Chapter 2: Formula Sheet

Adapted by Diana Bairaktarova

8 Total Energy of a system on a unit mass basis

$e=\frac{E}{m}(\frac{kJ}{kg})$

Kinetic Energy

$KE=m(kg)\frac{v^2(m/s)^2}{2}(kJ/kg)$

Kinetic Energy on a unit mass basis

$KE=\frac{v^2(\frac{m}{s})^2}{2}(kJ/kg)$

Potential Energy

$PE=mgz(kJ)$

Potential energy on a unit mass basis

$PE = mz (kJ/kg)$

Internal Energies

$E=U+KE+PE=U+\frac{v^2}{2}+gz(kJ/kg)$

Mechanical Energy

$e_{mech}=\frac{P}{\rho}+\frac{v^2}{2}+gz$

Mass flow rate

$\dot{m}=\rho \dot{V} = \rho A_c V_{avg} (kg/s) \ \text{where} \ A_c =\pi D^2/4$

Energy flow rate

$\dot{E}=\dot{m}e (kJ/s) \ or \ (kW)$

The work done per unit mass

$w = W/m \ (kJ/kg)$

Work

$W = F\Delta s (kJ) \ or \ W = Fs(kJ) \ \text{where} \ s \ \text{is shaft work}$

$s =(2\pi nT)$

Electric Work

$\dot{W_e} = VI(W)$

Torque T

$T=Fr$

Work Shaft

$\dot{W_{sh}}=2\pi \dot{n}T (kW)$

Conduction Heat Transfer

$Q_{cond}=-Ak_t\frac{\Delta T}{\Delta x} (W)$

Convection Heat Transfer

$Q_{conv}=hA(T_s-T_f)(W)$

Radiative Heat Transfer

$Q_{rad}=\mathcal{E} \sigma A(T_s^4-T_f^4) (W)$

License

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Thermodynamics Copyright © by Diana Bairaktarova (Adapted from Engineering Thermodynamics - A Graphical Approach by Israel Urieli and Licensed CC BY NC-SA 3.0) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Total Energy of a system on a unit mass basis