**3. Investigation of the hydraulic connections**

6

This chapter introduces calculations for pump/motor leakage flow in order to identify a suitable connection for the external leakage line. The DDH system can be divided into three separated fluid volumes: the piston side of the cylinders A1 and the rod side of the cylinders A3 with the hydraulic accumulator B and the hydraulic accumulator A between the pump/ motors. The accumulation of hydraulic fluid in any of these volumes will raise the pressure, as was shown in Figure 5 for the first prototype.

This chapter contains a simple model in which the volumetric efficiency of the pumps *η*v, p, determines the actual volume produced and the leakage flow of the pump/motor. For simplicity, it is assumed that this efficiency is the same for both pumps/motors. This model is a rough simplification as in reality, the leakage flow is distributed both as internal and external leakage flow [25] and volumetric efficiency varies as a function of the differential pressure and speed. This case study aims to find out how leakages should be controlled so that the system is as insensitive to fluctuations in the volumetric efficiency as possible.

The calculations were performed by determining the change in the volume of fluid during the lifting and lowering with the maximum stroke length of the cylinder *l*[26]. The volume changes are determined during a single lifting–lowering cycle.

Next, the theoretical flow *Q*Vt, the actual flow *Q*Vr and the leakage flow *Q*Vl for both pumps are determined. It is assumed that the speed of the electric motor is *n* in rev/s. In this case, the pumps produce theoretical flows of:

$$\mathcal{Q}\_{\rm Vt1} = n \cdot D\_{\rm 1} \tag{1}$$

Vt2 <sup>2</sup> *Q nD* = × , (2)

where *D*1 and *D*2 are the theoretical pump displacements for the pump/motors P1 and P2, respectively.

The actual flows *Q*Vr1 and *Q*Vr2 and leakage flows *Q*Vl1 and *Q*Vl2 in accordance with the volu‐ metric pump efficiency *η* (*D*, *p*) are

$$Q\_{\rm Vr1} = \eta\_{\rm V,p} \cdot n \cdot D\_{1\prime} \tag{3}$$

$$Q\_{\rm Vr2} = \eta\_{\rm V,p} \cdot \boldsymbol{n} \cdot \boldsymbol{D}\_{2} \,\prime \,\tag{4}$$

$$Q\_{\rm Vl1} = \left(1 - \eta\_{\rm V,p}\right) \cdot n \cdot D\_{1\prime} \tag{5}$$

$$Q\_{\rm V12} = \left(1 - \eta\_{\rm V,p}\right) \cdot n \cdot D\_2 \,. \tag{6}$$

During the lifting, the side flow *QV*c1*,up* of the cylinder A1 is equal to the actual flow *QV*r1 produced by the pump/motor P1. Theoretical flow *QV*t1 leaves the pump/motor P1 during the lowering from A3 cylinder's side. Therefore,

$$Q\_{\rm Vc1,up} = Q\_{\rm Vr1}.\tag{7}$$

$$
\mathbb{Q}\_{\text{Vc1,down}} = \mathbb{Q}\_{\text{Vt1}}.\tag{8}
$$

The cylinder losses are assumed to be negligible, so the required flows *Q*Vc2,up and *Q*Vc2,down are determined as

This chapter contains a simple model in which the volumetric efficiency of the pumps *η*v, p, determines the actual volume produced and the leakage flow of the pump/motor. For simplicity, it is assumed that this efficiency is the same for both pumps/motors. This model is a rough simplification as in reality, the leakage flow is distributed both as internal and external leakage flow [25] and volumetric efficiency varies as a function of the differential pressure and speed. This case study aims to find out how leakages should be controlled so that the system

The calculations were performed by determining the change in the volume of fluid during the lifting and lowering with the maximum stroke length of the cylinder *l*[26]. The volume changes

Next, the theoretical flow *Q*Vt, the actual flow *Q*Vr and the leakage flow *Q*Vl for both pumps are determined. It is assumed that the speed of the electric motor is *n* in rev/s. In this case, the

where *D*1 and *D*2 are the theoretical pump displacements for the pump/motors P1 and P2,

The actual flows *Q*Vr1 and *Q*Vr2 and leakage flows *Q*Vl1 and *Q*Vl2 in accordance with the volu‐

Vr1 V,p 1 *Q nD* = ×× h

Vr2 V,p 2 *Q nD* = ×× h

*Q*Vl1 = - ×× (1 , h

*Q*Vl2 = - ×× (1 . h

During the lifting, the side flow *QV*c1*,up* of the cylinder A1 is equal to the actual flow *QV*r1 produced by the pump/motor P1. Theoretical flow *QV*t1 leaves the pump/motor P1 during the

Vt1 1 *Q nD* = × , (1)

Vt2 <sup>2</sup> *Q nD* = × , (2)

, (3)

, (4)

V,p 1 ) *n D* (5)

V,p 2 ) *n D* (6)

is as insensitive to fluctuations in the volumetric efficiency as possible.

are determined during a single lifting–lowering cycle.

pumps produce theoretical flows of:

152 New Applications of Electric Drives

metric pump efficiency *η* (*D*, *p*) are

lowering from A3 cylinder's side. Therefore,

respectively.

$$Q\_{\rm Vc2,up} = R\_{\rm A} \cdot Q\_{\rm Vr1} = R\_{\rm A} \cdot \eta\_{\rm V,p} \cdot n \cdot D\_{1'} \tag{9}$$

$$\mathbf{Q}\_{\rm Vc2,down} = \mathbf{R}\_{\rm A} \cdot \mathbf{Q}\_{\rm Vt1} = \mathbf{R}\_{\rm A} \cdot \mathbf{n} \cdot \mathbf{D}\_{1} \,. \tag{10}$$

It is assumed that the cylinder is moved at a constant velocity *v*, which can be determined as follows for lifting and lowering, respectively:

$$\upsilon = \frac{Q\_{\rm Vt1}}{A\_1}, \text{ where } \upsilon\_{\rm up} = \frac{Q\_{\rm Vt1}}{A\_1}, \upsilon\_{\rm down} = \frac{Q\_{\rm Vt1}}{A\_1} \tag{11}$$

In this case, the raising (*t*up) and lowering (*t*down) times can be calculated by combining Equations (1), (3) and (11):

$$t\_{\rm up} = \frac{l}{\upsilon\_{\rm up}} = \frac{l \cdot A\_1}{Q\_{\rm vir1}} = \frac{l \cdot A\_1}{\eta\_{\rm v, p} \cdot n \cdot D\_1} \, \, \, \, \tag{12}$$

$$t\_{\text{down}} = \frac{l}{\upsilon\_{\text{down}}} = \frac{l \cdot A\_1}{Q\_{\text{Vt1}}} = \frac{l \cdot A\_1}{n \cdot D\_1} \,. \tag{13}$$

As was stated earlier, the external leakage line of the pump/motors that are used should be connected to the line under 0.3 MPa. The next section will determine the structure. The least sensitive case will be selected as the system structure. The review will be carried out for a total of four different alternative cases. Figure 7a illustrates Case I, where both the leakage lines are connected to the line with hydraulic accumulator A. In Case II in Figure 7b, both external leakage lines are connected to hydraulic accumulator B. Figure 7c shows Case III, where the external leakage lines of the pump/motors P2 and P1 are connected to the line with hydraulic accumulators A and B, respectively. Figure 7d illustrates Case IV, where the external leakage lines of the pump/motors P1 and P2 are connected to the lines with hydraulic accumulators A and B, respectively.

Figure 7. Cases to direct the pump/motor external leakage flows: (a) Case I, (b) Case II, (c) Case III and (d) Case IV. **Figure 7.** Cases to direct the pump/motor external leakage flows: (a) Case I, (b) Case II, (c) Case III and (d) Case IV.

#### **3.1. Case I**

**3.1 Case I** 

to remove the theoretical flow of *Q*Vt2. Thus, during the entire lifting motion, the volume Δ*V*up,I can be obtained with Equations (2), (9) and (12): ȟܸ୳୮ǡ୍ ൌ ݐ୳୮ ή ሺܳୡଶǡ୳୮ െ ܳ୲ଶሻൌ݈ήܣଵሺܴ െ ܴ୕ ǡ୮୴ߟ ሻǤ (14) In Case I, during the lifting, the A3 side of the cylinder should remove flow equal to *Q*Vc2,up, and the pump P2 is able to remove the theoretical flow of *Q*Vt2. Thus, during the entire lifting motion, the volume Δ*V*up,I can be obtained with Equations (2), (9) and (12):

$$
\Delta V\_{\rm up,1} = t\_{\rm up} \cdot \left( Q\_{\rm Vc2,up} - Q\_{\rm Vt2} \right) = l \cdot A\_1 \left( R\_A - \frac{R\_Q}{\eta\_{\rm v,p}} \right). \tag{14}
$$

During the lowering, the A3 chamber of the cylinder should receive the flow of *Q*Vc2,down. The pump/motor P2 is

In Case I, during the lifting, the A3 side of the cylinder should remove flow equal to *Q*Vc2,up, and the pump P2 is able

The entire cycle change of volume Δ*V*t,I in Case I is: ȟܸ୲ǡ୍ ൌ ȟܸ୳୮ǡ୍ ȟܸୢ୭୵୬ǡ୍ ൌ݈ήܣଵ ή ܴொሺߟ୴ǡ୮ െ <sup>ͳ</sup> ǡ୮୴ߟ ሻǤ (16) **3.2 Case II**  During the lowering, the A3 chamber of the cylinder should receive the flow of *Q*Vc2,down. The pump/motor P2 is able to produce the actual flow of *Q*Vr2. Thus, during the whole cycle in the A3 chamber of the cylinder, the following difference in volumes Δ*V*down,I occurs, which is obtained with Equations (4), (10) and (13):

$$
\Delta V\_{\rm down, I} = t\_{\rm down} \cdot \left( Q\_{\rm Vr2} - Q\_{\rm Vc2, down} \right) = l \cdot A\_1 \left( \eta\_{\rm v, p} \cdot R\_{\rm Q} - R\_{\rm A} \right). \tag{15}
$$

Figure 7b illustrates Case II, where both of the leakage lines are connected to the line with hydraulic accumulator B

The entire cycle change of volume Δ*V*t,I in Case I is:

$$
\Delta V\_{\rm t,l} = \Delta V\_{\rm up,l} + \Delta V\_{\rm down,l} = l \cdot A\_1 \cdot R\_Q \left(\eta\_{\rm v,p} - \frac{1}{\eta\_{\rm v,p}}\right). \tag{16}
$$

#### **3.2. Case II**

Figure 7b illustrates Case II, where both of the leakage lines are connected to the line with hydraulic accumulator B and A3 chamber branch. During the lifting, the chamber A<sup>3</sup> should remove the flow *Q*Vc2, up, as well as the leakage flows of the pumps *Q*Vl1 and *Q*Vl2. The pump/ motor P2 is able to remove the theoretical flow *Q*Vt2. In this case, the volume Δ*V*up, II is obtained with Equations (2), (5), (6), (9) and (12):

$$
\Delta V\_{\rm up, II} = t\_{\rm up} \cdot \left( Q\_{\rm Vc2,n} + Q\_{\rm Vl1} + Q\_{\rm Vl2} - Q\_{\rm Vl2} \right) = l \cdot A\_{\rm t} \left( R\_{\rm A} + \frac{1 - \eta\_{\rm v,p}}{\eta\_{\rm v,p}} - R\_{\rm Q} \right). \tag{17}
$$

During the lowering, the A1 chamber of the cylinder should remove the flow *Q*Vc2,down. The pump/motor P2 is able to provide the flow *Q*Vr2. In addition, the same line will also have the external leakage flows *Q*Vl1 and *Q*Vl2. In this case, the change in the volume Δ*V*down, II can be obtained with equations (4), (5), (6), (10) and (13):

$$
\Delta V\_{\text{down},\text{II}} = t\_{\text{down}} \cdot \left( Q\_{\text{Vr2}} + Q\_{\text{V1}} + Q\_{\text{V12}} - Q\_{\text{Vc2,down}} \right) = l \cdot A\_1 \left( R\_{\text{Q}} + 1 - \eta\_{\text{v,p}} - R\_{\text{A}} \right). \tag{18}
$$

The entire cycle change of volume Δ*V*t,II in Case II is:

$$
\Delta V\_{t,ll} = \Delta V\_{up,ll} + \Delta V\_{down,ll} = l \cdot A\_1 \left(\frac{1}{\eta\_{v,p}} - \eta\_{v,p}\right). \tag{19}
$$

#### **3.3. Case III**

9

(a) (b)

*Q*Vc1

*Q*Vc1

*Q*Vl1

In Case I, during the lifting, the A3 side of the cylinder should remove flow equal to *Q*Vc2,up, and the pump P2 is able to remove the theoretical flow of *Q*Vt2. Thus, during the entire lifting motion, the volume Δ*V*up,I can be obtained

ȟܸ୳୮ǡ୍ ൌ ݐ୳୮ ή ሺܳୡଶǡ୳୮ െ ܳ୲ଶሻൌ݈ήܣଵሺܴ െ ܴ୕

ȟܸ୲ǡ୍ ൌ ȟܸ୳୮ǡ୍ ȟܸୢ୭୵୬ǡ୍ ൌ݈ήܣଵ ή ܴொሺߟ୴ǡ୮ െ <sup>ͳ</sup>

Figure 7b illustrates Case II, where both of the leakage lines are connected to the line with hydraulic accumulator B and A3 chamber branch. During the lifting, the chamber A3 should remove the flow *Q*Vc2, up, as well as the leakage

h

following difference in volumes Δ*V*down,I occurs, which is obtained with Equations (4), (10) and (13):

( ) <sup>Q</sup>

During the lowering, the A3 chamber of the cylinder should receive the flow of *Q*Vc2,down. The pump/motor P2 is able to produce the actual flow of *Q*Vr2. Thus, during the whole cycle in the A3 chamber of the cylinder, the following difference in volumes Δ*V*down,I occurs, which is

æ ö = × - =× - ç ÷

Δ . *<sup>A</sup>*

Δ down,I down Vr2 Vc2,down ( ) 1 v,p ( <sup>A</sup> ) . *V t Q Q lA R R <sup>Q</sup>* = × - =× × -

During the lowering, the A3 chamber of the cylinder should receive the flow of *Q*Vc2,down. The pump/motor P2 is able to produce the actual flow of *Q*Vr2. Thus, during the whole cycle in the A3 chamber of the cylinder, the

ȟܸୢ୭୵୬ǡ୍ ൌ ݐୢ୭୵୬ ή ሺܳ୰ଶ െ ܳୡଶǡୢ୭୵୬ሻൌ݈ήܣଵሺߟ୴ǡ୮ ή ܴொ െ ܴሻǤ (15)

v,p

*R*

h

ç ÷ è ø

M

M

P1 P2

*Q*Vl2

*Q*Vc2

**A**

**A**

ሻǤ (14)

(14)

ሻǤ (16)

ǡ୮୴ߟ

ǡ୮୴ߟ

(15)

**B**

**B**

P1 P2

*Q*Vl1

*Q*Vl2

*Q*Vc2

**A**

**A**

Figure 7. Cases to direct the pump/motor external leakage flows: (a) Case I, (b) Case II, (c) Case III and

In Case I, during the lifting, the A3 side of the cylinder should remove flow equal to *Q*Vc2,up, and the pump P2 is able to remove the theoretical flow of *Q*Vt2. Thus, during the entire lifting

**Figure 7.** Cases to direct the pump/motor external leakage flows: (a) Case I, (b) Case II, (c) Case III and (d) Case IV.

*Q*Vl2

motion, the volume Δ*V*up,I can be obtained with Equations (2), (9) and (12):

up,I up Vc2,up Vt2 1

*V t Q Q lA R*

**B**

(c) (d)

**B**

M

M

P1 P2

*Q*Vl1

*Q*Vc2

*Q*Vl2 *Q*Vl1

P1 P2

*Q*Vc2

*Q*Vc1

154 New Applications of Electric Drives

*Q*Vc1

(d) Case IV.

**3.1. Case I**

**3.1 Case I** 

**3.2 Case II** 

obtained with Equations (4), (10) and (13):

with Equations (2), (9) and (12):

The entire cycle change of volume Δ*V*t,I in Case I is:

In Case III, the leakage flow of the pump/motor P2 is connected to the hydraulic accumulator A line and the P1 leakage flow to the A3 chamber line of the cylinder, as shown in Figure 7c. During the lifting, the A3 chamber of the cylinder should pump out the flow *Q*Vc2,up and the pump/motor P1 the leakage flow *Q*Vl1, but at the same time the pump/motor P2 can remove the flow of *Q*Vt2. Thus, during the whole stroke of the lift, the change in the volume is derived from Equations (2), (5), (9) and (12):

$$
\Delta V\_{\rm up,III} = t\_{\rm up} \cdot \left( Q\_{\rm Vc2,up} + Q\_{\rm Vl2} - Q\_{\rm V2t} \right) = l \cdot A\_1 \left( R\_{\rm A} + \frac{1 - \eta\_{\rm v,p}}{\eta\_{\rm v,p}} - \frac{R\_{\rm Q}}{\eta\_{\rm v,p}} \right). \tag{20}
$$

During the lowering, the A3 chamber of the cylinder should provide the flow *Q*Vc2,down. The pump/motor P2 is able to pump out a flow equal to *Q*Vr2. In this case, the total volume of Δ*V*down,III is obtained with Equations (4), (5), (10) and (13):

$$
\Delta V\_{\text{down,III}} = t\_{\text{down}} \cdot \left( Q\_{\text{Vr2}} + Q\_{\text{Vl2}} - Q\_{\text{Vc2,down}} \right) = l \cdot A\_1 \left( \eta\_{\text{v,p}} \cdot R\_{\text{Q}} + 1 - \eta\_{\text{v,p}} - R\_{\text{A}} \right). \tag{21}
$$

The entire cycle volume change Δ*V*t, III in Case III is:

$$
\Delta V\_{\rm t,III} = \Delta V\_{\rm up,III} + \Delta V\_{\rm down,III} = l \cdot A\_1 \left( R\_\odot - 1 \right) \left( \frac{1}{\eta\_{\rm v,p}} - \eta\_{\rm v,p} \right). \tag{22}
$$

#### **3.4. Case IV**

In Case IV, the leakage flow from the pump/motor P1 is directed to the hydraulic accumulator A and the leakage flow of the pump/motor P2 is directed to the hydraulic accumulator B. During the lifting, A3 chamber flow of the cylinder *Q*Vc2,up and leakage flow *Q*Vl2 should be removed by the pump/motor P2, but the pump/motor P2 is able to actually pump out the flow of *Q*Vt2. Thus, during the whole stroke, the change in the volume of the fluid chamber Δ*V*up,IV is obtained with Equations (2), (6), (9) and (12):

$$
\Delta V\_{\rm up,IV} = t\_{\rm up} \cdot \left( Q\_{\rm Vc2,up} + Q\_{\rm Vl2} - Q\_{\rm Vl2} \right) = l \cdot A\_1 \left( R\_\Lambda - R\_\mathbb{Q} \right). \tag{23}
$$

During the lowering, the A3 chamber of the cylinder should pump out the *Q*Vc2,down volume flow. The required pumping flow rate of P2 is *Q*Vr2 and the additional leakage flow is *Q*Vl2. In this case, the change in the volume Δ*V*down,IV is obtained with Equations (4), (6), (10) and (13):

$$
\Delta V\_{\text{down},\text{IV}} = t\_{\text{down}} \cdot \left( Q\_{\text{Vr2}} + Q\_{\text{Vl2}} - Q\_{\text{Vc2,down}} \right) = l \cdot A\_1 \left( R\_\odot - R\_\Lambda \right). \tag{24}
$$

The entire cycle change in the volume Δ*V*t, IV in Case IV is:

$$
\Delta V\_{\text{t,IV}} = \Delta V\_{\text{up,IV}} + \Delta V\_{\text{down,IV}} = 0 \,\,\,.\tag{25}
$$

## **4. Summary**

Four cases of the locations of pump/motor leakages were investigated. The different imple‐ mentation variants were determined by the changes in the volume of the cylinder as a function of the volumetric efficiency, and the results are shown in Figure 8. As a result, Case IV appears to be the most independent case [26]. However, this method did not investigate the pressure built up with various payloads. Neither did this investigation consider the practical aspects of realizing such a system. Therefore, the system structure will be investigated further in Matlab Simulink.

**Figure 8.** Results of the DDH system with four cases of external leakage connections [26].
