[熱力學]第五、六週筆記

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Ch 4 Energy Analysis of Closed System

For a closed system, first law of thermodynamics

Δ E = Q - w; Δ e = q - w

d E = δ Q - δ W; d e = δ q - δ w

\frac{d E}{d t} = δ \dot{Q} - δ \dot{W}; \frac{d e}{d t} = δ \dot{q} - δ \dot{w}

E = U + K E + P E; e = u + k e + p e

Special cases
1. Stationary System:
  $Δ K E = Δ P E \Rightarrow Δ E = Δ U$
2. Steady ~: 穩態
  $\Rightarrow Δ E = 0$
3. Undergoing a cycle:
  $Δ E = 0$
4. Adiabatic system:
  $Q = 0$
5. Rigid boundary:
  $W = 0$

4.1 Moving Boundary work/ Boundary work

δ W_{b} = F \cdot d x = \frac{F}{A} \cdot A d x = P d V

\Rightarrow p e r u n i t m a s s : δ w_{b} = P d v

w_{b} = \int_{1}^{2} δ w_{b} = \int_{1}^{2} P d v

\Rightarrow w_{b} = a r e a u n d e r p - v d i a g r a m)

Polytropic process

P = c V^{- n} o r P V^{n} = c n, c : c o n s t a n t

Special case
1. n=o –> P=c –> constant pressure
  $W_{b} = \int_{1}^{2} P d V = P (V_{2} - V_{1})$
2. n=1 –> isothermal process for an ideal gas
  $W_{b} = \int_{1}^{2} P d V = \int_{1}^{2} \frac{c}{V} d V = c \cdot l n (\frac{V_{2}}{V_{1}})$
  $\overset{i d e a l g a s}{\vec{P V = c = m R T}} {\begin{array}{r} W_{b} = m R T l n (\frac{V_{2}}{V_{1}}) \\ w_{b} = R T l n (\frac{V_{2}}{V_{1}}) \end{array}$
3. $n \neq 1$ –>
  $P = c V^{- n} \Rightarrow$ adiabatic process for an calorically perfect gas
  $W_{b} = \int_{1}^{2} P d V = c \int_{1}^{2} V^{- n} d V = \frac{c}{1 - n} \cdot V^{1 - n} |_{V_{1}}^{V_{2}} = \frac{1}{1 - n} \cdot [c V_{2}^{- n} - c V_{1}^{- n}]$
  $\Rightarrow W_{b} = \frac{P_{2} V_{2} - P_{1} V_{1}}{1 - n}$
  $\overset{i d e a l g a s}{\to} {\begin{array}{r} W_{b} = \frac{m R}{1 - n} (T_{2} - T_{1}) \\ w_{b} = \frac{R}{1 - n} (T_{2} - T_{1}) \end{array}$

4.2 Specific heat - A way to evaluate
$δ q$

Specific heat: energy required to raise the temperature of a unit mass of a substance by one degree. unit (
$\frac{k J}{k g \cdot^{o} C}$ )

Consider a stationary system with boundary work

{\begin{aligned} d u & = δ q - P d v \\ h & = u + P v \end{aligned}

\Rightarrow {\begin{aligned} d u & = δ q - P d v \\ d h & = d u + d (P v) = δ q + v d P \end{aligned}

\Rightarrow {\begin{array}{r} δ q |_{v}^{(f i x e d s p e c i f i c v o l u m e)} = d u |_{v} \\ δ q |_{P} = d h |_{P} \end{array}

From the postulate: u=u(T,v), h=h(T,P)
by calculus

{\begin{aligned} d u & = (\frac{\partial u}{\partial T})_{v} d T + (\frac{\partial u}{\partial v})_{T} d v \\ d h & = (\frac{\partial h}{\partial T})_{P} d T + (\frac{\partial h}{\partial P})_{T} d P \end{aligned}

\Rightarrow {\begin{aligned} d u |_{v} & = (\frac{\partial u}{\partial T})_{v} d T \\ d h |_{P} & = (\frac{\partial h}{\partial T})_{P} d T \end{aligned}

Define specific heat at constant volume
$C_{v} = \frac{δ q |_{v}}{d T} = \frac{d u |_{v}}{d T} = (\frac{\partial u}{\partial T})_{v}$
Define specific heat at constant pressure
$C_{P} = \frac{δ q |_{P}}{d T} = \frac{d h |_{P}}{d T} = (\frac{\partial h}{\partial T})_{P}$

1

{\begin{aligned} C_{P} & > C_{v} f o r a c o m p r e s s i b l e s u b s t a n c e \\ C_{P} & = C_{v} f o r a n i n c o m p r e s s i b l e s u b s t a n c e \end{aligned}

多一個壓縮能

會在Chapter 12證明

2

For a real gas

C_{v} = C_{v} (T, v)

;

C_{P} = C_{P} (T, P)

in general

$I f C_{v} = C_{v} (T), C_{P} = C_{P} (T) f o r a g a s \Rightarrow g a s i s t h e r m a l l y p e r f e c t 「熱完美氣體」$
$I f C_{v}, C_{P} = c o n s t a n t f o r a g a s \Rightarrow g a s i s c a l o r i c a l l y p e r f e c t 「恆比熱完美氣體」$

3

For a liquid and solid, since the substance nearly incompressible

\Rightarrow C_{v} \approx C_{P} = C

4.3 Internal energy, enthalpy, and specific heats of ideal gases

we can prove mathematically in Ch12 that for an ideal gas

(\frac{\partial u}{\partial v})_{T} = 0 \Rightarrow u = u (T) o n l y

Moreover, since

h = u + P v

e n t h a l p y f o r a n i d e a l g a s, h = u (T) + R T = h (T) o n l y

Therefore,

C_{v} = (\frac{\partial u}{\partial T})_{v} = \frac{d u}{d T}

C_{P} = (\frac{\partial h}{\partial T})_{P} = \frac{d h}{d T}

Both are function of Temperature

C_{v} = C_{v} (T), C_{P} = C_{P} (T)

Thus, an ideal gas is thermally perfect

\Rightarrow {\begin{array}{r} d u = C_{v} \cdot d T \\ d h = C_{P} \cdot d T \end{array}

\Rightarrow {\begin{array}{r} Δ u = \int_{T_{1}}^{T_{2}} C_{v} d T \\ Δ h = \int_{T_{1}}^{T_{2}} C_{P} d T \end{array}

實際上我們會用線性近似來求值

$F o r i d e a l g a s$
$Δ u \approx C_{v} (T_{a v g}) \cdot (T_{2} - T_{1})$
$Δ h \approx C_{P} (T_{a v g}) \cdot (T_{2} - T_{1})$
$T_{a v g} = \frac{T_{2} + T_{1}}{2}$

Specific heat ratio
$k (o r γ) \overset{d e f}{=} \frac{C_{P}}{C_{v}}$ 記法：大的（等壓比熱）在上面
- monatomic air, k=5/3
- Air, k=1.4記起來

大部分氣體都假設為理想氣體來計算

∵ h = u + R T, f o r i d e a l g a s

\Rightarrow Δ h = Δ u + R Δ T

lim_{Δ T \to 0} \frac{Δ h}{Δ T} = lim_{Δ T \to 0} \frac{Δ u}{Δ T} + R

C_{P} = C_{v} + R

\overset{\times M}{\Rightarrow} \overset{―}{C_{P}} = \overset{―}{C_{v}} + R_{u} (u n i v e r s a l g a s c o n s t a n t)

\overset{―}{C_{P}} : m o l a r s p e c i f i c \dots \overset{―}{C_{v}} : m o l a r s p e c i f i c \dots

4.4 Internal energy, enthalpy, and specific heats of liquids and solids

similar to ideal gas,

we can prove mathematically in Ch12

(\frac{\partial u}{\partial v})_{T} = 0 \Rightarrow u = u (T) o n l y

C_{v} = \frac{d u}{d T} = C_{P} = (\frac{\partial h}{\partial T})_{P} \overset{d e f}{=} C

enthalpy for an incompressible substance
$h \overset{d e f}{=} u + P v$
$d h = d u + d (p v)$
$d h = d u + v d p (∵ i n c o m p r e s s i b l e)$
$\Rightarrow Δ h = Δ u + v Δ P$
$F o r a n i n c o m p r e s s i b l e s u b s t a n c e$
$Δ u = \int_{T_{1}}^{T_{2}} C d T \approx C_{v} (T_{a v g}) \cdot (T_{2} - T_{1})$
$Δ h = \int_{T_{1}}^{T_{2}} C (T) d T + v Δ P$

1. constant pressure, isobaric

Δ h = \int_{T_{1}}^{T_{2}} C (T) d T = Δ u \approx C (T_{a v g} \cdot Δ T)

2. constant temperature, isothermal

以液體為例

$Δ h = v Δ P$
$Δ u = 0$
$Δ v = 0 (∵ i n c o m p r e s s i b l e) \Rightarrow v = v_{f}$

$h - h_{f} (T) = v_{f} (T) \cdot (P - P_{s a t} (T))$
$h = h_{f} (T) + v_{f} (T) \cdot (P - P_{s a t} (T))$

Guest Estrada2023/06/16 03:39:26

For a closed system, first law of thermodynamics $$\Delta E=Q-w;~\Delta e=q-w$$$$dE=\delta Q-\delta W;~de=\delta q-\delta w$$$$\frac{dE}{dt}=\delta \dot Q-\delta \dot W;~\frac{de}{dt}=\delta \dot q-\delta \dot w$$$$E=U+KE+PE;~e=u+ke+pe$$

第一行:Q(in,net)-W(out,net)=delE (Edited)