The circuit below is used to control the lifting and lowering of a 5 kN load. The hydraulic cylinder has
a blank side area of 1200 mm2 and a rod side area of 700 mm2. The circuit piping is of 25 mm
diameter and the equivalent lengths for the segments and fittings at this diameter are given in the
table below. The hydraulic fluid used has a density of 900 kg/m3 and a dynamic viscosity of 0.1 Pa.s
1. Assuming laminar flow is maintained up to ReD = 2000, find the maximum load speed, vc at
which laminar flow conditions are maintained in the piping.
2. Determine the pump power and the overall efficiency during the lifting, lowering and
holding states at a load speed of vc and of 2 vc
3. Suggest an alternative circuit design to improve efficiency during the lowering and the
holding stages.
Piping Segment / Fitting
Equivalent Length (m)
A–B
10
B–C
2
D–E
10
F-G
10
H–I
10
I–J
14
Pressure Relief Valve
20
Directional Control Valve
8
1
1. Maximum load speed for laminar flow:
The pipe speed, 𝑣𝑝 , which results in 𝑅𝑒𝐷 = 2000 is calculated from the definition of 𝑅𝑒𝐷 as:
𝑅𝑒𝐷 =
𝜌𝑣𝑝 𝐷
𝜇
= 2000, 𝑣𝑝 =
𝜇𝑅𝑒𝐷
𝜌𝐷
0.1×2000
= 900×25×10−3 = 8.89 m/s
We need to check that the 𝑅𝑒𝐷 ≤ 2000 in the pipe connected to the blank side of the cylinder and
to that connected to the rod side of the cylinder. Note that because the area of the blank side of the
cylinder is larger than that of the rod side, the flow speed in the piping connected to the blank side is
larger than that connected to the rod side for a given load speed. The load speed 𝑣𝑐 that produces
the value of 𝑣𝑝 = 8.89 m/s in the pipe connected to the blank-side of the cylinder is calculated using
continuity between the blank side of the cylinder and the piping connected to the blank side,
𝑣𝑝 𝐴𝑝 = 𝑣𝑐 𝐴𝑏 ,
𝜋
where 𝐴𝑏 = 1200 mm is the area of the blank side, and 𝐴𝑝 = 4 × 252 = 490.87 mm2 is the area of
2
the pipe. Thus,
𝐴𝑝
𝑣𝑐 = 𝑣𝑝 𝐴 = 8.89 ×
𝑏
490.87
1200
= 3.64 m/s.
Similarly, the value of 𝑣𝑐 which produce 𝑣𝑝 = 8.89 in the rod side piping is calculated using
continuity between the rod side of the cylinder and the piping connected to the rod:
𝑣𝑝 𝐴𝑝 = 𝑣𝑐 𝐴𝑟 ,
𝐴
𝑣𝑐 = 𝑣𝑐 𝐴𝑝 = 8.89 ×
𝑟
490.87
700
= 6.23 m/s
𝑣𝑐 = min(3.64 ,6.23) = 3.64 m/s
A lower limit on 𝑣𝑐 is necessary in order maintain laminar flow in the pipe connected to the blank
side. When the piston moves at a speed 𝑣𝑐 = 3.64 m/s, The speed of flow in the pipe connected to
the blank side is 𝑣𝑝−𝑏 = 8.89 m/s as calculated above. The speed of flow in the pipe connected to
the rod side, 𝑣𝑝−𝑟 is calculated from continuity based 𝑣𝑐 = 3.64 as:
𝐴
700
𝑣𝑝−𝑟 = 𝑣𝑐 𝐴𝑟 = 3.64 × 490.87 = 5.19 m/s
𝑝
2.a Pump power and circuit efficiency during load lifting:
Note that during lifting the pump flow is that of the rod-side of the cylinder. Pump power may be
calculated from the product of the head and flow rate as:
𝑃𝑝𝑢𝑚𝑝 = 𝑄𝑝𝑢𝑚𝑝 ∆𝑝
𝑄𝑝𝑢𝑚𝑝 = 𝑣𝑐 𝐴𝑟 = 3.64 × 700 × 10−6 = 2.55 × 10−3 m3/s
∆𝑝 = 𝑝𝑜𝑢𝑡 − 𝑝𝑖𝑛 , 𝑝𝑖𝑛 = 𝑝𝑎𝑡𝑚 = 0, 𝑝𝑜𝑢𝑡 = 𝑝𝑟 + 𝑝𝑙𝑜𝑠𝑠−𝑟
𝑝𝑙𝑜𝑠𝑠−𝑟 =
32𝜇𝐿𝑒𝑞−𝑟 𝑣𝑝−𝑟
𝐷𝑝2
where 𝑝𝑟 is the pressure in the rod side of the cylinder, 𝑝𝑙𝑜𝑠𝑠−𝑟 is the pressure loss in the pipe
connected to the rod side, and 𝐿𝑒𝑞−𝑟 is the equivalent length of the this pipe,
𝐿𝑒𝑞−𝑟 = 10 + 2 + 8 + 10 = 30 m
2
𝑝𝑙𝑜𝑠𝑠−𝑟 =
32×0.1×30×5.19
(25×10−3 )2
= 796.4 kPa
The value of 𝑝𝑟 , may be calculated from a force balance on the piston. Assuming that the piston is
moving with a constant speed upwards, we have
∑ 𝐹 = 0, 𝑝𝑟 𝐴𝑟 − 𝑝𝑏 𝐴𝑏 − 𝑊 = 0, 𝑝𝑟 =
𝑝𝑏 𝐴𝑏 +𝑊
𝐴𝑟
where 𝑝𝑏 is the pressure in the blank side of the piston, calculated as
𝑝𝑏 = 𝑝𝑎𝑡𝑚 + 𝑝𝑙𝑜𝑠𝑠−𝑏
where 𝑝𝑙𝑜𝑠𝑠−𝑏 is the pressure loss in the pipe connected to the blank side of the piston. Considering
gauge pressures, 𝑝𝑎𝑡𝑚 = 0, and calculating the equivalent length, 𝐿𝑒𝑞−𝑏 , of the blank side pipe:
𝐿𝑒𝑞−𝑏 = 10 + 8 + 10 + 14 = 42 m
𝑝𝑏 = 𝑝𝑙𝑜𝑠𝑠−𝑏 =
𝑝𝑟 =
𝑝𝑏 𝐴𝑏 +𝑊
𝐴𝑟
=
32𝜇𝐿𝑒𝑞−𝑏 𝑣𝑝−𝑏
𝐷𝑝2
=
32×0.1×42×8.89
(25×10−3 )2
1911.47 ×103 ×1200×10−6 +5×103
700×10−6
= 1911.5 kPa
= 10418.9 kPa
The outlet pressure of the pump can then be calculated as
𝑝𝑜𝑢𝑡 = 𝑝𝑟 + 𝑝𝑙𝑜𝑠𝑠−𝑟 = 10418.9 + 796.4 = 11215.3 kPa
Pump power is then:
𝑃𝑝𝑢𝑚𝑝 = 𝑄𝑝𝑢𝑚𝑝 ∆𝑝 = 2.55 × 10−3 × 11215.3 × 103 = 28.6 kW
Circuit efficiency during lifting is therefore,
𝜂=
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
𝑊×𝑣
=𝑃
𝑝𝑢𝑚𝑝
=
5000×3.64
28599
= 63.57%
Alternative method to calculate pump power:
𝑃𝑝𝑢𝑚𝑝 = 𝑃𝑜𝑢𝑡 + 𝑃𝑙𝑜𝑠𝑠−𝑏 + 𝑃𝑙𝑜𝑠𝑠−𝑟
Where 𝑃𝑙𝑜𝑠𝑠−𝑏 and 𝑃𝑙𝑜𝑠𝑠−𝑟 are the power losses in the piping connected to the blank-side and rodside of the cylinder, respectively, calculated as:
𝑃𝑙𝑜𝑠𝑠−𝑏 = 𝑝𝑙𝑜𝑠𝑠−𝑏 × 𝑄𝑏 = 1911.5 × 103 × 4.36 × 10−3 = 8.34 kW
𝑃𝑙𝑜𝑠𝑠−𝑟 = 𝑝𝑙𝑜𝑠𝑠−𝑟 × 𝑄𝑟 = 796.4 × 103 × 2.55 × 10−3 = 2.03 kW
𝑃𝑜𝑢𝑡 = 5000 × 3.64 = 18.2 kW
𝑃𝑝𝑢𝑚𝑝 = 18.2 + 8.34 + 2.0 = 28.6 kW
𝜂=
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
𝑊×𝑣
=𝑃
𝑝𝑢𝑚𝑝
=
5000×3.64
28599
= 63.57%
The same approach may be used to calculate the efficiency for load speed 𝑣𝑙 = 2𝑣𝑐 = 2 × 3.64 =
7.28 m/s. The outlet power in this case is
𝑃𝑜𝑢𝑡 = 𝐹. 𝑣 = 5000 × 7.28 = 36.4 kW
3
To calculate the power loss in the piping, the speed of flow in the blank-side and the rod-side piping
𝑣𝑝−𝑏 and 𝑣𝑝−𝑟 can be calculated from continuity as:
𝑣𝑙 𝐴𝑏 = 𝑣𝑝−𝑏 𝐴𝑝 ,
𝐴
1200
𝑣𝑝−𝑏 = 𝑣𝑙 𝐴𝑏 = 7.28 × 490.87 = 17.8 m/s
𝑝
and
𝑣𝑙 𝐴𝑟 = 𝑣𝑝−𝑟 𝐴𝑝 ,
𝑣𝑝−𝑟 =
𝐴
𝑣𝑙 𝑟
𝐴𝑝
= 7.28 ×
700
490.87
= 10.38 m/s
Reynolds numbers in the rod-side and the blank-side piping are calculated as:
𝑅𝑒𝐷−𝑏 =
𝜌𝑣𝑝−𝑏 𝐷 900 × 17.8 × 25 × 10−3
=
= 4000
𝜇
0.1
and
𝑅𝑒𝐷−𝑟 =
𝜌𝑣𝑝−𝑟 𝐷 900 × 10.38 × 25 × 10−3
=
= 2336
𝜇
0.1
Both 𝑅𝑒𝐷−𝑏 and 𝑅𝑒𝐷−𝑟 are above the critical value for laminar flow and the flow is turbulent in both
pipes. Assuming a relative roughness of 0.01, the friction factors in the blank-side and the rod-side
pipes are calculated using Haaland’s formula:
−2
1.11
6.9
𝜀 ⁄𝐷
𝑓 = (−1.8 log (
+(
)
𝑅𝑒
3.7
))
From which we obtain:
𝑓𝑏 = 0.05, 𝑓𝑟 = 0.06
The frictional losses in the blank-side and rod-side pipes are therefore:
𝑝𝑙𝑜𝑠𝑠−𝑏 = 𝑓𝑏
𝑝𝑙𝑜𝑠𝑠−𝑟 = 𝑓𝑟
2
𝐿𝑒𝑞−𝑏 𝜌𝑣𝑝−𝑏
𝐷𝑝
(
2
2
𝐿𝑒𝑞−𝑟 𝜌𝑣𝑝−𝑟
𝐷𝑝
(
2
) = 0.05 ×
42
900×17.82
(
)
−3
25×10
2
30
= 11977 kPa
900×10.382
)
2
) = 0.06 × 25×10−3 (
= 3491 kPa
The power loss in the blank side and rod-side piping is calculated as:
𝑃𝑙𝑜𝑠𝑠−𝑏 = 𝑝𝑙𝑜𝑠𝑠−𝑏 𝑄𝑏 = 𝑝𝑙𝑜𝑠𝑠−𝑏 (𝑣𝑝−𝑏 𝐴𝑝 ) = 11977 × 103 × (17.8 × 490.87 × 10−6 ) = 104.65 kW
𝑃𝑙𝑜𝑠𝑠−𝑟 = 𝑝𝑙𝑜𝑠𝑠−𝑟 𝑄𝑟 = 𝑝𝑙𝑜𝑠𝑠−𝑟 (𝑣𝑝−𝑟 𝐴𝑝 ) = 3491 × 103 × (10.38 × 490.87 × 10−6 ) = 17.79 kW
𝑃𝑝𝑢𝑚𝑝 = 𝑃𝑜𝑢𝑡 + 𝑃𝑙𝑜𝑠𝑠−𝑏 + 𝑃𝑙𝑜𝑠𝑠−𝑟
𝑃𝑝𝑢𝑚𝑝 = 36.4 + 104.65 + 17.79 = 158.8 kW
𝜂=
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
36.4
= 158.8 =
5000×3.64
28599
= 22.9 %
4
2.b Pump power and circuit efficiency during load lowering:
Using the expression above for the power produced by the pump, 𝑃𝑝𝑢𝑚𝑝 = 𝑄𝑝𝑢𝑚𝑝 ∆𝑝, and observing
that during the lowering of the load, pump flow is directed to the blank-side of the cylinder, 𝑄𝑝𝑢𝑚𝑝
and ∆𝑝 are calculated as:
𝑄𝑝𝑢𝑚𝑝 = 𝑣𝑐 𝐴𝑏 = 3.64 × 1200 × 10−6 = 4.37 × 10−3 m3/s
∆𝑝 = 𝑝𝑜𝑢𝑡 − 𝑝𝑖𝑛 , 𝑝𝑖𝑛 = 𝑝𝑎𝑡𝑚 = 0, 𝑝𝑜𝑢𝑡 = 𝑝𝑏 + 𝑝𝑙𝑜𝑠𝑠−𝑏
𝑝𝑙𝑜𝑠𝑠−𝑏 =
32𝜇𝐿𝑒𝑞−𝑏 𝑣𝑝−𝑏
𝐷𝑝2
where 𝑝𝑏 is the pressure in the blank side of the cylinder, 𝑝𝑙𝑜𝑠𝑠−𝑏 is the pressure loss in the pipe
connected to the rod side, and 𝐿𝑒𝑞−𝑏 is the equivalent length of the pipe connected to the blank
side, calculated as
𝐿𝑒𝑞−𝑏 = 10 + 2 + 8 + 10 = 30 m.
𝑝𝑙𝑜𝑠𝑠−𝑏 =
32×0.1×30×8.89
(25×10−3 )2
= 1365.5 kPa
The value of 𝑝𝑏 may be calculated using a force balance on the piston as above. Assuming that the
piston is moving with a constant speed downward, we have
∑ 𝐹 = 0, 𝑝𝑟 𝐴𝑟 − 𝑝𝑏 𝐴𝑏 − 𝑊 = 0, 𝑝𝑏 =
𝑝𝑟 𝐴𝑟 −𝑊
𝐴𝑏
where 𝑝𝑟 is the pressure in the rod side of the piston, calculated as
𝑝𝑟 = 𝑝𝑎𝑡𝑚 + 𝑝𝑙𝑜𝑠𝑠−𝑟 = 0 +
32𝜇𝐿𝑒𝑞−𝑟 𝑣𝑝−𝑟
𝐷𝑝2
Considering gauge pressures, 𝑝𝑎𝑡𝑚 = 0, as above. The value of 𝐿𝑒𝑞−𝑟 , is:
𝐿𝑒𝑞−𝑟 = 𝐹𝐺 + 𝐺𝐻 + 𝐻𝐼 + 𝐼𝐽 = 10 + 8 + 10 + 14 = 42 m.
Thus,
𝑝𝑟 =
𝑝𝑏 =
𝑝𝑟 𝐴𝑟 −𝑊
𝐴𝑏
=
32×0.1×42×5.19
(25×10−3 )2
= 1116.1 kPa
1116.1×103 ×700×10−6 −5000
1200×10−6
= −3515.6 kPa
𝑝𝑜𝑢𝑡 = 𝑝𝑏 + 𝑝𝑙𝑜𝑠𝑠−𝑏 = −3515.6 + 1365.5 = −2150.1 kPa
𝑃𝑝𝑢𝑚𝑝 = 𝑄𝑝𝑢𝑚𝑝 (𝑝𝑜𝑢𝑡 − 𝑝𝑖𝑛 ) = 4.37 × 10−3 × (−2150.1 × 103 − 0) = −9.4 kW
The ‘power output’ at the load side is:
𝑃𝑜𝑢𝑡 = 𝐹. 𝑣 = −5000 × 3.64 = −18.2 kW
The minus sign in the equation above indicates that the direction of the force is opposite to the load
speed. Note that since both signs are negative, the output power of the system is at the pump side,
and the input power is at the load. The efficiency therefore is:
𝜂=
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
9.4
= 18.2 = 51.6 %
5
Calculations of the pump power and circuit efficiency when the load speed 𝑣𝑙 = 2𝑣𝑐 = 7.28 m/s
may proceed by considering the input power (at the load) to be:
𝑃𝑙𝑜𝑎𝑑 = 𝐹. 𝑣𝑙 = 5000 × 7.28 = 36.4 kW
Power loss in the piping is calculated as above using 𝑣𝑝−𝑏 = 17.8 m/s and 𝑣𝑝−𝑟 = 10.38 m/s which
give 𝑅𝑒𝐷−𝑏 = 4000 𝑅𝑒𝐷−𝑟 = 2336, 𝑓𝑏 = 0.05, and 𝑓𝑟 = 0.06.
The frictional losses in the blank-side and rod-side pipes are therefore:
𝑝𝑙𝑜𝑠𝑠−𝑏 = 𝑓𝑏
𝑝𝑙𝑜𝑠𝑠−𝑟 = 𝑓𝑏
2
𝐿𝑒𝑞−𝑏 𝜌𝑣𝑝−𝑏
𝐷𝑝
(
2
2
𝐿𝑒𝑞−𝑟 𝜌𝑣𝑝−𝑟
𝐷𝑝
(
2
900×17.82
)
2
= 8554.7 kPa
900×10.382
)
2
= 4887.3 kPa
30
) = 0.05 × 25×10−3 (
42
) = 0.06 × 25×10−3 (
The power loss in the blank side and rod-side piping is calculated as:
𝑃𝑙𝑜𝑠𝑠−𝑏 = 𝑝𝑙𝑜𝑠𝑠−𝑏 𝑄𝑏 = 𝑝𝑙𝑜𝑠𝑠−𝑏 (𝑣𝑝−𝑏 𝐴𝑝 ) = 8554.7 × 103 × (17.8 × 490.87 × 10−6 ) = 74.75 kW
𝑃𝑙𝑜𝑠𝑠−𝑏 = 𝑝𝑙𝑜𝑠𝑠−𝑏 𝑄𝑏 = 𝑝𝑙𝑜𝑠𝑠−𝑏 (𝑣𝑝−𝑏 𝐴𝑝 ) = 4887.3 × 103 × (10.38 × 490.87 × 10−6 ) = 24.9 kW
The power at the pump side is now the output power and is calculated as
𝑃𝑝𝑢𝑚𝑝 = 𝑃𝑙𝑜𝑎𝑑 − 𝑃𝑙𝑜𝑠𝑠−𝑏 − 𝑃𝑙𝑜𝑠𝑠−𝑟
𝑃𝑝𝑢𝑚𝑝 = 36.4 − 74.75 − 24.9 = −63.25 kW
𝜂=
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
=
−63.25
36.4
=
5000×3.64
28599
= −174 %
The negative sign indicates that power is delivered to the system at both the load side and the pump
side during the lowering state. This power is lost in the piping due to turbulent friction.
2.c Pump power and circuit efficiency during load holding:
The power at the load side in this case is zero and so is the efficiency. The power consumed by the
pump is calculated using 𝑃𝑝𝑢𝑚𝑝 = 𝑄𝑝𝑢𝑚𝑝 ∆𝑝 = 𝑄𝑝𝑢𝑚𝑝 (𝑝𝑜𝑢𝑡 − 𝑝𝑖𝑛 ) where 𝑝𝑖𝑛 = 𝑝𝑎𝑡𝑚 = 0 and 𝑝𝑜𝑢𝑡
is estimated by assuming that pressure relief valve opens when the pressure at its inlet reached a
certain margin above the maximum pump value calculated during the lifting of the load. This margin
is called the shutoff margin for the pressure relief valve and we will assume it to be 110%. Assuming
that the system is set to operate at a maximum load speed 𝑣𝑙 = 2𝑣𝑐 and recalling that the rod side
pressure loss calculate in section 2.a for this case was 𝑝𝑙𝑜𝑠𝑠−𝑟 = 3491 kPa, 𝑝𝑜𝑢𝑡 is found to be
𝑝𝑜𝑢𝑡 = 1.1 × (𝑝𝑟 + 𝑝𝑙𝑜𝑠𝑠−𝑟 ) = 1.1 × (10418.9 + 3491) = 15301 kPa
Using 𝑄𝑝𝑢𝑚𝑝 = 2.55 × 10−3 m3/s as above, the power consumed by the pump in this case is:
𝑃𝑝𝑢𝑚𝑝 = 𝑄𝑝𝑢𝑚𝑝 ∆𝑝 = 2.55 × 10−3 × 15301 × 103 = 39kW
We need to check that the pressure generated upstream the pressure relief valve is sufficient to
overcome the resistance of the fully open pressure relief valve (equivalent length = 20 m) and
segment I-J (equivalent length = 14 m). The pressure upstream the relief valve is 𝑝𝑟𝑣 = 𝑝𝑜𝑢𝑡 −
𝑝𝑙𝑜𝑠𝑠−𝐴𝐵 where 𝑝𝑙𝑜𝑠𝑠−𝐴𝐵 is the pressure loss in segment AB, and is found to be:
6
𝑝𝑙𝑜𝑠𝑠−𝐴𝐵 = 𝑓𝑟
2
𝐿𝑒𝑞−𝐴𝐵 𝜌𝑣𝑝−𝑟
𝐷𝑝
(
2
) = 0.06 ×
10
900×10.382
(
)
25×10−3
2
= 1163.7 kPa
𝑝𝑟𝑣 = 𝑝𝑜𝑢𝑡 − 𝑝𝑙𝑜𝑠𝑠−𝐴𝐵 = 15301 − 1163.7 = 14137.3 kPa
The friction factor and the fluid speed in the rod-side piping were used above. The pressure loss in
the piping segment B-I-J pressure relief valve is calculated using the
𝑝𝑙𝑜𝑠𝑠−𝐵𝐼𝐽 = 𝑓𝑟
2
𝐿𝑒𝑞−𝐵𝐼𝐽 𝜌𝑣𝑝−𝑟
𝐷𝑝
(
2
20+14
900×10.382
)
2
) = 0.06 × 25×10−3 (
= 3956.4 kPa
As 𝑝𝑙𝑜𝑠𝑠−𝐵𝐼𝐽 < 𝑝𝑟𝑣 , the pressure loss in the piping B-I-J when the relief valve is fully open is less
than the pressure the valve is set to open. The relief valve will thus open partially and is considered
somehow oversized for this application.
3. Proposed Methods for Improving Efficiency:
The table below lists and compares a number of proposed methods for improving the efficiency of
the circuit.
1.
2.
3.
4.
5.
Method
Decrease the length of
the piping
Increase the piping area
Use an open center
directional control valve
Use a variable
displacement pump
Use an unloading valve
Notes
May not be feasible due to location constraints.
Reduces the critical load speed 𝑣𝑐 at which the flow
becomes turbulent. Cause an adverse effect if the
circuit is designed to operate in the laminar region.
Ok. But causes difficulties if more than one lifting
station is to be installed on the same pump.
Good solution. May increase the initial cost
Best solution. Little effect on the initial cost
7