To solve the energy balance, must calculate the enthalpy of the flow streams and system.
Enthalpy is affected by:
Temperature changes are the most common and obvious way to observe that the energy stored in matter has changed.
sensible heat is when only temperature changes enthalpy
Recall the general energy balance: \[\frac{d(m_{sys}H_{sys})}{dt} = \sum \dot{m}_{in}H_{in} - \sum \dot{m}_{out}H_{out} + \dot{Q} + \dot{W}\]
To solve it, \(H\) of each stream and the system is needed.
How much does the enthalpy change for a given temperature change?
The heat capacity at constant pressure is defined as: \[C_P \equiv \left(\frac{\partial H}{\partial T}\right)_P\]
Key Term
Heat capacity (\(C_P\)) — the enthalpy change per unit temperature change at constant pressure. Units: J kg\(^{-1}\) K\(^{-1}\) or J mol\(^{-1}\) K\(^{-1}\).
Enthalpy depends only very weakly on pressure.
Therefore can drop the partial derivative: \[C_P = \frac{dH}{dT}\]
For ideal gases, \(H^{ig}\) only depends on \(T\), therefore \(C_P^{ig}\) also only depends on \(T\)
\[\frac{dH}{dT}=C_P\]
\[dH = C_P dT\]
\[\int_{H_1}^{H_2} dH = \int_{T_1}^{T_2} C_P\,dT\]
\[\Delta H = \int_{T_1}^{T_2} C_P\,dT\]
Assuming \(C_P\) is constant over the temperature range: \[\Delta H = \int_{T_1}^{T_2} C_P dT = C_P\int_{T_1}^{T_2} dT = C_P(T_2 - T_1)\]
Important
\(C_P\) is never truly constant with temperature — but assuming so is a good first approximation, especially over small temperature ranges.
\(C_P\) can have mass or molar basis — make sure it matches the units in the balance equation.
Values at 25 \(^\circ\)C:
| Material | \(C_P\) (J g\(^{-1}\) K\(^{-1}\)) | \(C_P\) (J mol\(^{-1}\) K\(^{-1}\)) |
|---|---|---|
| Liquid Water | 4.18 | 75.3 |
| Water vapour | 2.0 | 36.0 |
| Ethanol | 2.44 | 112 |
| Air | 1.01 | 29.2 |
| Carbon dioxide | 0.84 | 36.9 |
| Iron | 0.45 | 25.1 |
| Aluminium | 0.87 | 23.5 |
Rule of thumb: almost any solid has a \(C_P \approx 25\) J mol\(^{-1}\) K\(^{-1}\)
In reality, \(C_P\) changes as the temperature changes.
Heat capacity data is usually tabulated as a polynomial in \(T\): \[\frac{C_P}{R} = A + BT + CT^2 + DT^{-2}\]
Another equation is: \[C_P = a + bT + cT^2 + dT^3\] Same idea, different constants.
Substitute the polynomial into \(\Delta H = \int C_P\,dT\) and integrate term by term: \[\Delta H = \int_{T_1}^{T_2}\left(a + bT + cT^2\right)dT\]
\[\Delta H = a(T_2 - T_1) + \frac{b}{2}\left(T_2^2 - T_1^2\right) + \frac{c}{3}\left(T_2^3 - T_1^3\right)\]
This is exact, but a hassle.
The algebra can be “hidden” by defining the integral-weighted mean heat capacity: \[\langle C_P \rangle = \frac{1}{T_2 - T_1}\int_{T_1}^{T_2} C_P\,dT\]
\[\Delta H = \langle C_P \rangle (T_2 - T_1)\]
Key Term
Mean (integral-weighted) heat capacity (\(\langle C_P \rangle\)) — the single constant \(C_P\) that gives the correct enthalpy change over a temperature interval.
To calculate an enthalpy relative to a reference temperature: \[H = H_{ref} + \int_{T_{ref}}^{T} C_P dT\]
Setting \(H_{ref} = 0\): \[H = \int_{T_{ref}}^{T} C_P\,dT = \langle C_P \rangle\,(T - T_{ref})\]
Use the same reference state for every term in the balance
Raw milk is pumped continuously through a heater on its way to a pasteuriser. The milk enters at 4 \(^\circ\)C and leaves at 72 \(^\circ\)C. The volumetric flow rate is 5000 L hr\(^{-1}\).
Calculate the rate of heat transfer required.
\(\rho_{milk} = 1030\) kg m\(^{-3}\), \(C_{P,milk} = 3.85\) kJ kg\(^{-1}\) K\(^{-1}\)
Hot flue gas leaving a furnace is passed through a heat exchanger to recover heat into a thermal oil loop.
The flue gas enters at 320 \(^\circ\)C at 2.5 kg s\(^{-1}\). The thermal oil enters at 60 \(^\circ\)C and must leave at 180 \(^\circ\)C.
Determine the heat transferred and the required thermal oil circulation rate.
\(C_{P,gas} = 1.15\) kJ kg\(^{-1}\) K\(^{-1}\), \(C_{P,oil} = 2.30\) kJ kg\(^{-1}\) K\(^{-1}\)
Canola oil (12,000 kg) at 15 \(^\circ\)C is heated to 60 \(^\circ\)C before a biodiesel reaction. The tank is well insulated.
(a) Calculate the heating time when heated by a 150kW electric immersion heater.
(b) Calculate the heating time when using a steam coil heater at 120 \(^\circ\)C where: \[\dot{Q} = UA\left(T_{steam} - T\right)\]
\(UA = 4.0\) kW K\(^{-1}\)
\(C_{P,oil} = 1.95\) kJ kg\(^{-1}\) K\(^{-1}\).
A stream of 200 kg/h of air is cooled from 150 \(^\circ\)C to 40 \(^\circ\)C at constant pressure with no phase change.
Determine the cooling requirement in kW.
\(C_P = 1.005\ \text{kJ}\,\text{kg}^{-1}\,\text{K}^{-1}\)