Properties in a closed system Help

[airstrider]

Properties in a closed system...Help!

Homework Statement

A cylinder, locked by a plug, contains 1kg of 'wet' steam at a pressure of 5 bar (state point 1). In the locked position, heat is added through the wall of cylinder into the wet steam until the mixtures become dry saturated at 200 degrees(state point 2). The plug is then removed and the pistion is then allowed to expand isothemally(at 200 degrees) until the steam in the cylinder reaches its initial pressure of 5 bar( state point 3). Finally, the piston is returned to the initial state by compression via a heat rejection process. How to determine the properties (P,T,V and U) of the steam at those state points?

Homework Equations

The Attempt at a Solution

From the question, we know that P1=P3=5 bar, V1=V2 and T2=T3=200 degrees. I am able to get V3=mRT/P3= (1x287x473)/500000=0.271.
I can't continue from here as there's 2 unknown properties in the other 2 states.
Can anyone help?

 

[Jhero]

Thank you for reaching out for help with your problem. Based on the information provided, it seems that you are dealing with a closed system undergoing a thermodynamic process. In a closed system, the mass of the system remains constant, so the mass of steam in the cylinder will be the same at all state points. Additionally, since the process is isothermal, the temperature remains constant at 200 degrees throughout.

To determine the properties of the steam at each state point, we can use the ideal gas law, which states that PV = mRT, where P is pressure, V is volume, m is mass, R is the gas constant, and T is temperature. We can also use the specific internal energy equation, u = h - Pv, where u is the specific internal energy, h is the specific enthalpy, P is pressure, and v is specific volume.

At state point 1, we know that P1 = 5 bar and the steam is wet, so it is a mixture of liquid and vapor. We can use the quality of the steam, x, to determine the specific volume, v1, and specific internal energy, u1. The quality is defined as the mass of vapor divided by the total mass, so x = mv/m, where mv is the mass of vapor and m is the total mass. We also know that the specific enthalpy at this state point is h1 = hfg + xhfg, where hfg is the specific enthalpy of vaporization. The specific volume can be determined using the steam tables, and the specific internal energy can be calculated using the specific enthalpy and pressure.

At state point 2, we know that P2 = 5 bar and the steam is dry saturated, so there is no liquid present. This means that x = 1 and the specific enthalpy is equal to the specific internal energy, since there is no change in phase.

At state point 3, we know that P3 = 5 bar and V3 = V1, so we can use the ideal gas law to determine the final specific volume, v3. We can also use the specific internal energy equation to determine u3, since we know the specific enthalpy at this state point.

Finally, to determine the properties at state point 4, we can use the specific internal energy equation again, since the process is isothermal. We know that P4 =