This notebook contains course material from CBE20255 by Jeffrey Kantor (jeff at nd.edu); the content is available on Github. The text is released under the CC-BY-NC-ND-4.0 license, and code is released under the MIT license.

# Basic Energy Computations¶

## Computing Enthalpy and Internal Energy Changes for Common Situations¶

Internal energy ($U$) and enthalpy ($H = U + PV$) are thermodynamic state variables. We can use this property to compute changes in internal energy or enthalpy due to changes in pressure, temperature, phase, composition, and mixing/solution. The following table presents basic formulas for these calculations.

Change in $\Delta\hat{H}=\Delta\hat{U} + P\Delta\hat{V}$ $\Delta\hat{U}$ Comments
Pressure ~ 0 (gas)
~$\hat{V}\Delta P$ (solid or liquid)
~ 0 Generally neglected except for large pressure changes.
Temperature $\int_{T_1}^{T_2} C_p(T) dT$

$\approx \bar{C}_p(T_2 - T_1)$
$\int_{T_1}^{T_2} C_v(T)dT$

$\approx \bar{C}_v(T_2 - T_1)$
Expressions available for $C_p(T)$
$C_p \approx C_v + R$ (gases)
$C_p \approx C_v$ (liquids and solids)
Phase $\Delta\hat{H}_{vap}$ (liquid to vapor)
$\Delta\hat{H}_{m}$ (solid to liquid)
$\Delta\hat{U}_{vap}\approx\Delta\hat{H}_{vap}-RT_b$
$\Delta\hat{U}_m\approx\Delta\hat{H}_m$
Composition due
to Reaction
$\Delta\hat{H}^\circ_r =\sum_i \nu_i \Delta\hat{H}^\circ_{f,i}$
$\Delta\hat{H}^\circ_r = -\sum_i \nu_i \Delta\hat{H}^\circ_{c,i}$
$\Delta\hat{U}_r \approx \Delta\hat{H}_r - RT \Delta n_r$
$\Delta\hat{U}_r \approx \Delta\hat{H}_r$ (solid or liquid)
$\Delta n_r$ is the cange in moles due to reaction
Standard conditions are 25$^\circ$C and 1 atm.
Be sure all data uses same standard conditions.
Composition due
to Mixing/Sol'n
$\Delta\hat{H}_{soln}$
$\Delta\hat{H}_{mix}$
$\Delta\hat{U}_{soln} \approx \Delta\hat{H}_{soln}$
$\Delta\hat{U}_{mix} \approx \Delta\hat{H}_{mix}$
Important for non-ideal mixtures.
Typical units are per mole of solute, not solution.

## Examples¶

### Pumping a Fluid¶

For a particular fire-fighting situation, it is determined that 1,250 gpm is required. The fire hydrant will supply sufficient water at a pressure of 35 psig. A pressure of 180 psig is needed to reach the top of the 212 foot bulding. What size engine (in Hp) is required to power the fire pump?

In :
Vdot = 1250/264.172/60           # flow in m**3/s
dP = (180 - 35)*101325/14.696    # pressure change in pascals (N/m**2)

P = Vdot*dP                      # power in N-m/sec = watts
print("fire pump requirement [watts] =", P)
print("fire pump requirement [hp] =", P/746)

fire pump requirement [watts] = 78841.96681903958
fire pump requirement [hp] = 105.68628259924876


## Exercises¶

### Vaporization of Phenol¶

Solid phenol at 25°C and 1 atm is converted to phenol vapor at 300°C and 3 atm. How much heat will be required?

In [ ]: