Answer to Question #257940 in Mechanics | Relativity for giel

Question

Assume 2 𝑙𝑏 of gas, for which 𝑅 = 26 𝑓𝑡.𝑙𝑏 𝑙𝑏.°𝑅 and 𝑘 = 1.10, undergoing a polytropic process from 𝑝1 = 15 𝑝𝑠𝑖𝑎 and 𝑡1 = 100℉ to 𝑝2 = 75 𝑝𝑠𝑖𝑎 and 𝑉2 = 3.72 𝑓𝑡3 . For both nonflow and steady-flow (∆𝑃 = 0, ∆𝐾 = 0) processes, determine (a) 𝑛 in 𝑝𝑉 𝑛 = 𝐶; (b) ∆𝐻, ∆𝑈, 𝑎𝑛𝑑 ∆𝑆; (c) ∫ 𝑝𝑑𝑉 𝑎𝑛𝑑 − ∫ 𝑉𝑑𝑝; (d) 𝑊 𝑎𝑛𝑑 𝑄.

Answer 1

$m= 2 \;lb \\ R = 26 \;ft \;lb /lb \;°R \\ k = 1.1 \\ p_1 = 15 \;psia = 15 \times 144 = 2160 \; lbf/ft^2\\ T_1 = 100 \;°F = 559.67 \;°R \\ p_2 = 75 \;psia = 75 \times 144 = 10800 \;lbf/ft^2\\ V_2 = 3.72 \;ft^3$

Using ideal gas equation:

$p_1V_1 = mRT_1 \\ V_1 = \frac{mRT_1}{p_1} \\ = \frac{2 \times 26 \times 559.67}{2160} \\ = 13.48 \;ft^3 p_2V_2 =mRT_2 \\ T_2 = \frac{p_2V_2}{mR} \\ = \frac{10800 \times 3.72}{2 \times 26} \\ = 772.62 \; °R$

Using relations

$\frac{c_p}{c_v} = k \\ c_p -c_v = R \\ c_p = \frac{kR}{k-1} \\ c_p = \frac{1.1 \times 26}{1.1-1} \\ c_p = 286 \; ft \; lbf/lb \;°R \\ c_v = \frac{R}{k-1} \\ c_v = \frac{28}{1.1-1} \\ c_v = 260 \; ft \;lbf/lb \; °R$

(a)

$p_1V_1^n = p_2V_2^n \\ \frac{p_1}{p_2} = (\frac{V_2}{V_1})^n \\ ln(\frac{p_1}{p_2}) = ln(\frac{V_2}{V_1})^n \\ n = \frac{ln(\frac{p_1}{p_2})}{ln(\frac{V_2}{V_1})} \\ n = \frac{ln(\frac{15}{75})}{ln(\frac{3.72}{13.48})} \\ n=1.25$

(b)

$ΔH = H_2 -H_1 =m(h_2-h_1) =m(u_2p_2V_2 -u_1 -p_1V_1) \\ ΔH = m(u_2-u_1 +p_2V_2 -p_1V_1) \\ =mc_v(T_2-T_1) + p_2V_2 -p_1V_1 \\ = 2 \times 260(772.62-559.67) + 10800 \times 3.72 – 2160 \times 13.48 \\ = 110734+40176 -29116.8 \\ = 121793.2 \;lbf \cdot ft \\ 1 \;BTU = 778 \;lbf \cdot ft \\ ΔH = \frac{1211793.2}{778}=156.5 \;BTU \\ ΔU = mc_v(T_2-T_1) = 2 \times 260 (772.62 -559.67) \\ = 110734 \;lbf \cdot ft \\ ΔU = \frac{110734}{778} = 142.3 \;BTU \\ ΔS = mc_p ln(\frac{V_2}{V_1}) + mc_v ln(\frac{p_2}{p_1}) \\ = 2 \times 286 \times ln(\frac{3.72}{13.48}) + 2 \times 260 \times ln(\frac{75}{15}) \\ = 100.47 \;lbf \cdot ft/ °R \\ ΔS = \frac{100.47}{778} = 0.129 \;BTU/°R$

(c)

$pV^n = c \\ p = \frac{c}{v^n} \\ \int^2_1 \frac{c}{v^n} dV = c \int^2_1 V^{-n} dV = c[\frac{V^{1-n}}{1-n}]^2_1 \\ = c[\frac{V_2^{1-n} -V_1^{1-n}}{1-n}] \\ = \frac{cV_2^{1-n} -cv_1^{1-n}}{1-n} \\ = \frac{p_2v_2^nv_2^{1-n} -p_1v_1^nV_1^{1-n}}{1-n} \\ = \frac{p_1V_1 -p_2V_2}{n-1} \\ = \frac{2160 \times 13.48 -10800 \times 3.72}{1.25-1} \\ = -44236.8 \; lbf \cdot ft \\ \int pdv = \frac{-44236.8}{778} = -56.86 \;BTU = -57 \;BTU \\ -\int Vdp = -\int^2_1 (\frac{c}{p})^{1/n} dp \\ = -c^{1/n} \int^2_1 p^{-1/n} dp \\ = -c^{1/n} [\frac{p^{1- 1/n}}{1- 1/n}]^2_1 \\ = -c^{1/n} [\frac{p_2^{1 – 1/n} -p_1^{1 -1/n}}{1 -1/n}] \\ = \frac{-c^{1/n} p_2^{1-1/n} +c^{1/n}p_1^{1-1/n}}{1-1/n} \\ = \frac{-p_2^{1/n}v_2p_2^{1-1/n} +p_1^{1/n}v_1p_1^{1-1/n}}{1-1/n} \\ = \frac{n(p_1V_1 -p_2V_2)}{n-1} \\ = 1.25(\frac{p_1V_1 -p_2V_2}{n-1}) \\ = 1.25 \int^2_1 pdV \\ = 1.25 \times -57 \\ = -71.25 \;BTU \\ -\int Vdp = -71.25 \;BTU$

(d) For a polytropic process:

For non-flow process

$W_{1-2} = \int pdV = -57 \;BTU$

For steady flow

$W_{1-2} = -\int Vdp = -71.3 \;BTU$

From the first law of thermodynamics

$Q=W+ΔU \\ = -57 +142.3 \\ = 85.3 \;BTU$