Question

A \(12-\mathrm{ft}^{3}\) rigid tank contains refrigerant- \(134 \mathrm{a}\) at 30 psia and 55 percent quality. Heat is transferred now to the refrigerant from a source at \(120^{\circ} \mathrm{F}\) until the pressure rises to 50 psia. Assuming the surroundings to be at \(75^{\circ} \mathrm{F}\), determine (a) the amount of heat transfer between the source and the refrigerant and ( \(b\) ) the exergy destroyed during this process.

Short answer

Question: The process of refrigerant-134a in a 12 ft^3 rigid tank, initially at 30 psia with 55% quality, gets heated by an external heat source which increases its pressure to 50 psia. Calculate the heat transfer (Q) between the source and the refrigerant and the exergy destroyed during the process. Assume that the reference state for exergy calculations is 75°F.

Step-by-step solution

Assess initial state and find initial properties

The initial state of the refrigerant is specified as 30 psia and 55% quality. Using the refrigerant-134a tables (available in any thermodynamics textbook or online), we can determine the initial properties: - Saturation temperature at 30 psia: \(T_1 = -9.41^\circ F\) - Initial quality: \(x_1 = 0.55\) - Initial specific volume: \(v_1 = x_1 * (v_g - v_f) + v_f\), where \(v_f\) and \(v_g\) are the saturated liquid and vapor specific volumes, respectively. - Initial internal energy: \(u_1 = x_1 * (u_g - u_f) + u_f\), where \(u_f\) and \(u_g\) are the saturated liquid and vapor internal energies, respectively.

Assess final state and find final properties

The final state is specified as 50 psia. The fluid is superheated at this point. Using the refrigerant-134a tables, we can determine the final properties: - Final temperature (at 50 psia): \(T_2 = 120^\circ F\) - Final specific volume: \(v_2 = v_1\) (due to rigid tank) - Final internal energy: \(u_2\) (at 50 psia and \(v_2\))

Calculate the mass of the refrigerant

We can determine the mass of the refrigerant using the relationship \(m = V/v\), where \(V\) is the volume of the tank and \(v\) is the specific volume. Mass, \(m = V/v_1 = 12\mathrm{ft}^3 / v_1\)

Calculate the heat transfer

We can now calculate the heat transfer (Q) between the source and the refrigerant using the first law of thermodynamics for a closed system: \(Q = m(u_2 - u_1)\)

Calculate the exergy destroyed

The exergy destroyed during the process can be calculated using the formula: \(E_x = T_0 S_{gen}\), where \(S_{gen}\) is the entropy generated We can find the entropy generated using the formula: \(S_{gen} = m(s_2 - s_1) - \frac{Q}{T_0}\), where \(s_1\) and \(s_2\) are the initial and final specific entropies. Using the refrigerant-134a tables, we can obtain the values for \(s_1\) and \(s_2\). Then, we can calculate the exergy destroyed: \(E_x = T_0 S_{gen}\) (with \(T_0 = 75^\circ F\) converted to an absolute scale, e.g., Rankine or Kelvin) These steps will allow us to determine both the heat transfer between the source and the refrigerant and the exergy destroyed during the process.