Energy Balances

Energy Balances – p.1/14

Energy Balances

One man’s magic is another man’s engineering. Lazarus Long Temperature and pressure are already fixed in the mass balance calculations of many unit operations due to the assumption of saturated streams.

An energy balance calculation is necessary to determine the heating and cooling duties in the flowsheet.

An estimate of the heat content of each stream can be rapidly determined by making the following assumptions:

1. The vapor and liquid streams have ideal properties.

2. Kinetic and potential energies of process streams are neglected and only enthalpy changes are considered.

3. Standard enthalpy reference (ΔH = 0) is at

P_{0} = 1 atm, T_{0} = 298 K and elemental species.

Energy Balances – p.3/14

Temperature and pressure are already fixed in the mass balance calculations of many unit operations due to the assumption of saturated streams.

An energy balance calculation is necessary to determine the heating and cooling duties in the flowsheet.

An estimate of the heat content of each stream can be rapidly determined by making the following assumptions:

1. The vapor and liquid streams have ideal properties.

2. Kinetic and potential energies of process streams are neglected and only enthalpy changes are considered.

3. Standard enthalpy reference (ΔH = 0) is at

P_{0} = 1 atm, T_{0} = 298 K and elemental species.

Energy Balances – p.3/14

Temperature and pressure are already fixed in the mass balance calculations of many unit operations due to the assumption of saturated streams.

An energy balance calculation is necessary to determine the heating and cooling duties in the flowsheet.

An estimate of the heat content of each stream can be rapidly determined by making the following assumptions:

1. The vapor and liquid streams have ideal properties.

2. Kinetic and potential energies of process streams are neglected and only enthalpy changes are considered.

3. Standard enthalpy reference (ΔH = 0) is at

P_{0} = 1 atm, T_{0} = 298 K and elemental species.

Energy Balances – p.3/14

Temperature and pressure are already fixed in the mass balance calculations of many unit operations due to the assumption of saturated streams.

An energy balance calculation is necessary to determine the heating and cooling duties in the flowsheet.

An estimate of the heat content of each stream can be rapidly determined by making the following assumptions:

1. The vapor and liquid streams have ideal properties.

2. Kinetic and potential energies of process streams are neglected and only enthalpy changes are considered.

3. Standard enthalpy reference (ΔH = 0) is at

P_{0} = 1 atm, T_{0} = 298 K and elemental species.

Energy Balances – p.3/14

Temperature and pressure are already fixed in the mass balance calculations of many unit operations due to the assumption of saturated streams.

An energy balance calculation is necessary to determine the heating and cooling duties in the flowsheet.

An estimate of the heat content of each stream can be rapidly determined by making the following assumptions:

1. The vapor and liquid streams have ideal properties.

2. Kinetic and potential energies of process streams are neglected and only enthalpy changes are considered.

3. Standard enthalpy reference (ΔH = 0) is at

P_{0} = 1 atm, T_{0} = 298 K and elemental species.

Energy Balances – p.3/14

Temperature and pressure are already fixed in the mass balance calculations of many unit operations due to the assumption of saturated streams.

An energy balance calculation is necessary to determine the heating and cooling duties in the flowsheet.

An estimate of the heat content of each stream can be rapidly determined by making the following assumptions:

1. The vapor and liquid streams have ideal properties.

2. Kinetic and potential energies of process streams are neglected and only enthalpy changes are considered.

3. Standard enthalpy reference (ΔH = 0) is at

P_{0} = 1 atm, T_{0} = 298 K and elemental species.

Energy Balances – p.3/14

Enthalpies of Vapor Streams

Using the ideal properties assumption, ΔH_{p} = 0. Thus, ΔH_{v} = ΔH_{f} + ΔH_{T} ΔH_{f} = y_{1}H_{f,1}(T_{0}) + ... + y_{n}H_{f,n}(T_{0}) = ∑ y_{k}H_{f,k}(T_{0}) where H_{f,k}(T_{0}) is the heat of formation of component k at temperature T_{0}.

ΔH_{T} = y_{1}∫_{T_{0}}^{T} c_{p,1}^{0}dT + ... + y_{n}∫_{T_{0}}^{T} c_{p,n}^{0}dT = ∑ y_{k}∫_{T_{0}}^{T} c_{p,k}^{0}dT

where c_{p,k}^{0} is the temperature dependent heat capacity of

component k.

Energy Balances – p.4/14

Using the ideal properties assumption, ΔH_{p} = 0. Thus, ΔH_{v} = ΔH_{f} + ΔH_{T} ΔH_{f} = y_{1}H_{f,1}(T_{0}) + ... + y_{n}H_{f,n}(T_{0}) = ∑ y_{k}H_{f,k}(T_{0}) where H_{f,k}(T_{0}) is the heat of formation of component k at temperature T_{0}.

ΔH_{T} = y_{1}∫_{T_{0}}^{T} c_{p,1}^{0}dT + ... + y_{n}∫_{T_{0}}^{T} c_{p,n}^{0}dT = ∑ y_{k}∫_{T_{0}}^{T} c_{p,k}^{0}dT

where c_{p,k}^{0} is the temperature dependent heat capacity of

component k.

Energy Balances – p.4/14

Using the ideal properties assumption, ΔH_{p} = 0. Thus, ΔH_{v} = ΔH_{f} + ΔH_{T} ΔH_{f} = y_{1}H_{f,1}(T_{0}) + ... + y_{n}H_{f,n}(T_{0}) = ∑ y_{k}H_{f,k}(T_{0}) where H_{f,k}(T_{0}) is the heat of formation of component k at temperature T_{0}.

ΔH_{T} = y_{1}∫_{T_{0}}^{T} c_{p,1}^{0}dT + ... + y_{n}∫_{T_{0}}^{T} c_{p,n}^{0}dT = ∑ y_{k}∫_{T_{0}}^{T} c_{p,k}^{0}dT

where c_{p,k}^{0} is the temperature dependent heat capacity of

component k.

Energy Balances – p.4/14

Using the ideal properties assumption, ΔH_{p} = 0. Thus, ΔH_{v} = ΔH_{f} + ΔH_{T} ΔH_{f} = y_{1}H_{f,1}(T_{0}) + ... + y_{n}H_{f,n}(T_{0}) = ∑ y_{k}H_{f,k}(T_{0}) where H_{f,k}(T_{0}) is the heat of formation of component k at temperature T_{0}.

ΔH_{T} = y_{1}∫_{T_{0}}^{T} c_{p,1}^{0}dT + ... + y_{n}∫_{T_{0}}^{T} c_{p,n}^{0}dT = ∑ y_{k}∫_{T_{0}}^{T} c_{p,k}^{0}dT

where c_{p,k}^{0} is the temperature dependent heat capacity of

component k.

Energy Balances – p.4/14

Using the ideal properties assumption, ΔH_{p} = 0. Thus, ΔH_{v} = ΔH_{f} + ΔH_{T} ΔH_{f} = y_{1}H_{f,1}(T_{0}) + ... + y_{n}H_{f,n}(T_{0}) = ∑ y_{k}H_{f,k}(T_{0}) where H_{f,k}(T_{0}) is the heat of formation of component k at temperature T_{0}.

ΔH_{T} = y_{1}∫_{T_{0}}^{T} c_{p,1}^{0}dT + ... + y_{n}∫_{T_{0}}^{T} c_{p,n}^{0}dT = ∑ y_{k}∫_{T_{0}}^{T} c_{p,k}^{0}dT

where c_{p,k}^{0} is the temperature dependent heat capacity of

component k.

Energy Balances – p.4/14

Using the ideal properties assumption, ΔH_{p} = 0. Thus, ΔH_{v} = ΔH_{f} + ΔH_{T} ΔH_{f} = y_{1}H_{f,1}(T_{0}) + ... + y_{n}H_{f,n}(T_{0}) = ∑ y_{k}H_{f,k}(T_{0}) where H_{f,k}(T_{0}) is the heat of formation of component k at temperature T_{0}.

ΔH_{T} = y_{1}∫_{T_{0}}^{T} c_{p,1}^{0}dT + ... + y_{n}∫_{T_{0}}^{T} c_{p,n}^{0}dT = ∑ y_{k}∫_{T_{0}}^{T} c_{p,k}^{0}dT

where c_{p,k}^{0} is the temperature dependent heat capacity of

component k.

Energy Balances – p.4/14

Using the ideal properties assumption, ΔH_{p} = 0. Thus, ΔH_{v} = ΔH_{f} + ΔH_{T} ΔH_{f} = y_{1}H_{f,1}(T_{0}) + ... + y_{n}H_{f,n}(T_{0}) = ∑ y_{k}H_{f,k}(T_{0}) where H_{f,k}(T_{0}) is the heat of formation of component k at temperature T_{0}.

ΔH_{T} = y_{1}∫_{T_{0}}^{T} c_{p,1}^{0}dT + ... + y_{n}∫_{T_{0}}^{T} c_{p,n}^{0}dT = ∑ y_{k}∫_{T_{0}}^{T} c_{p,k}^{0}dT

where c_{p,k}^{0} is the temperature dependent heat capacity of

component k.

Energy Balances – p.4/14

Heat Exchanger Calculation

Consider a heat exchanger with no phase change.

There is temperature change but no composition change. At steady state:

(μΔH)_{in} + Q = (μΔH)_{out}

Since composition does not change, ΔH_{f} at T_{0} is the same on both sides of heat balance and thus cancels out.

⧸ (μΔH_{f})_{in} + (μΔH_{T})_{in} + Q = ⧸ (μΔH_{f})_{in} + (μΔH_{T})_{out}

This implies:

There is temperature change but no composition change. At steady state:

(μΔH)_{in} + Q = (μΔH)_{out}

Since composition does not change, ΔH_{f} at T_{0} is the same on both sides of heat balance and thus cancels out.

⧸ (μΔH_{f})_{in} + (μΔH_{T})_{in} + Q = ⧸ (μΔH_{f})_{in} + (μΔH_{T})_{out}

This implies:

There is temperature change but no composition change. At steady state:

(μΔH)_{in} + Q = (μΔH)_{out}

Since composition does not change, ΔH_{f} at T_{0} is the same on both sides of heat balance and thus cancels out.

⧸ (μΔH_{f})_{in} + (μΔH_{T})_{in} + Q = ⧸ (μΔH_{f})_{in} + (μΔH_{T})_{out}

This implies:

There is temperature change but no composition change. At steady state:

(μΔH)_{in} + Q = (μΔH)_{out}

Since composition does not change, ΔH_{f} at T_{0} is the same on both sides of heat balance and thus cancels out.

⧸ (μΔH_{f})_{in} + (μΔH_{T})_{in} + Q = ⧸ (μΔH_{f})_{in} + (μΔH_{T})_{out}

This implies:

Reactor Calculation

Consider a gas phase reactor.

There is temperature change as well as composition change. At steady state:

[μ_{1}ΔH(T_{1}, y_{1})]_{in} + Q_{R} = [μ_{2}ΔH(T_{2}, y_{2})]_{out}

This implies

Q_{R} = μ_{2}ΔH(T_{2}, y_{2}) − μ_{1}ΔH(T_{1}, y_{1})

Q_{R} > 0 ⟹ reaction is endothermic. Q_{R} < 0 ⟹ reaction is exothermic.

Energy Balances – p.6/14

There is temperature change as well as composition change. At steady state:

[μ_{1}ΔH(T_{1}, y_{1})]_{in} + Q_{R} = [μ_{2}ΔH(T_{2}, y_{2})]_{out}

This implies

Q_{R} = μ_{2}ΔH(T_{2}, y_{2}) − μ_{1}ΔH(T_{1}, y_{1})

Q_{R} > 0 ⟹ reaction is endothermic. Q_{R} < 0 ⟹ reaction is exothermic.

Energy Balances – p.6/14

There is temperature change as well as composition change. At steady state:

[μ_{1}ΔH(T_{1}, y_{1})]_{in} + Q_{R} = [μ_{2}ΔH(T_{2}, y_{2})]_{out}

This implies

Q_{R} = μ_{2}ΔH(T_{2}, y_{2}) − μ_{1}ΔH(T_{1}, y_{1})

Q_{R} > 0 ⟹ reaction is endothermic. Q_{R} < 0 ⟹ reaction is exothermic.

Energy Balances – p.6/14

There is temperature change as well as composition change. At steady state:

[μ_{1}ΔH(T_{1}, y_{1})]_{in} + Q_{R} = [μ_{2}ΔH(T_{2}, y_{2})]_{out}

This implies

Q_{R} = μ_{2}ΔH(T_{2}, y_{2}) − μ_{1}ΔH(T_{1}, y_{1})

Q_{R} > 0 ⟹ reaction is endothermic. Q_{R} < 0 ⟹ reaction is exothermic.

Energy Balances – p.6/14

There is temperature change as well as composition change. At steady state:

[μ_{1}ΔH(T_{1}, y_{1})]_{in} + Q_{R} = [μ_{2}ΔH(T_{2}, y_{2})]_{out}

This implies

Q_{R} = μ_{2}ΔH(T_{2}, y_{2}) − μ_{1}ΔH(T_{1}, y_{1})

Q_{R} > 0 ⟹ reaction is endothermic. Q_{R} < 0 ⟹ reaction is exothermic.

Energy Balances – p.6/14

Enthalpies of Liquid Mixtures

Enthalpies of liquid mixtures are evaluated directly from the ideal vapor enthalpy and subtracting the heat of vaporization at saturation conditions.

ΔH_{L,k}(T) = ΔH_{f,k}^{0} + ∫_{T_{0}}^{T} c_{p,k}^{0}(T)dT − ΔH_{vap}^{k}

The dependence of ΔH_{vap}^{k} on temperature can be found through the Watson equation:

ΔH_{vap}^{k}(T) = ΔH_{vap}^{k}(T_{b}) [\frac{(T_{c}^{k} − T)}{(T_{c}^{k} − T_{b})}]^{0.38}

T_{c}^{k} : critical temperature T_{b}^{k} : boiling point at 1 atm ΔH_{vap}^{k}(T_{b}) heat of vaporization at T_{b}^{k}

Energy Balances – p.7/14

Enthalpies of Liquid Mixtures

Enthalpies of liquid mixtures are evaluated directly from the ideal vapor enthalpy and subtracting the heat of vaporization at saturation conditions.

ΔH_{L,k}(T) = ΔH_{f,k}^{0} + ∫_{T_{0}}^{T} c_{p,k}^{0}(T)dT − ΔH_{vap}^{k}

The dependence of ΔH_{vap}^{k} on temperature can be found through the Watson equation:

ΔH_{vap}^{k}(T) = ΔH_{vap}^{k}(T_{b}) [\frac{(T_{c}^{k} − T)}{(T_{c}^{k} − T_{b})}]^{0.38}

T_{c}^{k} : critical temperature T_{b}^{k} : boiling point at 1 atm ΔH_{vap}^{k}(T_{b}) heat of vaporization at T_{b}^{k}

Energy Balances – p.7/14

Enthalpies of Liquid Mixtures

Enthalpies of liquid mixtures are evaluated directly from the ideal vapor enthalpy and subtracting the heat of vaporization at saturation conditions.

ΔH_{L,k}(T) = ΔH_{f,k}^{0} + ∫_{T_{0}}^{T} c_{p,k}^{0}(T)dT − ΔH_{vap}^{k}

The dependence of ΔH_{vap}^{k} on temperature can be found through the Watson equation:

ΔH_{vap}^{k}(T) = ΔH_{vap}^{k}(T_{b}) [\frac{(T_{c}^{k} − T)}{(T_{c}^{k} − T_{b})}]^{0.38}

T_{c}^{k} : critical temperature T_{b}^{k} : boiling point at 1 atm ΔH_{vap}^{k}(T_{b}) heat of vaporization at T_{b}^{k}

Energy Balances – p.7/14

Enthalpies of Liquid Mixtures

Enthalpies of liquid mixtures are evaluated directly from the ideal vapor enthalpy and subtracting the heat of vaporization at saturation conditions.

ΔH_{L,k}(T) = ΔH_{f,k}^{0} + ∫_{T_{0}}^{T} c_{p,k}^{0}(T)dT − ΔH_{vap}^{k}

The dependence of ΔH_{vap}^{k} on temperature can be found through the Watson equation:

ΔH_{vap}^{k}(T) = ΔH_{vap}^{k}(T_{b}) [\frac{(T_{c}^{k} − T)}{(T_{c}^{k} − T_{b})}]^{0.38}

T_{c}^{k} : critical temperature T_{b}^{k} : boiling point at 1 atm ΔH_{vap}^{k}(T_{b}) heat of vaporization at T_{b}^{k}

Energy Balances – p.7/14

Enthalpies of Liquid Mixtures

Enthalpies of liquid mixtures are evaluated directly from the ideal vapor enthalpy and subtracting the heat of vaporization at saturation conditions.

ΔH_{L,k}(T) = ΔH_{f,k}^{0} + ∫_{T_{0}}^{T} c_{p,k}^{0}(T)dT − ΔH_{vap}^{k}

The dependence of ΔH_{vap}^{k} on temperature can be found through the Watson equation:

ΔH_{vap}^{k}(T) = ΔH_{vap}^{k}(T_{b}) [\frac{(T_{c}^{k} − T)}{(T_{c}^{k} − T_{b})}]^{0.38}

T_{c}^{k} : critical temperature T_{b}^{k} : boiling point at 1 atm ΔH_{vap}^{k}(T_{b}) heat of vaporization at T_{b}^{k}

Energy Balances – p.7/14

Enthalpies of Liquid Mixtures

Enthalpies of liquid mixtures are evaluated directly from the ideal vapor enthalpy and subtracting the heat of vaporization at saturation conditions.

ΔH_{L,k}(T) = ΔH_{f,k}^{0} + ∫_{T_{0}}^{T} c_{p,k}^{0}(T)dT − ΔH_{vap}^{k}

The dependence of ΔH_{vap}^{k} on temperature can be found through the Watson equation:

ΔH_{vap}^{k}(T) = ΔH_{vap}^{k}(T_{b}) [\frac{(T_{c}^{k} − T)}{(T_{c}^{k} − T_{b})}]^{0.38}

T_{c}^{k} : critical temperature T_{b}^{k} : boiling point at 1 atm ΔH_{vap}^{k}(T_{b}) heat of vaporization at T_{b}^{k}

Energy Balances – p.7/14

Liquid Mixtures and Two Phase Mixtures

The specific stream enthalpy for liquid mixtures is estimated as a weighted sum of respective mole fractions. ΔH_{L}(T, x) = x_{1}ΔH_{L}^{1} + x_{2}ΔH_{L}^{2} + ... + x_{n}ΔH_{L}^{n}

For a vapor-liquid mixture with vapor fraction ϕ, the stream enthalpy is given by the weighted sum of the liquid and vapor enthalpies. ΔH_{L}(T, x) = ϕΔH_{V} (T, y) + (1 − ϕ)ΔH_{L}(T, x)

Liquid Mixtures and Two Phase Mixtures

The specific stream enthalpy for liquid mixtures is estimated as a weighted sum of respective mole fractions. ΔH_{L}(T, x) = x_{1}ΔH_{L}^{1} + x_{2}ΔH_{L}^{2} + ... + x_{n}ΔH_{L}^{n}

For a vapor-liquid mixture with vapor fraction ϕ, the stream enthalpy is given by the weighted sum of the liquid and vapor enthalpies. ΔH_{L}(T, x) = ϕΔH_{V} (T, y) + (1 − ϕ)ΔH_{L}(T, x)

Flash Calculations

We considered the following scenarios in flash calculations

T_{V} , P_{V} specified  Isothermal flash
V/F = 0, P_{L} specified Bubble point temperature
V/F = 1, P_{V} specified Dew point temperature
T_{L} specified, V/F = 0 Bubble point pressure
T_{V} specified, V/F = 0 Dew point pressure
Q = 0, P_{V} specified  Adiabatic flash
Q, P_{V} specified  Nonadiabatic flash
V/F, P_{V} specified  Percent vaporization flash

We did not consider the flash calculations that involved energy balance. We will consider the adiabatic flash now.

Energy Balances – p.9/14

Flash Calculations

We considered the following scenarios in flash calculations

T_{V} , P_{V} specified  Isothermal flash
V/F = 0, P_{L} specified Bubble point temperature
V/F = 1, P_{V} specified Dew point temperature
T_{L} specified, V/F = 0 Bubble point pressure
T_{V} specified, V/F = 0 Dew point pressure
Q = 0, P_{V} specified  Adiabatic flash
Q, P_{V} specified  Nonadiabatic flash
V/F, P_{V} specified  Percent vaporization flash

We did not consider the flash calculations that involved energy balance. We will consider the adiabatic flash now.

Energy Balances – p.9/14

Flash Calculations

We considered the following scenarios in flash calculations

T_{V} , P_{V} specified  Isothermal flash
V/F = 0, P_{L} specified Bubble point temperature
V/F = 1, P_{V} specified Dew point temperature
T_{L} specified, V/F = 0 Bubble point pressure
T_{V} specified, V/F = 0 Dew point pressure
Q = 0, P_{V} specified  Adiabatic flash
Q, P_{V} specified  Nonadiabatic flash
V/F, P_{V} specified  Percent vaporization flash

We did not consider the flash calculations that involved energy balance. We will consider the adiabatic flash now.

Energy Balances – p.9/14

Nonadiabatic/Adiabatic Flash

\underline{Scenario} : Heat is added to a liquid mixture stream due to which part of it vaporizes. What is the flash temperature and what are the compositions of the liquid and vapor streams if the pressure is specified? \underline{Algorithm}:

For a given enthalpy specification (ΔH_{spec}) and pressure P, calculate the bubble point and dew point temperatures and the enthalpies associated with them. If ΔH_{spec} > ΔH_{dew}, then the mixture is all vapor, and we solve for T from ΔH_{v}(T) = ΔH_{spec}. If ΔH_{spec} < ΔH_{dew}, then the mixture is all liquid, and we solve for T from ΔH_{L}(T) = ΔH_{spec}.

\underline{Scenario} : Heat is added to a liquid mixture stream due to which part of it vaporizes. What is the flash temperature and what are the compositions of the liquid and vapor streams if the pressure is specified? \underline{Algorithm}:

For a given enthalpy specification (ΔH_{spec}) and pressure P, calculate the bubble point and dew point temperatures and the enthalpies associated with them. If ΔH_{spec} > ΔH_{dew}, then the mixture is all vapor, and we solve for T from ΔH_{v}(T) = ΔH_{spec}. If ΔH_{spec} < ΔH_{dew}, then the mixture is all liquid, and we solve for T from ΔH_{L}(T) = ΔH_{spec}.

\underline{Scenario} : Heat is added to a liquid mixture stream due to which part of it vaporizes. What is the flash temperature and what are the compositions of the liquid and vapor streams if the pressure is specified? \underline{Algorithm}:

For a given enthalpy specification (ΔH_{spec}) and pressure P, calculate the bubble point and dew point temperatures and the enthalpies associated with them. If ΔH_{spec} > ΔH_{dew}, then the mixture is all vapor, and we solve for T from ΔH_{v}(T) = ΔH_{spec}. If ΔH_{spec} < ΔH_{dew}, then the mixture is all liquid, and we solve for T from ΔH_{L}(T) = ΔH_{spec}.

\underline{Scenario} : Heat is added to a liquid mixture stream due to which part of it vaporizes. What is the flash temperature and what are the compositions of the liquid and vapor streams if the pressure is specified? \underline{Algorithm}:

For a given enthalpy specification (ΔH_{spec}) and pressure P, calculate the bubble point and dew point temperatures and the enthalpies associated with them. If ΔH_{spec} > ΔH_{dew}, then the mixture is all vapor, and we solve for T from ΔH_{v}(T) = ΔH_{spec}. If ΔH_{spec} < ΔH_{dew}, then the mixture is all liquid, and we solve for T from ΔH_{L}(T) = ΔH_{spec}.

\underline{Scenario} : Heat is added to a liquid mixture stream due to which part of it vaporizes. What is the flash temperature and what are the compositions of the liquid and vapor streams if the pressure is specified? \underline{Algorithm}:

For a given enthalpy specification (ΔH_{spec}) and pressure P, calculate the bubble point and dew point temperatures and the enthalpies associated with them. If ΔH_{spec} > ΔH_{dew}, then the mixture is all vapor, and we solve for T from ΔH_{v}(T) = ΔH_{spec}. If ΔH_{spec} < ΔH_{dew}, then the mixture is all liquid, and we solve for T from ΔH_{L}(T) = ΔH_{spec}.

\underline{Scenario} : Heat is added to a liquid mixture stream due to which part of it vaporizes. What is the flash temperature and what are the compositions of the liquid and vapor streams if the pressure is specified? \underline{Algorithm}:

For a given enthalpy specification (ΔH_{spec}) and pressure P, calculate the bubble point and dew point temperatures and the enthalpies associated with them. If ΔH_{spec} > ΔH_{dew}, then the mixture is all vapor, and we solve for T from ΔH_{v}(T) = ΔH_{spec}. If ΔH_{spec} < ΔH_{dew}, then the mixture is all liquid, and we solve for T from ΔH_{L}(T) = ΔH_{spec}.

If ΔH_{dew} ≥ ΔH_{spec} ≥ ΔH_{bub}, guess ξ_{n} (or ϕ).

Perform a flash calculation with ξ_{n} (or ϕ) and P specified to obtain y_{k}, x_{k} and T. Calculate ΔH(T) = ϕΔH_{v} + (1 − ϕ)ΔH_{L}

If f = ΔH_{spec} − ΔH(T) = 0 STOP. Otherwise, if f > 0 reguess a higher ξ_{n} (or ϕ), else guess a lower ξ_{n} (or ϕ) and go back to previous step.

Perform a flash calculation with ξ_{n} (or ϕ) and P specified to obtain y_{k}, x_{k} and T. Calculate ΔH(T) = ϕΔH_{v} + (1 − ϕ)ΔH_{L}

If f = ΔH_{spec} − ΔH(T) = 0 STOP. Otherwise, if f > 0 reguess a higher ξ_{n} (or ϕ), else guess a lower ξ_{n} (or ϕ) and go back to previous step.

Illustrative Example

Consider a 50-50 liquid mixture of benzene and toluene flowing at 100 gmol/s at 300 K and 1 bar. If heat is added to this stream at the rate of 860.42 kcal/s, what is the temperature of the mixture. \underline{Solution}

Compute the enthalpy, ΔH_{L}, for the incoming liquid stream.

ΔH_{L} = −847.5 kcal/s

Heat added to the liquid stream: 860.42 kcal/s. Thus the outlet stream enthalpy is:

Consider a 50-50 liquid mixture of benzene and toluene flowing at 100 gmol/s at 300 K and 1 bar. If heat is added to this stream at the rate of 860.42 kcal/s, what is the temperature of the mixture. \underline{Solution}

Compute the enthalpy, ΔH_{L}, for the incoming liquid stream.

ΔH_{L} = −847.5 kcal/s

Heat added to the liquid stream: 860.42 kcal/s. Thus the outlet stream enthalpy is:

Consider a 50-50 liquid mixture of benzene and toluene flowing at 100 gmol/s at 300 K and 1 bar. If heat is added to this stream at the rate of 860.42 kcal/s, what is the temperature of the mixture. \underline{Solution}

Compute the enthalpy, ΔH_{L}, for the incoming liquid stream.

ΔH_{L} = −847.5 kcal/s

Heat added to the liquid stream: 860.42 kcal/s. Thus the outlet stream enthalpy is:

860.42 − 847.50 = 12.868 kcal/s

Consider a 50-50 liquid mixture of benzene and toluene flowing at 100 gmol/s at 300 K and 1 bar. If heat is added to this stream at the rate of 860.42 kcal/s, what is the temperature of the mixture. \underline{Solution}

Compute the enthalpy, ΔH_{L}, for the incoming liquid stream.

ΔH_{L} = −847.5 kcal/s

Heat added to the liquid stream: 860.42 kcal/s. Thus the outlet stream enthalpy is:

860.42 − 847.50 = 12.868 kcal/s

It can be shown that the outlet stream enthalpy is between the enthalpy of saturated liquid and saturated vapor; thus the mixture has two phases.

We need to compute the split fractions of benzene and toluene in the vapor stream and the temperature so that the enthalpy of the outlet stream is 12.868 kcal/s

Assume toluene to be the key component and guess T = 370 K. Compute α_{B/T} = 2.453 at T = 370 K.

Guess ξ_{T} and solve for ξ_{B}, ϕ and T from:

\begin{aligned}ξ_{B} & = & \frac{α_{B/T} ξ_{T}}{1 + (α_{B/T} − 1)ξ_{T}} \\ ϕ & = & \frac{50(ξ_{B} + ξ_{T})}{100} \\ P_{T}^{0}(T) & = & \frac{P}{ᾱ} \end{aligned}

It can be shown that the outlet stream enthalpy is between the enthalpy of saturated liquid and saturated vapor; thus the mixture has two phases.

We need to compute the split fractions of benzene and toluene in the vapor stream and the temperature so that the enthalpy of the outlet stream is 12.868 kcal/s

Assume toluene to be the key component and guess T = 370 K. Compute α_{B/T} = 2.453 at T = 370 K.

Guess ξ_{T} and solve for ξ_{B}, ϕ and T from:

\begin{aligned}ξ_{B} & = & \frac{α_{B/T} ξ_{T}}{1 + (α_{B/T} − 1)ξ_{T}} \\ ϕ & = & \frac{50(ξ_{B} + ξ_{T})}{100} \\ P_{T}^{0}(T) & = & \frac{P}{ᾱ} \end{aligned}

It can be shown that the outlet stream enthalpy is between the enthalpy of saturated liquid and saturated vapor; thus the mixture has two phases.

We need to compute the split fractions of benzene and toluene in the vapor stream and the temperature so that the enthalpy of the outlet stream is 12.868 kcal/s

Assume toluene to be the key component and guess T = 370 K. Compute α_{B/T} = 2.453 at T = 370 K.

Guess ξ_{T} and solve for ξ_{B}, ϕ and T from:

\begin{aligned}ξ_{B} & = & \frac{α_{B/T} ξ_{T}}{1 + (α_{B/T} − 1)ξ_{T}} \\ ϕ & = & \frac{50(ξ_{B} + ξ_{T})}{100} \\ P_{T}^{0}(T) & = & \frac{P}{ᾱ} \end{aligned}

It can be shown that the outlet stream enthalpy is between the enthalpy of saturated liquid and saturated vapor; thus the mixture has two phases.

We need to compute the split fractions of benzene and toluene in the vapor stream and the temperature so that the enthalpy of the outlet stream is 12.868 kcal/s

Assume toluene to be the key component and guess T = 370 K. Compute α_{B/T} = 2.453 at T = 370 K.

Guess ξ_{T} and solve for ξ_{B}, ϕ and T from:

\begin{aligned}ξ_{B} & = & \frac{α_{B/T} ξ_{T}}{1 + (α_{B/T} − 1)ξ_{T}} \\ ϕ & = & \frac{50(ξ_{B} + ξ_{T})}{100} \\ P_{T}^{0}(T) & = & \frac{P}{ᾱ} \end{aligned}

With this information, calculate:

ΔH(T) = ϕΔH_{V} (T) + (1 − ϕ)ΔH_{L}(T)

and compare with ΔH_{spec} = 12.862. Adjust ξ_{T} till the computed ΔH converges to the specified value.

\begin{matrix}ξ_{T} & ξ_{B} & T & ϕ & ΔH(T) \\ 0.700 & 0.851 & 370.1 & 0.776 & 68.439 \\ 0.600 & 0.786 & 369.5 & 0.693 & 20.876 \\ 0.570 & 0.765 & 369.4 & 0.667 & 4.895 \\ 0.585 & 0.776 & 369.5 & 0.680 & 12.993 \end{matrix}

Thus, the stream is 68% vaporized at a temperature of 369.5 K.

Energy Balances – p.14/14

With this information, calculate:

ΔH(T) = ϕΔH_{V} (T) + (1 − ϕ)ΔH_{L}(T)

and compare with ΔH_{spec} = 12.862. Adjust ξ_{T} till the computed ΔH converges to the specified value.

\begin{matrix}ξ_{T} & ξ_{B} & T & ϕ & ΔH(T) \\ 0.700 & 0.851 & 370.1 & 0.776 & 68.439 \\ 0.600 & 0.786 & 369.5 & 0.693 & 20.876 \\ 0.570 & 0.765 & 369.4 & 0.667 & 4.895 \\ 0.585 & 0.776 & 369.5 & 0.680 & 12.993 \end{matrix}

Thus, the stream is 68% vaporized at a temperature of 369.5 K.

Energy Balances – p.14/14

With this information, calculate:

ΔH(T) = ϕΔH_{V} (T) + (1 − ϕ)ΔH_{L}(T)

and compare with ΔH_{spec} = 12.862. Adjust ξ_{T} till the computed ΔH converges to the specified value.

\begin{matrix}ξ_{T} & ξ_{B} & T & ϕ & ΔH(T) \\ 0.700 & 0.851 & 370.1 & 0.776 & 68.439 \\ 0.600 & 0.786 & 369.5 & 0.693 & 20.876 \\ 0.570 & 0.765 & 369.4 & 0.667 & 4.895 \\ 0.585 & 0.776 & 369.5 & 0.680 & 12.993 \end{matrix}

Thus, the stream is 68% vaporized at a temperature of 369.5 K.

Energy Balances – p.14/14

With this information, calculate:

ΔH(T) = ϕΔH_{V} (T) + (1 − ϕ)ΔH_{L}(T)

and compare with ΔH_{spec} = 12.862. Adjust ξ_{T} till the computed ΔH converges to the specified value.

\begin{matrix}ξ_{T} & ξ_{B} & T & ϕ & ΔH(T) \\ 0.700 & 0.851 & 370.1 & 0.776 & 68.439 \\ 0.600 & 0.786 & 369.5 & 0.693 & 20.876 \\ 0.570 & 0.765 & 369.4 & 0.667 & 4.895 \\ 0.585 & 0.776 & 369.5 & 0.680 & 12.993 \end{matrix}

Thus, the stream is 68% vaporized at a temperature of 369.5 K.

Energy Balances – p.14/14

With this information, calculate:

ΔH(T) = ϕΔH_{V} (T) + (1 − ϕ)ΔH_{L}(T)

and compare with ΔH_{spec} = 12.862. Adjust ξ_{T} till the computed ΔH converges to the specified value.

\begin{matrix}ξ_{T} & ξ_{B} & T & ϕ & ΔH(T) \\ 0.700 & 0.851 & 370.1 & 0.776 & 68.439 \\ 0.600 & 0.786 & 369.5 & 0.693 & 20.876 \\ 0.570 & 0.765 & 369.4 & 0.667 & 4.895 \\ 0.585 & 0.776 & 369.5 & 0.680 & 12.993 \end{matrix}

Thus, the stream is 68% vaporized at a temperature of 369.5 K.

Energy Balances – p.14/14

With this information, calculate:

ΔH(T) = ϕΔH_{V} (T) + (1 − ϕ)ΔH_{L}(T)

and compare with ΔH_{spec} = 12.862. Adjust ξ_{T} till the computed ΔH converges to the specified value.

\begin{matrix}ξ_{T} & ξ_{B} & T & ϕ & ΔH(T) \\ 0.700 & 0.851 & 370.1 & 0.776 & 68.439 \\ 0.600 & 0.786 & 369.5 & 0.693 & 20.876 \\ 0.570 & 0.765 & 369.4 & 0.667 & 4.895 \\ 0.585 & 0.776 & 369.5 & 0.680 & 12.993 \end{matrix}

Thus, the stream is 68% vaporized at a temperature of 369.5 K.

Energy Balances – p.14/14

With this information, calculate:

ΔH(T) = ϕΔH_{V} (T) + (1 − ϕ)ΔH_{L}(T)

and compare with ΔH_{spec} = 12.862. Adjust ξ_{T} till the computed ΔH converges to the specified value.

\begin{matrix}ξ_{T} & ξ_{B} & T & ϕ & ΔH(T) \\ 0.700 & 0.851 & 370.1 & 0.776 & 68.439 \\ 0.600 & 0.786 & 369.5 & 0.693 & 20.876 \\ 0.570 & 0.765 & 369.4 & 0.667 & 4.895 \\ 0.585 & 0.776 & 369.5 & 0.680 & 12.993 \end{matrix}

Thus, the stream is 68% vaporized at a temperature of 369.5 K.

Energy Balances – p.14/14