6.4 – Latent Heats

6.4.0 – Learning objectives

By the end of this section you should be able to:

Understand the concept of latent heat.
Differentiate between heat of melting, fusion, and vaporization.
Understand the concept of higher heating value and lower heating value.

6.4.1 – Introduction

The change in specific enthalpy of substance that occurs when that substance transitions from one phase to another is known as latent heat.

6.4.2 – Heat of Vaporization

The change in specific enthalpy for liquid water to steam at $100^{\circ }C$ and $1$ atm is $\Delta {\hat {H}}_{v}(T,P)=2,255.6{\text{ }}kJ/kg$ . This change specific enthalpy in is called the heat of vaporization. The condensation of this steam back to liquid water at $100^{\circ }C$ and $1$ atm would result in a change specific enthalpy of $\Delta {\hat {H}}_{v}(T,P)=-2,255.6{\text{ }}kJ/kg$ .

6.4.3 – Heat of Melting and Heat of fusion

The change in specific enthalpy for liquid water to ice at $0^{\circ }C$ and $1$ atm is $\Delta {\hat {H}}_{f}(T,P)=-333{\text{ }}kJ/kg$ . This change specific enthalpy in is called the heat of fusion. The melting of this ice back to liquid water at $100^{\circ }C$ and $1$ atm would result in a change specific enthalpy of $\Delta {\hat {H}}_{m}(T,P)=333{\text{ }}kJ/kg$ . This change specific enthalpy in is called the heat of melting.

6.4.4 – Higher Heat Value and Lower Heat Value

The heating value is the amount of energy released by the combustion of a material. it shares the same value as the heat of combustion but has the opposite sign.

The higher heating value (HHV) is the amount of energy to bring a material to combustion and turn any liquid products into vapor.

The lower heating value (LHV) is the amount of energy to bring a material to combustion and without turning liquid products into vapour.

To calculate the higher heating value or lower heating value we need the number of moles of water $n$ , and we can use the equation.

$HHV=LHV+n{\text{ }}\Delta {\hat {H}}_{v}{\text{ }}(H_{2}O,25^{\circ }C)$

6.4.5 – Problem Statement

Problem 1

Question

A sample of natural gas contains 75% methane, 15% ethane, and 10 % propane by volume. The heat of combustion of the methane, ethane, and propane at $25^{\circ }C$ and $1$ atm with water vapour as the product are shown below:

$CH_{4}{\text{(g)}}+2{\text{ }}O_{2}{\text{(g)}}\longrightarrow CO_{2}{\text{(g)}}+2{\text{ }}H_{2}O{\text{(v)}}\quad \Delta {\hat {H}}_{c}^{\circ }=-802{\text{ }}kJ/mol$

$C_{2}H_{6}{\text{(g)}}+{\frac {7}{2}}{\text{ }}O_{2}{\text{(g)}}\longrightarrow 2{\text{ }}CO_{2}{\text{(g)}}+3{\text{ }}H_{2}O{\text{(v)}}\quad \Delta {\hat {H}}_{c}^{\circ }=-1,428{\text{ }}kJ/mol$

$C_{3}H_{8}{\text{(g)}}+5{\text{ }}O_{2}{\text{(g)}}\longrightarrow 3{\text{ }}CO_{2}{\text{(g)}}+4{\text{ }}H_{2}O{\text{(v)}}\quad \Delta {\hat {H}}_{c}^{\circ }=-2,039{\text{ }}kJ/mol$

Calculate the higher heating value, in $MJ/kg$ , of the natural gas given

$\Delta {\hat {H}}_{v,H_{2}O}=44.013kJ/mol$

Answer

Since we want the higher heating value per unit mass, we must first convert the volume fraction to mass fraction. Assuming the gasses behave ideally, 1 mol of natural gas is:

$0.75mol{\text{ }}CH_{4}=12{\text{ }}g{\text{ }}CH_{4}$

$0.15mol{\text{ }}C_{2}H_{6}=4.5{\text{ }}g{\text{ }}C_{2}H_{6}$

$0.10mol{\text{ }}C_{3}H_{8}=4.4{\text{ }}g{\text{ }}C_{3}H_{8}$

Therefore,

$m_{total}=m_{CH_{4}}+m_{C_{2}H_{6}}+m_{C_{3}H_{8}}=(12+4.5+4.4){\text{ }}g=20.9{\text{ }}g$

$x_{CH_{4}}={\frac {m_{CH_{4}}}{m_{total}}}={\frac {12{\text{ }}g}{20.9{\text{ }}g}}=0.574$

$x_{C_{2}H_{6}}={\frac {m_{C_{2}H_{6}}}{m_{total}}}={\frac {4.5{\text{ }}g}{20.9{\text{ }}g}}=0.215$

$x_{C_{3}H_{8}}={\frac {m_{C_{3}H_{8}}}{m_{total}}}={\frac {4.4{\text{ }}g}{20.9{\text{ }}g}}=0.211$

Now we can solve for the higher heating value for each species

$(HHV)_{CH_{4}}=(LHV)_{CH_{4}}+n_{H_{2}O}{\Big (}\Delta {\hat {H}}_{v,H_{2}O}{\Big )}$

$(HHV)_{CH_{4}}={\Bigg [}802{\text{ }}{\frac {kJ}{mol{\text{ }}CH_{4}}}+{\frac {2{\text{ }}mol{\text{ }}H_{2}O}{mol{\text{ }}CH_{4}}}{\Bigg (}44.013{\frac {kJ}{mol{\text{ }}H_{2}O}}{\Bigg )}{\Bigg ]}\times {\frac {1{\text{ }}mol}{0.0160{\text{ }}kg{\text{ }}CH_{4}}}$

$(HHV)_{CH_{4}}=55.6{\text{ }}{\frac {MJ}{kg}}$

$(HHV)_{C_{2}H_{6}}=(LHV)_{C_{2}H_{6}}+n_{H_{2}O}{\Big (}\Delta {\hat {H}}_{v,H_{2}O}{\Big )}$

$(HHV)_{C_{2}H_{6}}={\Bigg [}1,428{\text{ }}{\frac {kJ}{mol{\text{ }}C_{2}H_{6}}}+{\frac {3{\text{ }}mol{\text{ }}H_{2}O}{mol{\text{ }}C_{2}H_{6}}}{\Bigg (}44.013{\frac {kJ}{mol{\text{ }}H_{2}O}}{\Bigg )}{\Bigg ]}\times {\frac {1{\text{ }}mol}{0.0300{\text{ }}kg{\text{ }}C_{2}H_{6}}}$

$(HHV)_{C_{2}H_{6}}=52.0{\text{ }}{\frac {MJ}{kg}}$

$(HHV)_{C_{3}H_{8}}=(LHV)_{C_{3}H_{8}}+n_{H_{2}O}{\Big (}\Delta {\hat {H}}_{v,H_{2}O}{\Big )}$

$(HHV)_{C_{3}H_{8}}={\Bigg [}2,039{\text{ }}{\frac {kJ}{mol{\text{ }}C_{3}H_{8}}}+{\frac {4{\text{ }}mol{\text{ }}H_{2}O}{mol{\text{ }}C_{3}H_{8}}}{\Bigg (}44.013{\frac {kJ}{mol{\text{ }}H_{2}O}}{\Bigg )}{\Bigg ]}\times {\frac {1{\text{ }}mol}{0.0440{\text{ }}kg{\text{ }}C_{3}H_{8}}}$

$(HHV)_{C_{3}H_{8}}=50.3{\text{ }}{\frac {MJ}{kg}}$

Now we can finally solve for the higher heating value of the mixture

$(HHV)=x_{CH_{4}}\cdot (HHV)_{CH_{4}}+x_{C_{2}H_{6}}\cdot (HHV)_{C_{2}H_{6}}+x_{C_{3}H_{8}}\cdot (HHV)_{C_{3}H_{8}}$

$(HHV)=[(0.574)\cdot (55.6)+(0.215)\cdot (52.0)+(0.211)\cdot (50.3)]$

$(HHV)=53.7{\frac {MJ}{kg}}$

Problem 2

Question

An equimolar liquid mixture of water and ethanol at 25 °C is fed continuously to a vessel in which the mixture is heated to 80 °C. The liquid product is 0.6 mole % E, and the vapor product is 99.4 mole % E. How much heat must be transferred to the mixture per g-mole of feed? (Bit unrealistic since almost all the ethanol would have been evaporated, but hey, question the world).

Answer

List all assumptions:

Basis: 1 mol feed

Let's figure out the mass balance of this process first:

${\text{Total balance:}}{\text{ }}1.00mol=n_{V}+n_{L}$

${\text{Ethanol balance:}}{\text{ }}0.500mol=0.006\cdot n_{V}+0.994\cdot n_{L}$

Next

Our energy balance has the form:

$Q=\Delta {H}$

And our enthalpy table for the process is:

Substance	$n_{in}$	${\hat {H_{in}}}$	$n_{out}$	${\hat {H_{out}}}$
$H2O_{(l)}$	0.5	0	3.35	${\hat {H_{1}}}$
$Ethanol_{(l)}$	0.5	0	3.35	${\hat {H_{2}}}$
$H2O_{(v)}$	-	-	63.55	${\hat {H_{3}}}$
$Ethanol_{(v)}$	-	-	33.1	${\hat {H_{4}}}$

Note: We do not know the feed-stream pressure and so we assume that $\Delta {H}$ for the change from 1 atm to $P_{feed}$ is negligible, and since the process is not running at an unusually low temperature or high pressure, we neglect the effects of pressure on the enthalpy calculations. The tables used in the calculations are referenced here [(Ethanol)](https://en.wikipedia.org/wiki/Ethanol_(data_page)) and here [(Water)](http://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Type=JANAFL&Plot=on=JANAFL).

${\hat {H_{1}}}=\int \limits _{25C}^{80C}(C_{p})_{H_{2}O_{(L)}}dT$

${\hat {H_{2}}}=\int \limits _{25C}^{80C}(C_{p})_{Ethanol_{(L)}}dT$

${\hat {H_{3}}}=\int \limits _{25C}^{100C}(C_{p})_{H_{2}O_{(L)}}dT+(\Delta {\hat {H}})_{V,H_{2}O}(100C)+\int \limits _{100C}^{80C}(C_{p})_{H_{2}O_{(V)}}dT$

${\hat {H_{4}}}=\int \limits _{25C}^{100C}(C_{p})_{Ethanol_{(L)}}dT+(\Delta {\hat {H}})_{V,Ethanol}(100C)+\int \limits _{100C}^{80C}(C_{p})_{Ethanol_{(V)}}dT$