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
p1=160kPa=1.6×105N/m2,V1=0.8m3,n=1.4p_1=160kPa=1.6\times 10^5 N/m^2, V_1=0.8m^3, n=1.4

C=p1V1nC=p_1V_1^n

C=1.6×105N/m2(0.8m3)1.41.17070×105Nm2.2C=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=V1V2pdV,where p=CVnWork\ done=W=\displaystyle\int_{V_1}^{V_2} pdV, where\ p={C \over V^n}

W=0.8m30.6m31.17070×105Nm2.2V1.4dV=W=\displaystyle\int_{0.8m^3}^{0.6m^3} {1.17070\times 10^5 N\cdot m^{2.2} \over V^{1.4}}dV=

=1.17070×105Nm2.2(10.4)[1V0.4]=0.6m30.8m3=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

1.17070×105Nm2.2(0.333353m1.2)39025.649J\approx1.17070\times 10^5 N\cdot m^{2.2}(-0.333353m^{1.2})\approx-39025.649 J

Simpson’s rule


Work done=W=V1V2pdV,where p=CVnWork\ done=W=\displaystyle\int_{V_1}^{V_2} pdV, where\ p={C \over V^n}

W=0.8m30.6m31.17070×105Nm2.2V1.4dV=W=\displaystyle\int_{0.8m^3}^{0.6m^3} {1.17070\times 10^5 N\cdot m^{2.2} \over V^{1.4}}dV=

=1.17070×105Nm2.20.8m30.6m31V1.4dV=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 0.8m30.6m31V1.4dV\displaystyle\int_{0.8m^3}^{0.6m^3} {1 \over V^{1.4}}dV using Simpson’s rule.


abf(V)dV\displaystyle\int_{a}^{b} f(V)dV\approxΔV3(f(V0)+4f(V1)+2f(V2)+...+2f(Vn2)+4f(Vn1)+f(Vn))\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))


ΔV=ban,n=4,a=0.6,b=0.8,ΔV=0.80.68=0.05\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(V0)=f(0.6)=10.61.42.044505f(V_0)=f(0.6)={1 \over 0.6^{1.4}}\approx2.044505

4f(V1)=4f(0.65)=40.651.47.3110764f(V_1)=4f(0.65)={4 \over 0.65^{1.4}}\approx7.311076

2f(V2)=2f(0.65)=20.71.43.2952832f(V_2)=2f(0.65)={2 \over 0.7^{1.4}}\approx3.295283

4f(V3)=4f(0.75)=40.751.45.9837614f(V_3)=4f(0.75)={4 \over 0.75^{1.4}}\approx5.983761

f(V5)=f(0.8)=10.81.41.366703f(V_5)=f(0.8)={1 \over 0.8^{1.4}}\approx1.366703

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


160(2.044505+7.311076+3.295283+5.983761+1.366703)\approx {1 \over 60}(2.044505+7.311076+3.295283+5.983761+1.366703)\approx

0.333355\approx0.333355

W1.17070×105Nm2.2(0.333355m1.2)39025.923JW\approx1.17070\times 10^5 N\cdot m^{2.2}(-0.333355m^{1.2})\approx-39025.923 J


Work done39026JWork\ done\approx-39026 J


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS