The state of liquid water is changed from 50 psia and 50^∘ F to 2000 psia and 100^∘ F. Determine

The state of liquid water is changed from 50 psia and 50^∘ F...

The state of liquid water is changed from 50 psia and $50^{\circ} \mathrm{F}$ to 2000 psia and $100^{\circ} \mathrm{F}$. Determine the change in the internal energy and enthalpy of water on the basis of the (a) compressed liquid tables, (b) in-compressible substance approximation and property tables, and (c) specific-heat model.

The state of liquid water is changed from 50 psia and $50^{\circ} \mathrm{F}$ to 2000 psia and $100^{\circ} \mathrm{F}$. Determine the change in the internal energy and enthalpy of water on the basis of the
(a) compressed liquid tables,
(b) in-compressible substance approximation and property tables, and (c) specific-heat model.

Key Concepts

Specific Heat Model

The specific heat model utilizes the specific heat capacity of a substance to determine how its internal energy and enthalpy change with temperature. By integrating the specific heat over the temperature range of interest, one can approximate the energy required for a temperature change. This method is particularly useful for analyzing processes where temperature effects dominate and precise property data might be less critical.

Thermodynamic Properties

This concept involves understanding internal energy and enthalpy as fundamental state functions in thermodynamics. Internal energy represents the energy contained within a substance due to both its molecular motion and intermolecular forces, while enthalpy is defined as the sum of the internal energy and the product of pressure and volume, often used to analyze energy changes during processes at constant pressure.

Compressed Liquid Properties

Compressed liquid properties refer to the thermodynamic characteristics of liquids under pressure conditions where they exist in a subcooled state, meaning they are not at their boiling point. Compressed liquid tables provide accurate data for properties like specific volume, internal energy, and enthalpy at various pressures and temperatures, which are essential for precise calculations when the liquid is not close to saturation conditions.

Incompressible Substance Approximation

The incompressible substance approximation is based on the assumption that the volume (or density) of a liquid changes negligibly with pressure. This simplification allows the use of constant specific volume in energy calculations, making it easier to estimate changes in internal energy and enthalpy when the liquid undergoes small volume changes, even under high pressure.

Transcript

00:01 In this problem, we have liquid water experiencing a process from one state to another.

00:08 We wish to calculate the internal energy and the enthalpy changes under different assumptions.

00:15 So for the first assumption in part a, we'll use the properties from the compressed liquids table.

00:33 And in our compressed liquids table, the internal energy of state one, which is u1, is equal to u .u .s.

00:42 Remember this is at 50 degrees fahrenheit and this is 18 .07 bt u per pound mass.

00:57 The enthalpy of state 1, h1 at 50 degrees fahrenheit is hf, hf actually plus vf into p minus p saturated temperature and that's all also at 50 degrees fahrenheit.

01:33 So these for more table are 18 .07.

01:38 That's btu per pound plus vf at this temperature is 0 .0162.

01:56 That's the specific volume, which is cubic feet per pound, multiplied by 50 psi minus 0 .17812 psi.

02:21 And we'll convert that to btu.

02:25 So this value for h1, the enthalpy of state 1, becomes 18 .21, btu per pound.

02:42 Now the pressure of state 2 p2 is equal to 2 ,000 psi absolute and the temperature of state 2 is 100 degrees fahrenheit.

02:56 So we go from 50 degrees fahrenheit to 100 degrees fahrenheit.

03:00 Using our tables again, this corresponds to an internal energy u2 of 67 .36.

03:10 Again that's btu per pound and an enthalpy h2 which is equal to 73 .3 btu per pound.

03:27 And so now we can find the change in internal energy using this method delta u is equal to u2 minus u1 and that's 67 .36 minus 18 .07.

03:47 And we get the change in internal energy to be 49 .29 .29 bt u per pound.

03:58 The change in enthalpy delta h is equal to h2 minus h1...

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