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$$
\begin{array}{l}{\text { An adiabatic closed system is accelerated from } 0 \mathrm{m} / \mathrm{s} \text { to } 30 \mathrm{m} / \mathrm{s} \text { . }} \\ {\text { Determine the specific energy change of this system, in } \mathrm{kJ} / \mathrm{kg} \text { . }}\end{array}
$$

$$
\Delta K_{e}=\frac{V_{2}^{2}-V_{1}^{2}}{2}
$$

$$
=\frac{(30)^{2}-0}{2}* \frac{1{\frac{kJ}{k y}}}{1000 \frac{m^{2}}{s^{2}}}=0.45 kJ/ky
$$

$$
\begin{array}{l}{\text { A classroom that normally contains } 40 \text { people is to be air-conditioned }} \\ {\text { with window air-conditioning units of } 5-\text { kW cooling capacity. }}\end{array}
$$

$$
\begin{array}{l}{\text { A person at rest may be assumed to dissipate heat at a rite of about } 360 \mathrm{kJ} / \mathrm{h}} \\ {\text { The rate are } 10 \text { light- bulbs in the classroom through the walls and }} \\ {\text { the windows is estimated to be } 15,000 \mathrm{kJ} / \mathrm{h} \text { . }}\end{array}
$$

$$
\begin{array}{l}{\text { If the room air is to be maintained at a constant temperature of } 21^{\circ} \mathrm{C}, \text { determine }} \\ {\text { the number of window air-conditioning units required }}\end{array}
$$

$$
\dot{Q}_{\text { cooling }}=\dot{Q}_{\text { lights }}+\dot{Q}_{\text { people }}+\dot{Q}_{\text { hendgim }}
$$

$$
\dot{Q}_{\text { lights }}=10 * 100=1 k w
$$

$$
Q_{\text { Peopl }}=40 * 360=4 \mathrm{Kw}
$$

$$
\dot{Q}_{\text { hendgim }}=15000 kJ/h=4.17 kw
$$

\(∴ \dot{Q}_{\text { cooling }}=1+4+4.17=9.17 \mathrm{kw} \)

$$
\# \text { units }=\frac{9.17}{5}=2 \text { units }
$$

$$
\begin{array}{l}{\text { An escalator in a shoping center is designed to move } 30 \text { people, } 75 \mathrm{kg}} \\ {\text { each, at a constant speed of } 0.8 \mathrm{m} / \mathrm{s} \text { at } 45^{\circ} \text { slope. }} \\ {\text { Determine the minimum power input needed to drive this escalator. }} \\ {\text { What would your answer be if the escalator velocity were to be doubled }}\end{array}
$$

$$
30 \text { peop. }
$$

$$
75 \mathrm{kg}/each
$$

$$
V=0.8 \mathrm{m/s}
$$

$$
\text { Mass }=30*75=2250 \mathrm{kg}
$$

$$
V_{\text { vert }}=V \sin 4 50.8^{\circ}=0.8 \sin 45=0.566 m/s
$$

$$
\dot{E}_{i n}-\dot{E}_{out}=
$$$$
\frac{d E _{system}}{d t}=0
$$

$$
\longrightarrow \quad \dot{E}_{i n}=
$$$$
\frac{d E _{system}}{d t}=
$$$$
\frac{\Delta E_{s y s}}{\Delta t}
$$

$$
\dot{W}_{i n}=\frac{\Delta P_{E}}{\Delta t}=
$$$$
\frac{m g \Delta Z}{\Delta t}=m g \ V_\text { vert. }
$$

$$
=(2250 * 9.81)*0.566 *\frac{1 kJ/kg}{1000 m^2/s^2}
$$

$$
=12.5 \mathrm{kJ} / \mathrm{s}={12.5 \mathrm{kw}}
$$

$$
V=0.8 \longrightarrow
$$$$
V_{2}=1.6 \mathrm{m} / \mathrm{s}
$$        ??

$$
\dot{W}_\text {in }=mg\text { vert }=2250*9.81*1.6 \sin 45* \frac{1}{1000}=25kJ/s
$$

\(∴ 25 kw\)

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