A 200-L tank (see Fig. P 6.117) initially contains water at 100 kPa and a quality of 1 % . Heat

step 1 A tank initially contains water at a given pressure and quality. Heat is then transferred to the

A 200-L tank (see Fig. P 6.117) initially contains water at...

A $200-\mathrm{L}$ tank (see Fig. $\mathrm{P} 6.117$) initially contains water at $100 \mathrm{kPa}$ and a quality of $1 \% .$ Heat is transferred to the water, thereby raising its pressure and temperature. At a pressure of $2 \mathrm{MPa}$ a safety valve opens and saturated vapor at $2 \mathrm{MPa}$ flows out. The process continues, maintaining $2 \mathrm{MPa}$ inside until the quality in the tank is $90 \%$ then stops. Determine the total mass of water that flowed out and the total heat transfer.

Key Concepts

Constant Pressure Operation with Safety Valve Action

In certain thermodynamic processes, external devices such as safety valves are used to regulate the internal pressure by releasing mass. Operating under constant pressure conditions, even in a system with fixed volume, requires understanding how the removal of mass (typically in the form of saturated vapor) influences the remaining fluid’s properties and the overall energy transfer in the system.

Energy Balance in a Control Volume with Mass Transfer

When analyzing closed systems like a rigid tank where mass may enter or leave (such as vapor exiting through a safety valve), the first law of thermodynamics is applied in a control volume formulation that accounts for changes in internal energy, work, heat transfer, and mass flow. This energy balance is essential for determining unknown quantities like the amount of mass discharged or the heat transfer required to achieve a specified state change.

Thermodynamic Property Data and Steam Tables

The use of steam tables or thermodynamic property data is crucial for obtaining the necessary state properties (such as specific volume, enthalpy, and quality) of water at various pressures and temperatures. These tables allow one to correctly evaluate the initial and final states of the system, as well as to interpolate between known values during phase changes.

Two?Phase Mixture and Quality

In thermodynamics, a two?phase mixture consists of a liquid phase and a vapor phase existing together, and the quality describes the mass fraction of the vapor within the mixture. Understanding quality is crucial for determining the overall state properties of the mixture, such as specific volume, enthalpy, and internal energy, which often require interpolation between saturated liquid and saturated vapor properties.

Phase Change and Saturation Properties

Phase change involves the transformation between liquid and vapor states at specific conditions known as saturation. Saturated properties (saturated liquid and saturated vapor values) are fundamental in analyzing processes where water undergoes heating or condensation. These properties guide the evaluation of the system’s state when energy is added or removed, particularly under constant pressure or constant volume conditions.

Transcript

00:02 A tank initially contains water at a given pressure and quality.

00:06 Heat is then transferred to the water and thus raising its temperature and pressure.

00:11 At a given pressure, a safety valve opens and saturated vapour flows out.

00:17 The process continues, maintaining a constant pressure until the quality reaches a different value, which is also given.

00:26 We want to calculate the total mass of water that has flowed out and the total heat transfer in the process.

00:31 So if we take the tank as our control volume, there is no work being done, and there is heat transfer in and flow out.

00:42 So we'll denote state one as our initial state, state two as the valve opens, and state three as the final state.

00:51 And we'll firstly write our continuity equation, m3 minus m1, the difference in masses between the initial and final state, must equal the mass of water that has exited minus m -e.

01:11 And then we can write the energy equation.

01:23 Now the energy equation is m3 u -3, the internal energy of the final state minus the internal energy of the initial state, m -1u -1, must equal to the energy of the water that is exited m -e -h -e plus the heat transfer out of the system, q.

01:43 So here we have our energy equation.

01:48 Now we look at our states one by one.

01:51 So individually, we'll first look at state one, and we know that the specific volume of state one is equal to vf plus the quality x1 times vfg.

02:07 And if we go to our tables, this is 0 .001043 plus the quality 0 .01 into 1 .6929.

02:28 And so we get our initial specific volume to be 0 .01 -797 and that's cubic meters per kg.

02:44 Now the initial internal energy, specific internal energy, u1, similarly is uf plus equality x1 times ufg and this is equal to that's uf again we go to our tables and find these values and this is 417 .33 plus 0 .01 into 2 .01 into 288 .7 2.

03:28 And so we get the internal energy of the initial state to be 438 .22 kilojoules per kg.

03:44 Now lastly for state 1, we need the mass, m1, and this is the total volume over the specific volume, capital v over little v.

03:58 We can put our values into this and that's 0 .2 cubic meters over the specific volume...

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