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$\begin{array}{l}{\text { Air at } 80 \mathrm{kPa} \text { and } 127^{\circ} \mathrm{C} \text { enters an adiabatic diffuser steadily at a rate of } 6000 \mathrm{kg} / \mathrm{h} \text { and leaves at } 100 \mathrm{kPa} \text { . }} \\ {\text { The velocity of the air stream is decreased from } 230 \text { to } 30 \mathrm{m} / \mathrm{s} \text { as it passes through the diffuser. }} \\ {\text { Find (a) the exit temperature of the air and }} \\ {\text { (b) the exit area of the diffuser. }}\end{array}$

$R=0.287 \quad k P a . m^{3} / k g. k$

$\longrightarrow A-17 \quad \longrightarrow \quad h_{1}= 400.98 kJ/kg$

$\dot{E}_{i n}-E_{out}=\Delta E_{\text {system }}=0$

$\dot{E}_{i n}=\dot{E}_{out}$

$\dot m\left(h_{1}+\frac{v_{1}^{2}}{2}\right)=\dot m (h_2+ \frac{V_{2}^{2}}{2})=0$

$h_{2}-h_{1}+\frac{V_{2}^{2}-V_{1}^{2}}{2}=0$

$h_{2}=h_{1}-\frac{V_{2}^{2}-V_{1}^{2}}{2}=400.98-$$\frac{(30)^{2}-(230)^{2}}{2} * \frac{1}{1000}$

$∴ h_{2}=426.98 \mathrm{kJ} / \mathrm{kg}$

$A-17 \longrightarrow T_{2}=426.6 \mathrm{K}$

$V_{2}=\frac{R T_{2}}{P_{2}}=$$\frac{0.287 *(426.6)}{100}=1.221 \mathrm{m}^{3} / \mathrm{kg}$

$\dot{m}=\frac{1}{v_{2}} A V_{2} \rightarrow A_{2}=$$\frac{\dot m V_{2}}{V_{2}}=\frac{\frac{6000}{3600} * 1.221}{30}$

$∴ A_ 2=0.00678 \mathrm{m}^{2}$

$\begin{array}{l}{\text { Steam enters a nozzle at } 400^{\circ} \mathrm{C} \text { and } 800 \mathrm{kPa} \text { with a velocity of } 10 \mathrm{m} / \mathrm{s} \text { , }} \\ {\text { and leaves at } 300^{\circ} \mathrm{C} \text { and } 200 \mathrm{kPa} \text { while losing heat at at ate of } 25 \mathrm{kW} \text { . }}\end{array}$

$\begin{array}{l}{\text { For an inlet area of } 800 \mathrm{cm} 2, \text { determine the velocity and the volume }} \\ {\text { flow rate of the steam at the nozzle exit. }}\end{array}$

$\dot{E}_{i n}=\dot{E}_{out}$            $\left(\Delta E_{s ys}\right)=0$

$\dot m\left(h_{1}+\frac{v_{1}^{2}}{2}\right)$$=\dot m\left(h_{2}+\frac{v_{2}^{2}}{2}\right)+Q_{out}$

$h_{1}+\frac{V_{1}^{2}}{2}=h_{2}+\frac{V_{2}^{2}}{2}+\frac{Q_{out}}{\dot m}$

$\left.\begin{array}{l}{P_{1}=800 K P{a}} \\ {T_{1}=400 \mathrm{c^{\circ}}}\end{array}\right] \rightarrow A - 6\quad$$\begin{array}{l}{V_{1}=0.38429{m^3/kg}} \\ {h_{1}=3267.7{kJ/kg}}\end{array}$

$\left.\begin{array}{l}{P_{2}=200 \mathrm{KPa}} \\ {T_{2}=300 \mathrm{c^{\circ}}}\end{array}\right] \rightarrow A-6\quad \begin{array}{l}{v_{2}=1.31623 \mathrm{m^3/kg}} \\ {h_{2}=3072.1 \mathrm{kJ/kg}}\end{array}$

$\dot{m}=\frac{1}{v_1} A_{1} V_{1}=\frac{1}{0.38429}(0.08)(10)=2.082 kg/s$

$3267 \cdot 7+\frac{(10)^{2}}{2}(\frac{1}{1000})=3072.1+\frac{V_{2}^{2}}{2}(\frac{1}{1000})+\frac{25}{2.082}$

$V_{2}=606 \mathrm{m} / \mathrm{s}$

$\dot{U}_{2}=\dot{m} v_{2} \quad=2.082 * 1.31623=2.74 \mathrm{m}^{3} / \mathrm{s}$