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=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 \times 10^{5} \pm 0.5 \times 10^{5} N / m ^{2},n = 1.3 ± 0.05,
D=4 \times 10^{-2} \pm 25 \times 10^{-6} mR=2 \times 10^{-2} \pm 50 \times 10^{-6} m
M=10.5 \times 10^{-2} \pm 50 \times 10^{-6} m,
H=2 \times 10^{-2} \pm 25 \times 10^{-6} m,
L=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.
\frac{\partial W}{\partial R}=-2599.46.\frac{\partial W}{\partial M}=2888.39.
\frac{\partial W}{\partial H}=-2888.39.
\frac{\partial W}{\partial L}=-2888.39.
Note: It may be useful to recall that if y=a^{x}, then d y / d x=x \log _{e}(a).
The sure-fit “extreme” limit of the work done is given by
\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 nThe average values of C, Q, and W are \bar{C}=0.005, \bar{Q}=0.1111 \text {, and } \bar{W}=20.22.
\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} .
\Delta P=2 \times 0.5 \times 10^{5}=1 \times 10^{5}.
\left|\frac{\partial W}{\partial P}\right| \Delta P=1.011035 .
\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
\Delta D=50 \times 10^{-6}
\left|\frac{\partial W}{\partial D}\right| \Delta D=0.0505517
\frac{\partial W}{\partial M}=2888.39, \quad \Delta M=100 \times 10^{-6}
\left|\frac{\partial W}{\partial M}\right| \Delta M=0.288839.
\frac{\partial W}{\partial R}=-2599.46, \quad \Delta R=100 \times 10^{-6}.
\left|\frac{\partial W}{\partial R}\right| \Delta R=0.259946.
\frac{\partial W}{\partial L}=-2888.39, \quad \Delta L=100 \times 10^{-6}.
\left|\frac{\partial W}{\partial L}\right| \Delta L=0.288839.
\frac{\partial W}{\partial H}=-2888.39, \quad \Delta H=50 \times 10^{-6} .
\left|\frac{\partial W}{\partial H}\right| \Delta H=0.1444195.
Let
W=K_{3}\left(\frac{1-Q^{n-1}}{n-1}\right), \quad K_{3}=\frac{P \pi D^{2} C}{4}.\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 K_3 gives
=\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)\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.
\Delta n=2 \times 0.05=0.1\left|\frac{\partial W}{\partial n}\right| \Delta n=1.979.
\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=\bar{W} \pm(\Delta W / 2)=20.22 \pm 2.011 J.
Statistical tolerance limits are
\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}=5.196ΔW = 2.2796.
So W=\bar{W} \pm(\Delta W / 2)=20.22 \pm 1.1398 J.
If required, \sigma_{W}=1.1398 / 3=0.38.