5. Mathematical Diagnostics of the Six-Cylinder Pump
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5. Mathematical Diagnostics of the Six-Cylinder Pump
The previous section showed how the pump works in general. Now let us move further to the calculations. We start from the delivered water.
Because this pump has many mechanical parts, it is divided into small sections as shown in figure 7. Each system is studied separately and the results from each are processed and then integrated for the whole pump.
Figure 7: The pump divided into three sections. Source: A. Y. Al-Hassan and D. R. Hill, Islamic Technology: An Illustrated History (Cambridge, 1986, p. 51).
Now that the pump has become a combination of three small mechanical systems, each system can be studied separately and then the results from each system can be processed to give final results for the whole pump.
5.1. Pipes, Cylinders and Pistons
Section 1 of the pump consists of the collective pipe, the delivery pipes, the cylinder block, the suction pipes, the piston cylinders and the lead weight. The dimensions of the pump that were given by Taqī al-Dīn were too few and were not enough to analyse the pump. Only the length of the cylinder block and the height of the piston cylinders were given by him. All other dimensions are to be as much as needed, according to what he said. Let us see if it is possible to estimate the other dimensions by knowing those two dimensions only. We first list the knowns and the unknowns and develop the equations needed to determine the unknowns.
- Height of delivery pipes
- Hole diameter
- Distance between holes
- Weight of lead
- Piston rod diameter.
There are more unknowns than equations. Hence some independent dimensions have to be guessed guided by the facts provided from Taqī al-Dīn's description and the historical research. In this case, most dimensions are flexible except the two mentioned above. However, some other dimensions can be assumed that they are fixed for this particular analysis. Taqī al-Dīn said in his description that the cylinder block is to be divided into six equal parts so each part will be:
a = 1400 / 6 = 233mm
Taqī al-Dīn said that the length of the cylinder block is to be about two arms, so for the ease of calculations let us round a to 250mm and change the cylinder block's length to:
LCB = (6). (0.25) = 1500mm
By looking at Taqī al-Dīn's drawing in appendix 7.1 (Chester Beatty Library in Dublin, Arabic MS 5232, p. 37), the diameter of the holes can be estimated. Setting the diameter at 100mm makes this part of the pump Wilmore like the one in his drawing. Now assume that the cylinder block is divided into two equal horizontal parts and the distance between the six holes in a row is the same as the distance between the two rows which is 250mm. Then the width of the cylinder block will be:
WCB = (2) . (0.25) = 500mm
Figure 8: Top view of the pump. Source: A. Y. Al-Hassan and D. R. Hill, Islamic Technology: An Illustrated History (Cambridge, 1986, p. 51).
The height of the cylinder block depends on the hole diameter as well. Let us assume that it is 180mm. Because the cylinder block will be placed on a water stream , the suction pipes do not have to be that long. Let us assume that they are 100mm long. The height of the delivery pipes should be flexible, because it totally depends on where the pump is going to be placed. But this height should be limited so that the maximum output of the pump can be determined. So let us think of the minimum and the maximum heights the pump can deliver water to. As it can be seen from figures 8 and 9, the lower the collective pipe the smaller the slope is needed to fit it in the conic pipe. The horizontal area will get bigger as well as it depends on the angle as shown in figure 7. As a result of this, the head force will increase. Besides, if the angle is too small, it is too difficult to join the delivery pipes with the collective pipes. So let us say that the minimum angle is 60°. This will be the angle of pipe 3. To find the angles of the other two pipes, the "flow chart" method is to be used to make it easier.
Figure 9: Front view of the pump. Adapted from A. Y. Al-Hassan and D. R. Hill, Islamic Technology: An Illustrated History (Cambridge, 1986, p. 51).
To find the distances a1, a2 and a3 the method of how the delivery pipes are going to be joined must be specified. Figure 10a-b shows a top view of a guess of how the pipes can be joined.
Because the three sides of both triangles abd and bcd in figure 10a are the same, then the triangles are equilateral. This means that all the angles equal 60°. In Figure 10b the angle g is:
g = b1 + b2
Using the law of cosine, ce can be obtained:
ce = Ö (eb)2 + (bc)2 - 2 • eb • bc • cos g
and the diameter of the collective pipe will be:
dcone = ce + dD / 2 = 182.288 mm
a1 = a / 2 - dD / 2
a2 = dD / 2 + a1 + a
a3 = a2 - dD + a
tan 60° = hD / a3 Þ hD = a3 • tan 60°
a1 = tan-1 hD / a1
a2 = tan-1 hD / a2
Assuming that the diameter of the collective pipe to be the same as that of any delivery pipe, the height of the collective pipe, after adding 50mm, will be 150mm, as shown in figure 11.
Now what happens if this height increased? According to Taqī al-Dīn, only the height of the collective pipe is to be increased. As it is known that the head force is proportional to the height as well as to the pipe diameter. So does the diameter of the collective pipe that was guessed earlier cause high forces? Let us think of the answer. As it was mentioned previously, the objective of this investigation is not to improve the pump but to analyse it. Because this pump lifts one piston at a time, according to Taqī al-Dīn, only the pipe that requires more force to lift the water through should be considered. After selecting this pipe, let us check if the assumed value for the collective pipe's diameter was acceptable.
We need to ensure that:
- (FD + Fcone + FC)1 < (FD + Fcone + FC)2
where subscript 1 means case 1, where the collective pipe's height increases and subscript 2 means case 2, where the delivery pipe's height increases.
The head force is:
FW = ρW • g • V
where V is the volume.
The volumes of pipes are:
Volume of conic pipe
Vcone = π / 3 • hcone • [(dcone / 2)2 + dcone • dC / 2 + (dC / 2)2]
Volume of collective pipe
VC = π • (dC / 2)2 • hC
Volume of delivery pipe
VD = π • (dD / 2)2 • hD
Volume of piston cylinder
Vp = π • (dp / 2)2 • hp
Upon substitution, equation (A*) becomes:
ρ • g • [VD / sin a3 + Vcone + VC]1 < ρ • g • [VD / sin a3 + Vcone + VC]2
[hD • (dD / 2)2 / sin a3 + hC • (dC / 2)2]1 < [hD • (dD / 2)2 / sin a3 + hC • (dC / 2)2]2
dC < [(hD)2 - (hD)1] • (dD / 2)2 / sin a3 + (hC)2 • (dC / 2)2
The diameter of the collective pipe can be anything but must satisfy the condition above.
The force required to lift the water will be:
FW = ρw • g • [VP + VD + Vcone + VC]
Taqī al-Dīn said that the weight of the lead must be greater than weight of the water (i.e. the force exerted by the water), which is true.
WL > FW
where WL is the lead weight and FW is the total water force. The lead weight will be calculated in the next section as some other dimensions needed.
Results from section 1:
MathCAD was used to perform the calculations. The input data and output results were obtained using this software. They are tabulated as follows:
|ρwater=994.1 kg/m3|| a1=0.125m|
| hcone=100mm|| a2=0.425m|
| dcone=188.507|| a3=0.575m|
Note: The density of water is at 35°, assuming that the region where the pump was intended to work is a hot region.
The graph shows that the force required to lift the water through pipe 3 is the highest.
A plot of collective pipe diameter verses height shows that a diameter of 90mm satisfies the condition above for all the heights up to 10m.
5.2. The Connecting Rod, the Pivot and the Cam
Section 2 consists of the connecting rod, the pivot and the cam. First of all let us start listing the unknowns.
- Cam length and diameter
- Connecting rod length and diameter
- Pivot location
- Exact weight of the lead
- Diameter of the piston rod
- Diameter of the camshaft (consider it a solid bar).
By looking at Taqī al-Dīn's drawing again, some necessary dimensions can be guessed such as the length of the cams, the diameter of the cams, the length of the connecting rod and the diameter of the camshaft. As an initial guess, any reasonable dimensions can be given to those parts. Let us say that initially the dimensions are as follows:
- Length of cams = 50mm
- Diameter of cams = 50mm
- Length of connecting rod = 1m
- Diameter of camshaft = 150mm
Let us start by calculating the vertical distance that the cam pushes the connecting rod:
ac = Ö (lcam + rCS)2 + (rcam)2
a = tan-1 (rcam / lcam + rCS)
b = 60° - 2a / 2
sin (b) = Ycam / 2 / lcam + rCS Þ Ycam = 2 • (lcam + rCS) • sin (b)
x1 / lCR - x1 = Ycam / 2 / hp / 2 = Ycam / hp
x1 = lCR • Ycam - x1 • Ycam / hp
x1 (1 - Ycam / hp) = lCR • Ycam / hp Þ x1 = lCR • Ycam / hp(1 + Ycam / hp) = lCR / hp / Ycam + 1
and so x2 will be x2 = lCR - x1
To calculate the total force FL some data from our study of section 1 of the pump (see above 5.1. Pipes, Cylinders and Pistons) will be needed. We know that the time the camshaft takes to rotate 300° is the same as the time the piston must take to return to the bottom. Assuming that the centre of camshaft and piston cylinder is in the same level, the length of the wheel lever will be:
llever = hCB + hp - rCS
ω = uriver / rCS + llever
t = 60° / ω
By equating potential and kinetic energies we get
hp • FW = 1 / 2 • m • u2
The linear motion under gravity gives u = u0 + g • t
hp • Fw = 1 / 2 • m • g2 • t2 Þ Fw = 1 / 2 • hp • (FL - Fw) • g • t2
FL = FW + 2 • hp • FW / g • t2
This force has to equal the weight of the lead ball hence
4 / 3 • π • (dlead / 2)3 = FL / (ρ • g)lead
and the diameter of the lead weight will be
dlead = 2 • 3 Ö3 / 4 • π • FL / (ρ • g)lead
For the piston rod not buckle under this load we need
F3 = 4 • π2 • E • I / lPR2
where I = lPR2 • FL / 4 • π2 • E
hence π / 2 • rPR4 = lPR2 • FL / 4 • π2 • E
and the diameter of the piston rod must be larger than
dPR = 2 • 4 Ö 2 • lPR2 • FL / 4 • π3 • E
Figure 15 shows a sketch of the connecting rod as a beam pivoting on O.
To find a suitable diameter for the connecting rod we use moment equilibrium:
åMo = Fcamx1 - FLx2 = 0
Fcam = FLx2 / x1
M = FL • x2
The diameter required to withstand the moment M
dCR = 2 • 3Ö 4 / π • M / siron
The diameter required to withstand the shear force
dCR = 2 • ÖFcam / π / 2 • sironThe diameter of the connecting rod cannot be less than any of those diameters. Now Fcam must be higher than the one calculated so that it can lift the weight. Using the same concept in calculating the weight of the lead, but the angle here should be 60° :
Fcam = FL + 2 • hp • FL / g • t2
and so the diameters will be:
The diameter due to moment: dcam = 2 • 3Ö 4 / π • M / swood
Where: M = Fcam • lcam
the diameter due to shear force: dcam = 2 • ÖFcam / π / 2 • swood
Again the cam diameter cannot be less than those diameters.
Results from section 2
|siron=160Mpa||dCR=0.036m (due to moment)|
|swood=110MPa||dCR=0.005m (due to shear)|
| vriver=0.5m/s|| dcam=0.02m (due to moment)|
| Safety factor=2|| dcam=0.006 (due to shear)|
5.3. Complementary Parts
- The Unknowns are:
- Camshaft length
- Diameter of wheel lever
- Area of the scoop.
To find the force exerted by the river the toque has to be calculated first. Knowing that the torque is constant for the wheel and camshaft:
Fcam • (FCS + llever) = Friver • (rCS + lcam)
Friver = Fcam • (rCS + llever) / (rCS + lcam)
From this force the diameter of wheel lever can be determined. Using the same equations used for the cam:
The diameter due to moment:
dlever = 2 • 3 Ö4 / π • M / swood
Where: M = Friver • llever
The diameter due to shear force: dlever = 2 • ÖFriver / π / 2 • swood
Of course, the diameter of the lever cannot exceed those diameters. Now the camshaft has two main forces acting on it. One of them is its own weight and the other is that of the wheel's weight. Those weights can be calculated as:
WCS = ρwood • VCS
Wwheel = 12 • ρwood • Vlever • Vscoop
The volumes of the lever and the camshaft are
VCS = π • (dCS / 2)2 • lCS
Vlever = π • (dlever / 2)2 • llever
The scoop can be assumed as a half ball, so its volume will be:
Vscoop = 2 / 3 • π • r3scoop
To find the reactions RA and RB:
RA = WCS • lCS / 2 + Wwheel • lwheel / lCS
RB = ECS + Wwheel - RA
so the diameters of camshaft will be:
The diameter due to moment:
dCS = 2 • 3Ö4 / π • M / swood
Where: M=RB • (lCS - lwheel)
The diameter due to shear force: dCS = 2 • ÖRA / π / 2 • swood
We can make sure of the diameter of the scoop by finding the drag force. To find it we need the friction force, which is:
Ff = µ • (RA + RB)
Re = ρwater • uriver • dscoop / μwater
D = 1 / 2CDρU2A
So the diameter will be: dscoop = 2 • D / CD • ρwater • u2river
To test the camshaft under torsion:
twood = T • rCS / J Þ J = T • rCS / twood
π / 2 rCS4 = T • rCS / twood
rCS3 = 2 • T / π • twood
rCS = 3Ö2T / π twood
Results from section 3:
| lcs=1.65||dlever=0.035m (due to moment)|
| µwood=0.38||dlever=0.007m (due to shear)|
| Wwood=470 kg/m3|| dcs=0.017m (due to moment)|
| Cd=1.4||dcs=0.002m (due to shear)|
| Lwheel=1.5375||dcs=0.038m (due to torsion)|
Note: µwood is the static coefficient of friction for wood against wood because the value for the one between wood and brick was not found in any reference.
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by: FSTC, Mon 21 July, 2008