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$$
\begin{array}{l}{\text { An air compressor compresses } 6 \mathrm{L} \text { of air at } 120 \mathrm{kPa} \text { and } 20^{\circ} \mathrm{C} \text { to } 1000 \mathrm{kPa} \text { and } 400^{\circ} \mathrm{C} \text { . Determine the flow work, }} \\ {\text { in } \mathrm{kJ} / \mathrm{kg} \text { , required by the compressor. }}\end{array}
$$

$$
A-1 \rightarrow R=0.287 \frac{kPa*m^3}{kg.k}
$$

$$
W_{flow}=P_{2} v_{2}-P_{1} v_{1}
$$

$$
=R\left(T_{2}-T_{1}\right)
$$

$$
=0.287(400-20)
$$

$$
=109 \mathrm{kJ} / \mathrm{kg}
$$

$$
\begin{array}{l}{\text { Air flows steadily in a pipe at } 300 \mathrm{kPa}, 77^{\circ} \mathrm{C}, \text { and } 25 \mathrm{m} / \mathrm{s} \text { at a rate of } 18 \mathrm{kg} / \mathrm{min} \text { . }} \\ {\text { Determine (a) the diameter of the pipe, }(\mathrm{b}) \text { the rate of flow energy, }}\end{array}
$$

$$
\begin{array}{l}{\text { (c) the rate of energy transport by mass, and (d) the error involved in part (c) }} \\ {\text { if the kinetic energy is neglected. }}\end{array}
$$

$$
R=0.287 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}
$$

$$
C_{P}=1.008 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{k}
$$

(a) $$
P{v}=R T
$$

$$
\rightarrow v=\frac{R T}{p}
$$$$
=\frac{0.287 *(77+273)}{300}=0.3349 \mathrm{m}^{3} / \mathrm{kg}
$$

$$
A=\frac{\dot{m} v}{V} \rightarrow \dot{m}=\frac{AV}{v} \rightarrow \frac{18/60*0.3349}{25}
$$

$$
=0.004018 \mathrm{m}^{2}
$$

$$
A=\frac{\pi D^{2}}{4} \rightarrow D=\sqrt{\frac{4 A}{\pi}}=\sqrt{\frac{4 * 0.004018}{\pi}}=0.0715 m
$$

(b) $$
\dot{W}_{flow}=\dot{m} P_{V}=\frac{18}{60}(300)(0.3349)=30.14kw
$$

(c) $$
\dot{E}_{\text {mass }}=\dot{m}\left(h+k{c}\right)=\dot{m}\left(C_{p} T+\frac{1}{2} V^{2}\right)
$$

$$
=\left(\frac{18}{60}\right)\left(1.008 *(77+273)+\frac{1}{2}(25)^{2} \frac{1}{1000}\right)
$$

$$
=105.94 \mathrm{Kw}
$$

$$
\dot{E}_{mass }=\dot{m } h=\frac{18}{60}\left(C_{P} T\right)=\frac{18}{60} * 1.005*(77+273)=105.84 kw
$$

$$
\text { Error } \% =\frac{|105.84-105.94|}{105 .94} * 100=0.09 \%
$$

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