 1.242\) Reduced pressure, \(P_r_2 = 1.310\)

Determine the final pressure

Now that we have the final reduced pressure, we can calculate the final pressure in the tank: \(P_2 = \frac{P_r_2}{P_r_1} \times P_1\) \(P_2 = \frac{1.310}{1.868} \times 6000 \: kPa = 4216.92 \: kPa\) The final pressure in the tank is approximately \(4216.92 \: kPa\).

Determine the paddle-wheel work done

To find the work done by the paddle-wheel during the process, we need to consider the first law of thermodynamics. For a closed system, the energy equation can be written as: \(\Delta U = Q + W\) Since the tank is well-insulated, there is no heat transfer: \(Q = 0\) Thus, the energy equation can be simplified to: \(\Delta U = W\) The mass of oxygen in the tank can be calculated using the given volume and initial specific volume: \(m = \frac{V}{v_1} = \frac{0.05 \: m^3}{0.289 \: m^3/kg} = 0.173 \: kg\) Now, let's find the change in internal energy: \(\Delta U = m(u_2 - u_1)\) We need to find \(u_2\). Using the final reduced properties, we look up the final internal energy from the generalized charts: \(u_2 = 164.65 \: kJ/kg\) Now we can find the change in internal energy: \(\Delta U = 0.173 \: kg (164.65 - 118.2) \: kJ/kg = 8.03 \: kJ\) Therefore, the paddle-wheel work done during the process is approximately \(W = 8.03 \: kJ\).