Two rigid tanks are filled with water (Fig. P 3.214 ). Tank A is 0.2 m^3 at 100 kPa, 150^∘ C an

Two rigid tanks are filled with water (Fig. P 3.214 ). Tank...

Two rigid tanks are filled with water (Fig. $\mathrm{P} 3.214$ ). Tank $A$ is $0.2 \mathrm{~m}^{3}$ at $100 \mathrm{kPa}, 150^{\circ} \mathrm{C}$ and $\operatorname{tank} B$ is $0.3 \mathrm{~m}^{3}$ at saturated vapor of $300 \mathrm{kPa}$. The tanks are connected by a pipe with a closed valve. We open the valve and let all the water come to a single uniform state while we transfer enough heat to have a final pressure of $300 \mathrm{kPa}$. Give the two property values that determine the final state and find the heat transfer.

Key Concepts

Property Relationships and Steam Tables

Determining the state of water under different conditions relies on using steam tables and property diagrams, which provide essential values such as saturation pressure, specific volume of saturated liquid and vapor, and internal energy. By knowing the overall specific volume and the final pressure or temperature, one can deduce the state, including whether the water is superheated, saturated, or a mixture of two phases.

Phase Equilibrium and Quality Determination

For a pure substance like water, the final state when a mixture of vapor and liquid is present is determined by two independent intensive properties, such as pressure and specific volume or temperature and quality. The quality (or vapor fraction) is a critical property that indicates the proportion of the mass that is in the vapor phase, and it is determined using property data along with the system’s overall specific volume.

Heat Transfer in Constant-Volume Systems

In processes involving rigid tanks, the amount of heat transfer is directly related to the change in the internal energy of the system because there is no work done by boundary movement. The overall energy balance, combined with the specific property data of the substances involved, allows for the calculation of the heat transfer required to achieve a final state with a given pressure, taking into account any phase changes that may occur.

Rigid Tank (Constant Volume Process)

In a rigid tank, the volume remains constant regardless of the processes occurring within the system. This constraint means that any changes in the state of the fluid (whether it is heating, cooling, or phase-change) occur at a fixed volume, which significantly affects the energy balance and the determination of properties such as pressure and temperature.

Closed System Energy Analysis

A closed system does not exchange matter with its surroundings, though energy (as heat or work) can be transferred. The first law of thermodynamics is applied by accounting for the total internal energy of the system, where the system’s energy change is balanced by the net heat transfer and work done. In constant-volume processes, no boundary work is performed, so energy transfer is purely in the form of heat.

Transcript

00:01 All right, guys, we are given the following data for two rooms of water, that is ta is equal to 150 degree centigrade, and b .a.

00:10 Is equal to 100 kioskal, and we've got bb is equal to 300 kilopascal, and we've got p2 is equal to 300 kilopascal, and we've got vb is equal to 0 .3 cubic meters, and we have va, is equal to 0 .2 cubic meters.

00:35 So the energy equation for this problem is q is equal to ma into u2 minus u1a plus mb into u2 minus u1b plus w.

00:50 So this is constant volume process, so the work done is equal to zero.

00:55 That is w is equal to zero.

00:57 And the heat transfer is q is equal to ma into u2 minus u1a plus mb into u2 minus u1b.

01:11 And from superheated water table corresponding to temperature ta, that is 150 degrees centigrade, and p .a, that is 100 kilopascal, we can obtain specific volume and internal energy, that is va, is equal to 1 .93636 cubic meters per cubic, and then we've got ua is equal to 258 .75 kilojoules per kg so the mass of the mass of water from room a can be calculated as m a into v a divided by v a that is 0 .2 divided by 1 .93636 is equal to 0 .1 kg so from saturated water table corresponding to pb that is 300 kiosk we can obtain specific volume and internal energy that is vb is equal to vf that is equal to 60582 cubic meters per kg and we've got ub is equal to uf is equal to 2543 0 .55 kilojoules per kg so the mass of water from room b is mb is equal to vb divided by small vb and that is equal to 0 .3 0 .3 divided by 0 .682 is equal to 0 .6582 is equal to 0 .5kg and the total mass is given as m2 is equal to m a plus mb and that is 0 .1 plus 0 .5 is equal to 0 .6 kg now the total volume is equal to v1 is equal to is equal to va plus vb that is 0 .2 plus 0 .3 is equal to 0 .5 cubic meters so the total specific volume is equal to v2 is equal to v2 divided by m2, that is 0 .5 over 0 .6 is equal to 0 .8333 cubic meters per k...

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