ME 200 – Thermodynamics I – Spring 2020

Homework 27: Evaluating Entropy Changes

Part (i): Entropy Change Assumptions

Part (ii): Methods for Calculating Entropy Change of a Two Phase Fluid

Part (iii): Entropy Change for an Incompressible Fluid

Part (iv): Constant Pressure Ideal Gas Entropy Change

Part (v): Constant Volume Ideal Gas Entropy Change

Part (vi): Entropy Change: Constant Volume vs. Constant Process with Ideal Gas

Part (vii): Methods for Calculating Entropy Change Changes of Ideal Gases

Part (i): Entropy Change Assumptions

List the minimum set of assumptions required to employ each of the following equations. Possible assumptions include:

Closed system, open system, steady state, adiabatic, no work, negligible changes in potential energy, negligible changes in kinetic energy, isothermal, constant pressure, incompressible, ideal gas, constant specific heats, internally reversible. Show how any simplified equations are developed from basic equations that appear on your equation sheet when applying the specified assumptions.

(a) $\Delta s_{12} = c \cdot ln(\frac{T_{2}}{T_{1}})$

(b) $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{P}\cdot\frac{dT}{T} - R \cdot ln(\frac{P_{2}}{P_{1}})$

(c) $\Delta s_{12} = s^{o}_{2} - s^{o}_{1} - R \cdot ln(\frac{P_{2}}{P_{1}})$

(d) $\Delta s_{12} = \frac{1}{m}\int_{1}^{2} \frac{\delta Q}{T}$

Solution:

(a) $\Delta s_{12} = c \cdot ln(\frac{T_{2}}{T_{1}})$

$\underline{Assumptions}$: incompressible, constant specific heat

This relation comes from $ds=c\frac{dT}{T}$ which is the basic equation for evaluating specific entropy changes of an incompressible substance. You should know that this is not a relation for an ideal gas because there is no differentiation between $c_p$ and $c_v$ for an incompressible. The integral of this basic equation is $\Delta s=\int {c\frac{dT}{T}}$, the solution for which matches the provided equation.

(b) $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{P}\cdot\frac{dT}{T} - R \cdot ln(\frac{P_{2}}{P_{1}})$

$\underline{Assumptions}$: ideal gas

This comes from the second T-ds equation, $Tds = dh - vdP$. The substitutions $dh = c_{P}dT$ and $\frac{v}{T} = \frac{R}{P}$ are made to produce $ds = c_{P}\frac{dT}{T} - R\frac{dP}{P}$. This result in is on the ME 200 basic equation sheet and its integral is the equation shown in (b). Note, because $c_{P}$ is still within the integral we are not assuming constant specific heats.

(c) $\Delta s_{12} = s^{o}_{2} - s^{o}_{1} - R \cdot ln(\frac{P_{2}}{P_{1}})$

$\underline{Assumptions}$: ideal gas

This comes directly from assessing change in specific entropy for an ideal gas with variable specific heats. Starting with $ds = c_{P}\frac{dT}{T} - R\frac{dP}{P}$, substitute $\int_{T_{1}}^{T_{2}} c_{P}\cdot\frac{dT}{T} = s^{o}(T_{2}) - s^{o}(T_{1})$ into the final solution for the initial integral. The resulting expression is the equation shown in (c).

(d) $\Delta s_{12} = \frac{1}{m}\int_{1}^{2} \frac{\delta Q}{T}$

$\underline{Assumptions}$: closed system and internally reversible

This comes from an entropy balance without mass flow and a single location for heat transfer, such that $dS/dt = \frac{\dot Q}{T_{boundary}} + \dot \sigma$. For an internally reversible process, $\dot \sigma = 0$. Additionally, $S = m\cdot s$. Therefore, $m \cdot ds = \frac{\dot Q \cdot dt}{T_{boundary}} = \frac{\delta Q}{T_{boundary}}$. Then, dividing by the mass and integrating leads to the equation provided in (d).

Part(ii): Methods for Calculating Entropy Change of a Two Phase Fluid

Which of the following is applicable for determining the specific entropy change of a two-phase mixture between two known states that has undergone a quasi-equilibrium, constant temperature heat addition process with no frictional effects in a closed system? Explain

(a) $\Delta s_{12} = \frac{1}{m}\int_{1}^{2} \frac{\delta Q}{T}$

(b) $\Delta s_{12} = \frac{1}{m}\cdot\frac{Q_{12}}{T}$

(c) property tables

(d) a, b and c give the same result

Solution:

(d) because (a) and (b) will give the same result, which also matches that of the property tables which would result from (c). A key concept here is that entropy change is path independent. It only depends on the beginning and end states.

Part(iii): Entropy Change for an Incompressible Fluid

Liquid water treated as an incompressible substance undergoes a process whereby its pressure is raised but its temperature remains constant. Does its entropy increase, decrease, or remain the same? Justify using appropriate equations!

Solution:

The entropy of an incompressible substance is only a function of temperature. Therefore, the entropy will remain the same if the temperature remains constant. You can also clearly see that by using the basic equation, $ds=c\frac{dT}{T}$, for evaluating specific entropy changes of an incompressible substance. If $dT = 0$, then $ds = 0$.

Part(iv): Constant Pressure Ideal Gas Entropy Change

Given:

A fixed mass of an ideal gas is heated from 40 °C to 60 °C at a constant pressure of (a) 100 kPa and (b) 300 kPa.

Find:

For which case do you think the entropy change will be greater? Use thermodynamic relations to explain.

Solution:

The entropy change is the same for both situations because entropy change of an ideal gas is not dependent on the absolute pressure, but the change in pressure. This relationship is shown by $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{P}\cdot\frac{dT}{T} - R \cdot ln(\frac{P_{2}}{P_{1}})$. If $P_{2} = P_{1}$, then the $ln(1) = 0$, resulting in $\Delta s_{12}$ being dependent on temperature only.

Part(v): Constant Volume Ideal Gas Entropy Change

Given:

A fixed mass of an ideal gas is heated from 40 °C to 60°C at a constant volume of (a) 1 m$^{3}$ and (b) 3 m$^{3}$.

Find:

For which case will the entropy change be greater? Use thermodynamic relations to explain.

Solution

The entropy change is the same for both situations because entropy change of an ideal gas is not dependent on the volume, but the change in volume. This relationship is shown by $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{v}\cdot\frac{dT}{T} + R \cdot ln(\frac{v_{2}}{v_{1}})$. If $v_{2} = v_{1}$, then the $ln(1) = 0$, resulting in $\Delta s_{12}$ being dependent on temperature only.

Part(vi): Entropy Change: Constant Volume vs. Constant Process with Ideal Gas

Given

A fixed mass of an ideal gas is heated from 40 °C to 60°C (a) at constant volume and (b) at constant pressure.

Find:

For which case will the entropy change be greater? Use thermodynamic relations to explain.

Solution:

The constant pressure process was discussed in Part (iv) and the constant volume process was discussed in Part (v) of these solutions.

For the constant volume process, $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{v}\cdot\frac{dT}{T} + R \cdot ln(\frac{v_{2}}{v_{1}})$ reduces to $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{v}\cdot\frac{dT}{T}$.

Similarly, for the constant pressure process, $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{P}\cdot\frac{dT}{T} - R \cdot ln(\frac{P_{2}}{P_{1}})$ reduces to $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{P}\cdot\frac{dT}{T}$.

Therefore, this question becomes a matter of the relative magnitudes of $c_{P}$ and $c_{v}$. Given that $c_{P} = c_{v} + R$ for an ideal gas, $c_{P} > c_{v}$.

This brings the conclusion that the change in specific entropy is greater for the constant pressure process.

Part(vii): Methods for Calculating Entropy Change Changes of Ideal Gases

Given

Air (treated as an ideal gas) is heated from 300 K to 500 K in a rigid, closed container.

Find

Compare the specific entropy change of the air as determined using 3 different approaches:

a) Variable specific heat (use ideal gas tables for air)

b) Constant specific heat for air evaluated at the average temperature

c) Constant specific heat for air evaluated at the initial temperature

Solution

a) Variable specific heat for air

From the ME 200 basic equation sheet,

$$\Delta s_{12} = s^{o}_{2} - s^{o}_{1} - R \cdot ln(\frac{P_{2}}{P_{1}})$$

We can use the ideal gas law to replace the pressure ratio with a temperature ratio since both the mass and volume of this closed system are constant. Since $Pv = RT$ and $v$ and $R$ are constant, then $\frac{P}{T} = constant$. Therefore, $\frac{P_{2}}{P_{1}} = \frac{T_{2}}{T_{1}}$, resulting in

$$\Delta s_{12} = s^{o}_{2} - s^{o}_{1} - R \cdot ln(\frac{T_{2}}{T_{1}})$$

import math

# Solution

R_bar = 8.314           # universal gas constant [kJ/(kmol-K)]
M_air = 28.97           # molar mass of air [kg/kmol]
R_air = R_bar/M_air     # gas constant for air [kJ/(kg-K)]
T_1 = 300               # initial temperature for air [K]
T_2 = 500               # final temperature for air [K]
T_ratio = T_2/T_1       # ratio of final to initial air temperatures [K]

s_o_300 = 1.703         # tabulated specific entropy at 300 K [kJ/(kg-K)]
s_o_500 = 2.221         # tabulated specific entropy at 500 K [kJ/(kg-K)]

# Changes in specific entropy for part a
delta_s_a = s_o_500 - s_o_300 - R_air*math.log(T_ratio) # [kJ/(kg-K)]

print('delta_s_a = ',round(delta_s_a,3),'(kJ/(kg-K))')

delta_s_a =  0.371 (kJ/(kg-K))

b) Constant specific heat for air evaluated at the average temperature

Since the specific volume is constant, then $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{v}\cdot\frac{dT}{T} + R \cdot ln(\frac{v_{2}}{v_{1}})$ reduces to $\Delta s_{12} = \int_{T_{1}}^{T_{2}} c_{v}\cdot\frac{dT}{T}$. Also, since specific heat is assumed to be constant and evaluated at the average temperature, then

$$\Delta s_{12} = c_{v, 400 K}\cdot ln(\frac{T_{2}}{T_{1}})$$

c_v_400 = 0.726                             # specific heat at constant volume at 400 K [kJ/(kg-K)]

delta_s_b = c_v_400*math.log(T_ratio)       # change in specific entropy, spec heat at T_avg [kJ/(kg-K)]

print('delta_s_b = ',round(delta_s_b,3),'(kJ/(kg-K))')

delta_s_b =  0.371 (kJ/(kg-K))

c) Constant specific heat for air evaluated at the initial temperature

$$\Delta s_{12} = c_{v, 300 K}\cdot ln(\frac{T_{2}}{T_{1}})$$

c_v_300 = 0.718                             # specific heat at constant volume at 300 K [kJ/(kg-K)]

delta_s_c = c_v_300*math.log(T_ratio)       # change in specific entropy, spec heat at T_i [kJ/(kg-K)]

print('delta_s_c = ',round(delta_s_c,3),'(kJ/(kg-K))')

delta_s_c =  0.367 (kJ/(kg-K))

The three approaches give almost the same results!