Answer to Question #88477 in Quantitative Methods for Thasser

Question #88477
A new design of nitrogen charged accumulator, for use in a hydraulic circuit, is being factory tested for certification purposes.(http://www.accumulators.co.uk/). The following data has been collected by a technician:
o Gas volume: V = 0.8 m3
o Gas pressure: p = 160 kPa
o Compression index: n = 1.4
o Pressure/volume relationship: pV ⁿ = C
o Work done =
i. Evaluate C.
ii. Calculate the work done in compressing the nitrogen from 0.8 m3 to 0.6 m3 using:
o integral calculus
o Simpson’s rule
1
Expert's answer
2019-04-24T09:07:51-0400
"p_1=160kPa=1.6\\times 10^5 N\/m^2, V_1=0.8m^3, n=1.4"

"C=p_1V_1^n"

"C=1.6\\times 10^5 N\/m^2(0.8m^3)^{1.4}\\approx1.17070\\times 10^5 N\\cdot m^{2.2}"

"Work\\ done=W=\\displaystyle\\int_{V_1}^{V_2} pdV, where\\ p={C \\over V^n}"

"W=\\displaystyle\\int_{0.8m^3}^{0.6m^3} {1.17070\\times 10^5 N\\cdot m^{2.2} \\over V^{1.4}}dV="

"=1.17070\\times 10^5 N\\cdot m^{2.2}(-{1 \\over 0.4})[{1 \\over V^{0.4}}]\\begin{matrix}=-\n 0.6m^3 \\\\\n 0.8m^3\n\\end{matrix}\\approx"

"\\approx1.17070\\times 10^5 N\\cdot m^{2.2}(-0.333353m^{1.2})\\approx-39025.649 J"

Simpson’s rule


"Work\\ done=W=\\displaystyle\\int_{V_1}^{V_2} pdV, where\\ p={C \\over V^n}"

"W=\\displaystyle\\int_{0.8m^3}^{0.6m^3} {1.17070\\times 10^5 N\\cdot m^{2.2} \\over V^{1.4}}dV="

"=1.17070\\times 10^5 N\\cdot m^{2.2}\\displaystyle\\int_{0.8m^3}^{0.6m^3} {1 \\over V^{1.4}}dV"

Evaluate "\\displaystyle\\int_{0.8m^3}^{0.6m^3} {1 \\over V^{1.4}}dV" using Simpson’s rule.


"\\displaystyle\\int_{a}^{b} f(V)dV\\approx""\\approx{\\Delta V \\over 3}(f(V_0)+4f(V_1)+2f(V_2)+...+2f(V_{n-2})+4f(V_{n-1})+f(V_n))"


"\\Delta V ={b-a \\over n}, n=4, a=0.6, b=0.8, \\Delta V ={0.8-0.6 \\over 8}=0.05"

"f(V_0)=f(0.6)={1 \\over 0.6^{1.4}}\\approx2.044505"

"4f(V_1)=4f(0.65)={4 \\over 0.65^{1.4}}\\approx7.311076"

"2f(V_2)=2f(0.65)={2 \\over 0.7^{1.4}}\\approx3.295283"

"4f(V_3)=4f(0.75)={4 \\over 0.75^{1.4}}\\approx5.983761"

"f(V_5)=f(0.8)={1 \\over 0.8^{1.4}}\\approx1.366703"

"\\displaystyle\\int_{a}^{b} f(V)dV\\approx"


"\\approx {1 \\over 60}(2.044505+7.311076+3.295283+5.983761+1.366703)\\approx"

"\\approx0.333355"

"W\\approx1.17070\\times 10^5 N\\cdot m^{2.2}(-0.333355m^{1.2})\\approx-39025.923 J"


"Work\\ done\\approx-39026 J"


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