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• Notes

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

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

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

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

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

$\begin{array}{l}{\text { If the room air is to be maintained at a constant temperature of } 21^{\circ} \mathrm{C}, \text { determine }} \\ {\text { the number of window air-conditioning units required }}\end{array}$ $\dot{Q}_{\text { cooling }}=\dot{Q}_{\text { lights }}+\dot{Q}_{\text { people }}+\dot{Q}_{\text { hendgim }}$

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

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

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

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

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

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

$30 \text { peop. }$

$75 \mathrm{kg}/each$

$V=0.8 \mathrm{m/s}$ $\text { Mass }=30*75=2250 \mathrm{kg}$

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

$\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}$        ??

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

$∴ 25 kw$