Question 2.3: Variation of Density with Temperature and Pressure Consider ...
Variation of Density with Temperature and Pressure
Consider water initially at 20°C and 1 atm. Determine the final density of the water (a) if it is heated to 50°C at a constant pressure of 1 atm, and (b) if it is compressed to 100-atm pressure at a constant temperature of 20°C. Take the isothermal compressibility of water to be α = 4.80 × 10^{−5} atm^{−1}.
Water at a given temperature and pressure is considered.
The densities of water after it is heated and after it is compressed are to be determined.
Assumptions 1 The coefficient of volume expansion and the isothermal compressibility of water are constant in the given temperature range. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes.
Properties The density of water at 20°C and 1 atm pressure is \rho_1 = 998.0 kg/m^3.
The coefficient of volume expansion at the average temperature of (20 + 50)/2 = 35°C is \beta = 0.337 × 10^{−3} K^{−1}. The isothermal compressibility of water is given to be \alpha = 4.80 × 10^{−5} atm^{−1}.
Analysis When differential quantities are replaced by differences and the properties α and β are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as (Eq. 2.23)
\frac{\Delta V}{V}=-\frac{\Delta \rho}{\rho} \cong \beta \Delta T – \alpha \Delta P (2.23)
\Delta \rho = \alpha \rho \Delta P – \beta \rho \Delta T
(a) The change in density due to the change of temperature from 20°C to 50°C at constant pressure is
\Delta \rho = -\beta \rho \Delta T = -(0.337 \times 10^{-3} K^{-1})(998 kg/m^3)(50 – 20) K = -10.0 kg/m^3
Noting that Δ\rho = \rho_2 − \rho_1, the density of water at 50°C and 1 atm is
\rho _2 = \rho _1 + \Delta \rho = 998.0 + (-10.0) = 988.0 kg/m^3
which is almost identical to the listed value of 988.1 kg/m³ at 50°C in Table A–3.
This is mostly due to 𝛽 varying with temperature almost linearly, as shown in Fig. 2–16.
(b) The change in density due to a change of pressure from 1 atm to 100 atm at constant temperature is
\Delta \rho = \alpha \rho \Delta P = (4.80 \times 10^{-5} atm^{-1})(998 kg/m^3)(100 – 1) atm = 4.7 kg/m^3
Then the density of water at 100 atm and 20°C becomes
ρ_2 = ρ_1 + Δρ = 998.0 + 4.7 = 1002.7 kg/m^3
Discussion Note that the density of water decreases while being heated and increases while being compressed, as expected. This problem can be solved more accurately using differential analysis when functional forms of properties are available.
