## Tuesday, January 29, 2013

### Work of a Ski Lift

I wanted to post this example because I used to forget that when the ski lift accelerates to its operating speed it also rises the chairs. Hence, the total power required at start-up equals the sum of the power needed for acceleration as well as the power needed for the upward lift. It was such a silly mistake to make and, though harmless in problem-solving at school, can cause a lot of damage in real life situations. It is extremely important to consider the problem thoroughly before attempting to solve it.
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Problem 2-32 (Thermodynamics - An Engineering Approach 7th Edition)
A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart. The lift is operating at a steady speed of 10 km/h. Neglecting friction and air drag and assuming that the average mass of each loaded chair is 250 kg. Estimate the power required to accelerate this ski lift in 5 s to its operating speed when it is first turned on.

Power is needed for (1) acceleration and (2) lifting the loaded chairs. This two parts can be calculated separately and then added together.
(1) Power for acceleration:
The final speed of the lift is
$V=\left( 10\ km/h \right)\left( \frac{1\ h \times 1000\ m}{60\ sec \times 1\ km}\right)=2.887\ m/s$.
Then the power needed is
$P_a=\frac{1}{2}m\left(V^2-V_0^2\right)/\Delta t=\frac{1}{2}\left(50\times 250\ kg\right)\left(2.778\ m/s\right)^2=9.6\ kW$.
(2) Power for lift
Assume that the acceleration is constant (i.e. power supply is constant), its value will be
$a=\frac{\Delta V}{\Delta t}=\frac{2.778\ m/s}{5\ s}=0.556\ m/s^2$.
Then the vertical lift during acceleration will be
$\left(\frac{1}{2}at^2\right)\times\left(\frac{200}{1000}\right)=1.36\ m$.
Hence, the power needed to increase the potential energy of the lift is
$P_g=\frac{mg\Delta h}{\Delta t}=\left(50 \times 250\ kg \right)\left(9.89\ m/s^2\right) \left(1.36\ m \right)/ \left(5\ s \right)=3.41\ kW$.
Then the total Power required is
$P_{total}=P_a+P_g=9.6+34.1=43.7\ kW$.

## Saturday, January 26, 2013

### Newton-Raphson Method: Application

This is an application of the Newton-Raphson method for finding the joint angles of a mechanism. This was adapted from a note I made in a class called Mechanics of Machines in my sophomore year. It has proven useful to me for numerous times since then. The Newton-Raphson method can be used on any closed-loop mechanisms, including non-inverted and inverted slider-crank mechanisms. It also converges relatively fast in many common cases which makes it such a convenient tool. Throughout the years the Newton-Raphson method one of my favorite tools as a ME student.
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Consider the double-loop mechanism:
Link 1 is the input, i.e. the rotation angle of link 1 is specified. Since this mechanism have one degree of freedom the positions of all other links can be found. Realizing that the two loops shown in the figure are in fact two four-bar linkages, we can solve them independently.

First of all, to find the joint angles of the 4-bar linkage made by links 1, 2, 3, and the ground link,notice that the vectors R1, R2, R3, and R6' have the following relationships:
$R_1+R_2-R_3-R_6=0$.
This is the loop equation for Loop 1. One must be aware of the direction of the vectors. Rewriting this equation in terms of link lengths r1, r2, r3, r6', and joint angles $\theta_1,\ \theta_2,\ \theta_3,$ and $\phi_6'$, it becomes:
$r_1\cos\theta_1+r_2\cos\theta_2-r_3\cos\theta_3-r_6\cos\phi_6=0$
$r_1\sin\theta_1+r_2\sin\theta_2-r_3\sin\theta_3-r_6\sin\phi_6=0$.
There are two unknowns, $\theta2$ and $\theta3$. The goal here is to find solutions to these two equations:
$F_1(\theta_2,\ \theta_3)=r_1\cos\theta_1+r_2\cos\theta_2-r_3\cos\theta_3-r_6\cos\phi_6$
$F_2(\theta_2,\ \theta_3)=r_1\sin\theta_1+r_2\sin\theta_2-r_3\sin\theta_3-r_6\sin\phi_6$
Here we will apply the Newton-Raphson method.
1. Find estimated values $\theta_2$ and $\theta_3$. Let's call them $\theta_2'$ and $\theta_3'$.
2. The partial differivative of F1 and F2 with respect to $\theta1$ and $\theta2$ are
$\frac{\partial F_1}{\partial \theta_2}=-r_2\sin\theta_2$,    $\frac{\partial F_1}{\partial \theta_3}=r_3\sin\theta_3$

$\frac{\partial F_2}{\partial \theta_2}=r_2\cos\theta_2$,   $\frac{\partial F_2}{\partial \theta_2}=-r_3\cos\theta_3$.
3. Since the values are estimated, it is highly unlikely that they will be the exact solution to F1 and F2. That is, F1 and F2 will not be zero when substituting these values into the equations. We write:
$F_1(\theta_2',\ \theta_3')=\epsilon_1'$
$F_2(\theta_2',\ \theta_3')=\epsilon_2'$.
4. Then Newton-Raphson method yields that
$-r_2\sin\theta2'\times(\Delta\theta_2)+r_3\sin\theta_3\times(\Delta\theta_3)=\epsilon_1-\epsilon_1'$
$r_2\cos\theta2'\times(\Delta\theta_2)-r_3\cos\theta_3\times(\Delta\theta_3)=\epsilon_2-\epsilon_2'$
where $\epsilon_1=\epsilon_2=0$ in this case (the cost of F1 and F2).
5. Solving the equations in 4. gives the change for $\theta_2$ and $\theta_3$.
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6. The new estimation for the two values are $\theta_2''=\theta_2'+\Delta\theta_2$ and $\theta_3''=\theta_3'+\Delta\theta_3$, respectively. Using these new values we can start the process from step 3 and find new values again. The values will converge to the exact value of  $\theta_2$ and $\theta_3$. Once the values are withing tolerance, the procedure is terminated and we had our answer.
The procedure for Loop 2 is the same as that described above. Just use the Newton-Raphason method again to find  $\theta_4$ and $\theta_5$.