Question

A \(0.2-\mathrm{m}^{3}\) rigid tank equipped with a pressure regulator contains steam at \(2 \mathrm{MPa}\) and \(300^{\circ} \mathrm{C}\). The steam in the \(\operatorname{tank}\) is now heated. The regulator keeps the steam pressure constant by letting out some steam, but the temperature inside rises. Determine the amount of heat transferred when the steam temperature reaches \(500^{\circ} \mathrm{C}\).

Short answer

Question: Determine the amount of heat transferred when heating steam in a rigid tank from an initial pressure of 2 MPa and temperature of 300°C to a final temperature of 500°C. The volume of the rigid tank is 0.2 m³.

Step-by-step solution

Find Initial and Final Specific Volumes

Using the initial pressure and temperature of the steam (\(2 \mathrm{MPa}\) and \(300^{\circ} \mathrm{C}\)), look up the initial specific volume, \(v_1\), and initial specific internal energy, \(u_1\), in the steam table. Similarly, look up the final specific volume, \(v_2\), and final specific internal energy, \(u_2\), at the final temperature of \(500^{\circ} \mathrm{C}\) and the same pressure of \(2 \mathrm{MPa}\).

Determine the Mass of the Steam

Given the volume of the rigid tank (\(0.2 \ \mathrm{m}^3\)) and the initial specific volume (\(v_1\)), calculate the mass of the steam \(m\): $$ m = \frac{V}{v_1} $$

Find the Initial and Final Internal Energies

Find the internal energies at the start and end of the process, respectively: $$ U_1 = m\times u_1 $$ $$ U_2 = m\times u_2 $$

Apply the First Law of Thermodynamics

Using the First Law of Thermodynamics for a closed system with no work done on the system: $$ Q - W = \Delta U $$ Since the pressure is constant, no work is done on the system, \(W = 0\). $$ Q = \Delta U = U_2 - U_1 $$ Calculate the amount of heat transferred, \(Q\), using the initial and final internal energies found in Step 3.