The following equation is the entropy balance equation for an open system fraction numerator d S subscript C V end subscript over denominator d t end fraction equals sum for j of fraction numerator stack Q subscript j with dot on top over denominator T subscript j end fraction plus sum for i of m with dot on top subscript i s subscript i minus sum for e of m with dot on top subscript e s subscript e plus sigma with dot on top subscript C V end subscript Simplify the above equation for a control volume with a single inlet and two outlets, and adiabatic in the CV.
Added by Deborah C.
Step 1
Step 1: Since the control volume has a single inlet and two outlets, the equation can be simplified as follows: dS_CV/dt = Q_in/T_in - Q_out1/T_out1 - Q_out2/T_out2 Show more…
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A closed system consists of an ideal gas with constant specific heat ratio $k$. (a) The gas undergoes a process in which temperature increases from $T_{1}$ to $T_{2}$. Show that the entropy change for the process is greater if the change in state occurs at constant pressure than if it occurs at constant volume. Sketch the processes on $p-v$ and $T-s$ coordinates. (b) Using the results of (a), show on $T-s$ coordinates that a line of constant specific volume passing through a state has a greater slope than a line of constant pressure passing through that state. (c) The gas undergoes a process in which pressure increases from $p_{1}$ to $p_{2}$. Show that the ratio of the entropy change for an isothermal process to the entropy change for a constant-volume process is $(1-k)$. Sketch the processes on $p-v$ and $T-s$ coordinates.
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If the initial temperature is $T_{0}$ and volume is $V_{0}$ then in adiabatic expansion. so, $$ \begin{gathered} T V^{\gamma-1}=T_{0} V_{0}^{\gamma-1} \\ T=T_{0} n^{1-\gamma}=T_{1} \text { where } n=\frac{V_{1}}{V_{0}} \end{gathered} $$ $V_{1}$ being the volume at the end of the adiabatic process. There is no entropy change in this process. Next the gas is compressed isobarically and the net entropy change is $$ \Delta S=\left(\frac{m}{M} C_{p}\right) \ln \frac{T_{f}}{T_{1}} $$ But $\quad \frac{V_{1}}{T_{1}}=\frac{V_{0}}{T_{f}}, \quad$ or $\quad T_{f}=T_{1} \frac{V_{0}}{V_{1}}=T_{0} n^{-\gamma}$ So $\quad \Delta S=\left(\frac{m}{M} C_{p}\right) \ln \frac{1}{n}=-\frac{m}{M} C_{p} \ln n=-\frac{m}{M} \frac{R \gamma}{\gamma-1} \ln n=-9 \cdot 7 \mathrm{~J} / \mathrm{K}$
Thermodynamics And Molecular Physics
The Second Law of Thermodynamics. Entropy
Consider the situation shown in the diagram below: In state A, the system consists of two separate chambers of equal volume with 1 mole of A molecules on the left and 1 mole of B molecules on the right. The A and B molecules do not interact with each other. In State B, the partition between the two halves of the box is removed. Starting from the equation for the change in entropy for the expansion of an ideal gas, show that the change in entropy on going from State A to State B (the entropy of mixing) is given by: ΔS = SB - SA = -NkB(XA ln XA + XB ln XB) In this equation: N = NA + NB (number of molecules) XA = NA / (NA + NB) (mole fraction of A) XB = NB / (NA + NB) (mole fraction of B)
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