A 10- ft ^3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure inside rises to 20 psia. During the process 20 Btu of heat is lost to the surroundings. Neglecting the energy stored in the paddle wheel, determine the work done by the

Question

A [latex] 10- ft ^{3}[/latex] tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure inside rises to 20 psia. During the process 20 Btu of heat is lost to the surroundings. Neglecting the energy stored in the paddle wheel, determine the work done by the paddle wheel.

Accepted Answer

A paddle wheel in an oxygen tank is rotated until the pressure inside rises to 20 psia while some heat is lost to the surroundings. The paddle wheel work done is to be determined.

Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -181°F and 736 psia. 2 The kinetic and potential energy changes are negligible, [latex] \Delta k e \cong \Delta p e \cong 0.3 [/latex] The energy stored in the paddle wheel is negligible. 4 This is a rigid tank and thus its volume remains constant.

Properties The gas constant of oxygen is [latex] R=0.3353 ext { psia.ft }^{3} / 1 bm. R =0.06206 Btu / lbm . R [/latex] (Table A-1E).

Analysis We take the oxygen in the tank as our system. This is a closed system since no mass enters or leaves. The energy balance for this system can be expressed as

[latex] \underbrace{E_{i n} - E_{ ext {out }}}_{\begin{array}{c} ext { Net energy transfer } \ ext { by heat, work, and mass }\end{array}}=\underbrace{\Delta E_{ ext {system}}}_{\begin{array}{c} ext { Change in intemal, kinetic, } \ ext { potential, etc. energies } \end{array}}[/latex]

[latex] W_{p w, ext { in }} - Q_{ ext {out }}=\Delta U [/latex]

[latex] W_{p w, i n}=Q_{ ext {out }}+m\left(u_{2} - u_{1} ight) \cong Q_{ ext {out }}+m C_{v}\left(T_{2} - T_{1} ight) [/latex]

The final temperature and the number of moles of oxygen are

[latex] \frac{P_{1} V}{T_{1}}=\frac{P_{2} V}{T_{2}} \quad \longrightarrow \quad T_{2}=\frac{P_{2}}{P_{1}} T_{1}=\frac{20 ext { psia }}{14.7 psia }(540 R )=735 [/latex]

[latex] m=\frac{P_{1} V}{R T_{1}}=\frac{(14.7 psia )\left(10 ft ^{3} ight)}{\left(0.3353 psia \cdot ft ^{3} / lbmol \cdot R ight)(540 R )}=0.812 lbm [/latex]

The specific heat ofoxygen at the average temperature of [latex] T_{ ext {ave }}=(735+540) / 2=638 R =178^{\circ} F[/latex] is [latex] C_{v, ave }=C_{p} - R=0.2216 - 0.06206=0.160 Btu / lbm \cdot R[/latex] Substituting,

[latex] W_{ ext {pwin }}=(20 Btu )+(0.812 lbm )(0160 Btu / lbm . R )(735 - 540) Btu / lbmol = 4 5 . 3 Btu[/latex]

Discussion Note that a “cooling” fan actually causes the internal temperature of a confined space to rise. In fact, a 100-W fan supplies a room as much energy as a 100-W resistance heater.

Question 1.32.E: A 10- ft ^3 tank contains oxygen initially at 14.7 psia and ...

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