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Question 19.8: In the design work for a model single cylinder reciprocating...

In the design work for a model single cylinder reciprocating engine, the work done during the expansion stroke is to be represented as:

W=PπD2C(1−Qn−1)/(4(n−1))W=P \pi D^{2} C\left(1-Q^{n-1}\right) /(4(n-1))

where

Q = C/(C + 2R)

and

C = M – R – L – H

The quantities are defined as follows:

W is the work done during the expansion stroke,
P is the cylinder pressure at start of stroke,
D is the piston diameter,
C is the length of the clearance volume,
n is the expansion stroke polytropic exponent,
R is the crank radius,
M is the distance from the crank center to the underside of the cylinder head,
L is the connecting rod length,
H is the distance from gudgeon pin to piston crown,
Q is the ratio of clearance to clearance-plus-swept volumes.

It is recognized that the independent quantities are subject to variability, and it is desired to determine the consequent extreme and probable variability in the work done.

If

P=20×105±0.5×105N/m2,P=20 \times 10^{5} \pm 0.5 \times 10^{5} N / m ^{2},

n = 1.3 ± 0.05,

D=4×10−2±25×10−6mD=4 \times 10^{-2} \pm 25 \times 10^{-6} m

 

R=2×10−2±50×10−6mR=2 \times 10^{-2} \pm 50 \times 10^{-6} m

 

M=10.5×10−2±50×10−6m,M=10.5 \times 10^{-2} \pm 50 \times 10^{-6} m,

 

H=2×10−2±25×10−6m,H=2 \times 10^{-2} \pm 25 \times 10^{-6} m,

 

L=6×10−2±50×10−6mL=6 \times 10^{-2} \pm 50 \times 10^{-6} m,

determine the extreme and probable (normal model) limits of the work (Ellis, 1990).

To help with your solution, certain partial derivatives have already been calculated.

∂W∂R=−2599.46.\frac{\partial W}{\partial R}=-2599.46.

 

∂W∂M=2888.39.\frac{\partial W}{\partial M}=2888.39.

 

∂W∂H=−2888.39.\frac{\partial W}{\partial H}=-2888.39.

 

∂W∂L=−2888.39.\frac{\partial W}{\partial L}=-2888.39.

Note: It may be useful to recall that if y=axy=a^{x}, then dy/dx=xlog⁡e(a).d y / d x=x \log _{e}(a).

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The sure-fit “extreme” limit of the work done is given by

ΔWsure-fit =∣∂W∂P∣ΔP+∣∂W∂D∣ΔD+∣∂W∂M∣ΔM+∣∂W∂R∣ΔR+∣∂W∂L∣ΔL+∣∂W∂H∣ΔH+∣∂W∂n∣Δn\Delta W_{\text {sure-fit }}=\left|\frac{\partial W}{\partial P}\right| \Delta P+\left|\frac{\partial W}{\partial D}\right| \Delta D+\left|\frac{\partial W}{\partial M}\right| \Delta M+\left|\frac{\partial W}{\partial R}\right| \Delta R+\left|\frac{\partial W}{\partial L}\right| \Delta L+\left|\frac{\partial W}{\partial H}\right| \Delta H+\left|\frac{\partial W}{\partial n}\right| \Delta n

The average values of C, Q, and W are Cˉ=0.005,Qˉ=0.1111, and Wˉ=20.22.\bar{C}=0.005, \bar{Q}=0.1111 \text {, and } \bar{W}=20.22.

 

∂W∂P=πD2C(1−Qn−1)4(n−1)=π0.0420.0051−0.11110.34×0.3=1.011035×10−5.\frac{\partial W}{\partial P}=\frac{\pi D^{2} C\left(1-Q^{n-1}\right)}{4(n-1)}=\pi 0.04^{2} 0.005 \frac{1-0.1111^{0.3}}{4 \times 0.3}=1.011035 \times 10^{-5} .

 

ΔP=2×0.5×105=1×105.\Delta P=2 \times 0.5 \times 10^{5}=1 \times 10^{5}.

 

∣∂W∂P∣ΔP=1.011035.\left|\frac{\partial W}{\partial P}\right| \Delta P=1.011035 .

 

∂W∂D=2PπDC1−Qn−14(n−1)=2π20×105×0.04×0.0051−0.11110.34×0.3=1011.035\frac{\partial W}{\partial D}=2 P \pi D C \frac{1-Q^{n-1}}{4(n-1)}=2 \pi 20 \times 10^{5} \times 0.04 \times 0.005 \frac{1-0.1111^{0.3}}{4 \times 0.3}=1011.035

 

ΔD=50×10−6\Delta D=50 \times 10^{-6}

 

∣∂W∂D∣ΔD=0.0505517\left|\frac{\partial W}{\partial D}\right| \Delta D=0.0505517

 

∂W∂M=2888.39,ΔM=100×10−6\frac{\partial W}{\partial M}=2888.39, \quad \Delta M=100 \times 10^{-6}

 

∣∂W∂M∣ΔM=0.288839.\left|\frac{\partial W}{\partial M}\right| \Delta M=0.288839.

 

∂W∂R=−2599.46,ΔR=100×10−6.\frac{\partial W}{\partial R}=-2599.46, \quad \Delta R=100 \times 10^{-6}.

 

∣∂W∂R∣ΔR=0.259946.\left|\frac{\partial W}{\partial R}\right| \Delta R=0.259946.

 

∂W∂L=−2888.39,ΔL=100×10−6.\frac{\partial W}{\partial L}=-2888.39, \quad \Delta L=100 \times 10^{-6}.

 

∣∂W∂L∣ΔL=0.288839.\left|\frac{\partial W}{\partial L}\right| \Delta L=0.288839.

 

∂W∂H=−2888.39,ΔH=50×10−6.\frac{\partial W}{\partial H}=-2888.39, \quad \Delta H=50 \times 10^{-6} .

 

∣∂W∂H∣ΔH=0.1444195.\left|\frac{\partial W}{\partial H}\right| \Delta H=0.1444195.

Let

W=K3(1−Qn−1n−1),K3=PπD2C4.W=K_{3}\left(\frac{1-Q^{n-1}}{n-1}\right), \quad K_{3}=\frac{P \pi D^{2} C}{4}.

 

∣∂W∂n∣=K3(−(n−1)∂∂nQn−1−(1−Qn−1)1(n−1)2)=K3(−(n−1)Qn−1ln⁡Q−(1−Qn−1)(n−1)2),\left|\frac{\partial W}{\partial n}\right|=K_{3}\left(\frac{-(n-1) \frac{\partial}{\partial n} Q^{n-1}-\left(1-Q^{n-1}\right) 1}{(n-1)^{2}}\right)=K_{3}\left(\frac{-(n-1) Q^{n-1} \ln Q-\left(1-Q^{n-1}\right)}{(n-1)^{2}}\right),

which on substituting for K3K_3 gives

=W(n−1)1−Qn−1(−(n−1)Qn−1ln⁡Q−(1−Qn−1)(n−1)2)=\frac{W(n-1)}{1-Q^{n-1}}\left(\frac{-(n-1) Q^{n-1} \ln Q-\left(1-Q^{n-1}\right)}{(n-1)^{2}}\right)

 

∂W∂n=−Wn−1((n−1)Qn−1ln⁡Q1−Qn−1+1)=20.220.3(0.3×0.11110.3ln⁡0.11111−0.11110.3+1)\frac{\partial W}{\partial n}=-\frac{W}{n-1}\left(\frac{(n-1) Q^{n-1} \ln Q}{1-Q^{n-1}}+1\right)=\frac{20.22}{0.3}\left(\frac{0.3 \times 0.1111^{0.3} \ln 0.1111}{1-0.1111^{0.3}}+1\right)

= 19.7909.

Δn=2×0.05=0.1\Delta n=2 \times 0.05=0.1

 

∣∂W∂n∣Δn=1.979.\left|\frac{\partial W}{\partial n}\right| \Delta n=1.979.

 

ΔWsure-fit =1.011,035+0.0,505,505517+0.288,839+0.259,946+0.288,839+0.1,444,444195+1.979=4.022\Delta W_{\text {sure-fit }} = 1.011,035 + 0.0,505,505517 + 0.288,839 + 0.259,946 + 0.288,839 + 0.1,444,444195 + 1.979 = 4.022

 

W=Wˉ±(ΔW/2)=20.22±2.011J.W=\bar{W} \pm(\Delta W / 2)=20.22 \pm 2.011 J.

Statistical tolerance limits are

ΔWnormal 2=(∂W∂P)2ΔP2+(∂W∂D)2ΔD2+(∂W∂M)2ΔM2+(∂W∂R)2ΔR2+(∂W∂L)2ΔL2+(∂W∂H)2ΔH2+(∂W∂n)2Δn2=5.196\Delta W_{\text {normal }}^{2}=\left(\frac{\partial W}{\partial P}\right)^{2} \Delta P^{2}+\left(\frac{\partial W}{\partial D}\right)^{2} \Delta D^{2}+\left(\frac{\partial W}{\partial M}\right)^{2} \Delta M^{2}+\left(\frac{\partial W}{\partial R}\right)^{2} \Delta R^{2}+\left(\frac{\partial W}{\partial L}\right)^{2} \Delta L^{2} +\left(\frac{\partial W}{\partial H}\right)^{2} \Delta H^{2}+\left(\frac{\partial W}{\partial n}\right)^{2} \Delta n^{2}=5.196

ΔW = 2.2796.

So  W=Wˉ±(ΔW/2)=20.22±1.1398JW=\bar{W} \pm(\Delta W / 2)=20.22 \pm 1.1398 J.

If required, σW=1.1398/3=0.38.\sigma_{W}=1.1398 / 3=0.38.

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