Answer to Question #88477 in Quantitative Methods for Thasser

Question

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

Answer 1

$$p_1=160kPa=1.6\times 10^5 N/m^2, V_1=0.8m^3, n=1.4$$

$$C=p_1V_1^n$$

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

$$\approx0.333355$$

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

$$Work\ done\approx-39026 J$$