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$$
\begin{array}{l}{\text { A } 30 \text { -kg aluminum block initially at } 140^{\circ} \mathrm{C} \text { is brought into contact with a } 40-\mathrm{kg} \text { block of iron. }} \\ {\text { at } 60^{\circ} \mathrm{C} \text { in an insulated enclosure. Determine the final equilibrium. }} \\ {\text { temperature and the total entropy change for this process. }}\end{array}
$$

$$
\begin{array}{l}{C_{P}=0.949kJ/kg.c^{\circ}} \\ {C_{P}=0.45 kJ/kg.c^{\circ}}\end{array}]\rightarrow A-3
$$

$$
E_{\text {in }}-E_{\text {out }}=\Delta E_{\text {sys }}
$$

Or

$$
\Delta U_{a lu m}+\Delta U_{i r o n}=0  \quad (\Delta U=0)
$$

$$
\left[m c\left(T_{2}-T_{1}\right)\right]_{a lu m}+\left[m c\left(T_{2}-T_{1}\right)\right]_{iron}=0
$$

$$
\left.[30*0.949 *\left(T_{2}-140\right)\right]+\left[40 * 0.45\left(T_{2}-60\right)\right]=0
$$

\(∴ T_{2}=109 \mathrm{c^{\circ}}=382 \mathrm{k} \)

$$
\Delta S_{i r o n}=m c_{avrg} \ln (\frac{T_2}{T_1})=40*0.45*\ln (\frac{382}{333}=2.472 kJ/k
$$

$$
\Delta S_{alum }=mc_{avrg} \ln (\frac{T_2}{T_1})=30*0.949*\ln (\frac{382}{413})=-2.221 kJ/k
$$

\(∴ \Delta S_{total}=\Delta S_{iron}+\Delta S_{alum}=2.472-2.221=0.251 kJ/k \)

$$
\begin{array}{l}{\text { An adiabatic pump is to be used to compress saturated liquid water at } 10 \text { kPa }} \\ {\text { to a pressure to } 15 \text { MPa in a reversible manner. }} \\ {\text { Determine the work input using (a) entropy data from the compressed liquid table, }} \\ {\text { (b) inlet the volume and presure values, (c) average specific volume and pressure values. }} \\ {\text { Also, determine the errors involved in parts (b) and (c). }}\end{array}
$$

$$
\left.\begin{array}{l}{P_{1}=10 \mathrm{kPa}} \\ {X_{1}=0}\end{array}\right] \rightarrow \ \begin{array}{l}{h_{1}=191.81 \mathrm{kJ/kg}} \\ {S_{1}=0.6492 kJ/kg} \\ {V_1=0.001010 m^{3}/kg}\end{array}
$$

$$
\left.\begin{array}{rl}{P_{2}} & {=15 \mathrm{MPa}} \\ {\mathrm{S}_{2}} & {=\mathrm{S}_{1}=0.6492}\end{array}\right]\rightarrow \ \begin{array}{l}{h_{2}=206.9 kJ/kg} \\ {V_2=0.001004 m^{3}/kg}\end{array}
$$

(a) $$
W_{p}=h_{2}-h_{1}=206.9- 191.81=15.1 \mathrm{kJ} / \mathrm{kg}
$$

(b) $$
W_{p}=v_{1}\left(P_{2}-P_{1}\right)=0.001010(15000-10)=15.14 \mathrm{kJ} / \mathrm{kg}
$$

$$
W_ p=V_{\text {avg }}\left(P_{2}-P_{1}\right)=(\frac{0.001010+0.001004}{2})(15000-10)=15.1 kJ/kg
$$

$$
\text { Emor } \%=\frac{|15.14-15.1|}{15.1}*100=0.3 \%
$$

 

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