Page:International Library of Technology, Volume 93.djvu/52

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Solution. — Using formula 3 and substituting,

V = ⁠p₁v₁ + p₂v₂/P⁠ = ⁠⁠26+1/2⁠ × 6 + 18 × ⁠4+1/2⁠/30⁠ = ⁠150 + 81/30⁠ = 8 cu. ft. Ans.

44. Mixture of Two Quantities of Air Having: Unequal Pressures, Volumes, and Temperatures. If a body of air having a temperature $\scriptstyle t_{1}$ , a pressure $\scriptstyle p_{1}$ , and a volume $\scriptstyle v_{1}$ is mixed with another volume of air having a temperature $\scriptstyle t_{2}$ , a pressure $\scriptstyle p_{2}$ , and a volume $\scriptstyle v_{2}$ to form a volume $\scriptstyle V$ having a pressure $\scriptstyle P$ and a temperature $\scriptstyle t$ , either the new temperature $\scriptstyle t$ , the new volume $\scriptstyle V$ , or the new pressure $\scriptstyle P$ may be found, if the other two quantities are known, by the following formula, in which $\scriptstyle T_{1}$ , $\scriptstyle T_{2}$ , and $\scriptstyle T$ are the absolute temperatures corresponding to $\scriptstyle t_{1}$ , $\scriptstyle t_{2}$ , and $\scriptstyle t$ :

$PV=\left({\frac {p_{1}v_{1}}{T_{1}}}+{\frac {p_{2}v_{2}}{T_{2}}}\right)T$ (1)

It follows from this formula that

$P=\left({\frac {p_{1}v_{1}}{T_{1}}}+{\frac {p_{2}v_{2}}{T_{2}}}\right){\frac {T}{V}}$ (2)

$V=\left({\frac {p_{1}v_{1}}{T_{1}}}+{\frac {p_{2}v_{2}}{T_{2}}}\right){\frac {T}{P}}$ (3)

and $T={\frac {PV}{\left({\frac {p_{1}v_{1}}{T_{1}}}+{\frac {p_{2}v_{2}}{T_{2}}}\right)}}$ (4)

Example 1. — Five cubic feet of air having a pressure of 30 pounds per square inch and a temperature of 80° F. is to be compressed, together with 11 cubic feet of air having a pressure of 21 pounds per square inch and a temperature of 45° F., in a vessel whose cubical contents is 8 cubic feet; the new pressure is required to be 45 pounds per square inch. What is the temperature of the mixture?

Solution. — Substituting the given values in formula 4,

$T={\frac {45\times 8}{\left({\frac {30\times 5}{540}}+{\frac {21\times 11}{505}}\right)}}={\frac {360}{.7352}}=489.7$ ° ,nearly

Then, $t=T-460=489.7-460=28.7$ ° F. Ans.

Example 2. — Fourteen cubic feet of air at 80 pounds pressure, and at a temperature of 100° F., is compressed with 26 cubic feet of air at 60 pounds pressure and 60° F. into a volume of 20 cubic feet; what is the resultant pressure, if the final temperature is 140° F.?