diff --git a/uni/mmme/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md b/uni/mmme/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md index 65fa9c9..20e17f2 100755 --- a/uni/mmme/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md +++ b/uni/mmme/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md @@ -42,6 +42,84 @@ The system is not in equilibrium if parts of the system are at different conditi A process in which does not cross the system boundary +## Perfect (Ideal) Gasses + +A perfect gas is defined as one in which: + +- all collisions between molecules are perfectly elastic +- there are no intermolecular forces + +Perfect gases do not exist in the real world and they have two requirements in thermodynamics: + +### The Requirements of Perfect Gasses + +#### Obey the Perfect Gas Equation + +$$pV = n \tilde R T$$ + +where $n$ is the number of moles of a substance and $\tilde R$ is the universal gas constant + +or + +$$pV =mRT$$ + +where the gas constant $R = \frac{\tilde R}{\tilde m}$, $\tilde m$ is molecular mass + +or + +$$pv = RT$$ + +(using the specific volume) + +#### $c_p$ and $c_v$ are constant + +This gives us the equations: + +$$u_2 - u_1 = c_v(T_2-T_1)$$ + +$$h_2 - h_1 = c_p(T_2-T_1)$$ + +### Relationship Between Specific Gas Constant and Specific Heats + +$$c_v = \frac{R}{\gamma - 1}$$ + +$$c_p = \frac{\gamma}{\gamma -1} R$$ + +
+ + +#### Derivation + + + +We know the following are true (for perfect gases): + +$$\frac{c_p}{c_v} = \gamma$$ + +$$u_2 - u_1 = c_v(T_2-T_1)$$ + +$$h_2 - h_1 = c_p(T_2-T_1)$$ + +So: + +\begin{align*} +h_2 - h_1 &= u_2 - u_1 + (p_2v_2 - p_1v_1) \\ +c_p(T_2-T_1) &= c_v(T_2-T_1) + R(T_2-T_1) \\ +c_p &= c_v + R \\ +\\ +c_p &= c_v \gamma \\ +c_v + R &= c_v\gamma \\ +c_v &= \frac{R}{\gamma - 1} \\ +\\ +\frac{c_p}{\gamma} &= _v \\ +c_p &= \frac{c_p}{\gamma} + R \\ +c_p &= \frac{\gamma}{\gamma -1} R +\end{align*} + + +
+ + ## Properties of State *State* is defined as the condition of a system as described by its properties. @@ -79,7 +157,7 @@ In thermodynamics we distinguish between *intensive*, *extensive*, and *specific - Specific (extensive) --- extensive properties which are reduced to unit mass of substance (essentially an extensive property divided by mass) (e.g. specific volume) -## Units +### Units Property | Symbol | Units | Intensive | Extensive --------------- | ------ | --------------- | --------- | --------- @@ -87,15 +165,128 @@ Pressure | p | Pa | Yes | Temperature | T | K | Yes | Volume | V | m$^3$ | | Yes Mass | m | kg | | Yes -Specific Volume | $\nu$ | m$^3$ kg$^{-1}$ | Yes | +Specific Volume | v | m$^3$ kg$^{-1}$ | Yes | Density | $\rho$ | kg m$^{-3}$ | Yes | Internal Energy | U | J | | Yes Entropy | S | J K$^{-1}$ | | Yes Enthalpy | H | J | | Yes +### Density + +For an ideal gas: + +$$\rho = \frac{p}{RT}$$ + +### Enthalpy and Specific Enthalpy + +Enthalpy does not have a general physical interpretation. +It is used because the combination $u + pv$ appears naturally in the analysis of many +thermodynamic problems. + +The heat transferred to a closed system undergoing a reversible constant pressure process is equal +to the change in enthalpy of the system. + +Enthalpy is defined as: + +$$H = U+pV$$ + +and Specific Enthalpy: + +$$h = u + pv$$ + +### Entropy and Specific Entropy + +Entropy is defined as the following, given that the process s reversible: + +$$S_2 - S_1 = \int\! \frac{\mathrm{d}Q}{T}$$ + +### Heat Capacity and Specific Heat Capacity + +Heat capacity is quantity of heat required to raise the temperature of a system by a unit +temperature: + +$$C = \frac{\mathrm{d}Q}{\mathrm{d}T}$$ + +Specific heat capacity is the quantity of heat required to raise the temperature of a unit mass +substance by a unit temperature: + +$$c = \frac{\mathrm{d}q}{\mathrm{d}T}$$ + +
+ + +#### Heat Capacity in Closed Systems and Internal Energy + +The specific heat transfer to a closed system during a reversible constant **volume** process is +equal to the change in specific **internal energy** of the system: + +$$c_v = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}u}{\mathrm{d}T}$$ + + + +This is because if the change in volume, $\mathrm{d}v = 0$, then the work done, $\mathrm{d}w = 0$ +also. +So applying the (1st Corollary of the) 1st Law to an isochoric process: + +$$\mathrm{d}q + \mathrm{d}w = \mathrm{d}u \rightarrow \mathrm{d}q = \mathrm{d}u$$ + +since $\mathrm{d}w = 0$. + +
+ +
+ + +#### Heat Capacity in Closed Systems and Enthalpy + +The specific heat transfer to a closed system during a reversible constant **pressure** process is +equal to the change in specific **enthalpy** of the system: + +$$c_p = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}h}{\mathrm{d}T}$$ + + + +This is because given that pressure, $p$, is constant, work, $w$, can be expressed as: + +$$w = -\int^2_1\! p \,\mathrm{d}v = -p(v_2 - v_1)$$ + +Applying the (1st corollary of the) 1st law to the closed system: + +\begin{align*} + q + w &= u_2 - u_1 \rightarrow q = u_2 - u_1 + p(v_2 - v_1) \\ + q &= u_2 + pv_2 - (u_1 + pv_1) \\ + &= h_2 - h_1 = \mathrm{d}h \\ + \therefore \mathrm{d}q &= \mathrm{d}h +\end{align*} + +
+ +
+ + +#### Ratio of Specific Heats + +$c_p > c_v$ is always true. + + + +Heating a volume of fluid, $V$, at a constant volume requires specific heat $q_v$ where + +$$q_v = u_2 - u_1 \therefore c_v = \frac{q_v}{\Delta T}$$ + +Heating the same volume of fluid but under constant pressure requires a specific heat $q_p$ where + +$$q_p =u_2 - u_1 + p(v_2-v_1) \therefore c_p = \frac{q_p}{\Delta T}$$ + +Since $p(v_2-v_1) > 0$, $\frac{q_p}{q_v} > 1 \therefore q_p > q_v \therefore c_p > c_v$. + +The ratio $\frac{c_p}{c_v} = \gamma$ + +
+ ## Thermodynamic Processes and Cycles -When a thermodynaic system changes from one state to another it is said to execute a *process*. +When a thermodynamic system changes from one state to another it is said to execute a *process*. An example of a process is expansion (volume increasing). A *cycle* is a process or series of processes in which the end state is identical to the beginning. @@ -130,7 +321,8 @@ Heat and Work are different forms of enery transfer. They are both transient phenomena and systems never possess heat or work. Both represent energy crossing boundaries when a system undergoes a change of state. -By convention, the transfer of energy into the system from the surroundings is positive. +By convention, the transfer of energy into the system from the surroundings is positive (work is +being done *on* the system *by* the surroundings). ### Heat @@ -146,8 +338,36 @@ By convention, the transfer of energy into the system from the surroundings is p $$W = \int\! F \mathrm{d}x$$ (the work, $W$, done by a force, $F$, when the point of application of the force undergoes a -displacement, $\mathrm dx$) +displacement, $\mathrm{d}x$) + +## Thermally Insulated and Isolated Systems + +In thermally insulated systems and isolated systems, heat transfer cannot take place. + +In thermally isolated systems, work transfer cannot take place. # Process and State Diagrams Reversible processes are represented by solid lines, and irreversible processes by dashed lines. + +# 1st Law of Thermodynamics + +The 1st Law of Thermodynamics can be thought of as: + +> When a closed system is taken through a cycle, the sum of the *net* work transfer ($W$) and *net* +> heat transfer ($Q$) equals zero: +> +> $$W_{net} + Q_{net} = 0$$ +> + +## 1st Corollary + +> The change in internal energy of a closed system is equal to the sum of the heat transferred +> and the work done during any change of state +> +> $$W_{12} + Q_{12} = U_2 - U_1$$ + +## 2nd Corollary + +> The internal energy of a closed system remains unchanged if it +> [thermally isolated](#thermally-insulated-and-isolated-systems) from its surroundings