Hydraulic calculations Mobile Machines IHA‐2307
Janne Uusi‐Heikkilä janne.uusi‐
[email protected]
Behaviour of simplified hydraulic power transmission The figure presents simplified hydraulic power transmission. The figure shows two main parts and auxiliary pump. Parameters of the system: 3 6 Hydraulic pump: Vp 80 10 m /rev η vol 90 % Electric motor: n m 1000 rev/min
η tot 89 %
Hydraulic motor: η mec 87%
η vol 93% Overall pressure loss in pipelines is 8 bar. Total efficiency of the system is 60 % (including electric motor).
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Behaviour of simplified hydraulic power transmission State of the system: 1. Relative displacement of the pump is se to be 50 %. 2. Constant load of the hydraulic motor is 12kW. 3. Efficiencies is assumed to be constant on the operating range. Calculate: a) Outlet pressure of main pump b) Flow rate Q1 c) Load when pressure relief valve opens d) Displacement of the motor when rotational speed of the motor is 200 rpm e) Supply power of electric motor f) Mechanical efficiency of the main pump
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Behaviour of..: a) Outlet pressure of main pump Q p 50 % n Vp ηvol rev 80 l cm³ l 0.9 36 600 Q p 0.5 1000 s min min 1000 rev
Pressure at inlet of the motor: p inlet
P Q p η mec η vol
12 103W 247 bar m ³ 600 10 -6 0.93 0.87 s
Outlet pressure of pump: 247 bar + 8 bar = 255 bar IHA‐2307
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Behaviour of..: b) Flow rate Q1 Leak flow Q1: Leakage of the pump: Q pl (1 η vol ) 50% n Vp cm³ rev (1 0.9) 0.5 1000 80 rev min l 4 min
Leakage of the motor: Q ml (1 η vol ) Q p (1 0.93) 36
l l 2.52 min min
Total leakage: Ql = Qpl + Qml = (4 + 2.52) l/min = 6.52 l/min IHA‐2307
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Behaviour of..: c) Load when pressure relief valve opens
Maximum of the pressure on the out let of the pump is 270 bar. ‐> pressure difference over the motor can be: 270 bar – 8 bar = 262 bar
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Behaviour of..: d) Displacement of the motor when rotational speed of the motor is 200 rpm
Flow equation m C m nm Vm Q p η vol Vm
Q p η vol nm
36 l
0.93 % min Cm 200 rev min 167 cm³ rev
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Behaviour of..: e) Supply power of electric motor Total efficiency of the hydraulic system was 60 % 12 kW Pp 20 kW 60 %
20 kW Pem 22.5 kW 89 %
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Behaviour of..: f) Mechanical efficiency of the main pump
Equation of power and torque P M ω,
M
and to numbers Mp
η mec
Pp ωp
0.5 Δp
Vp
p C η mec 1
2π η mec
0.5 Δp Vp ω p P 2π 0.5 Δp Vp n p 2π P 2π 0.5 255 10 5 Pa 80 10 6 m ³
85%
20 103W 60 s
rev
1000 rev
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Variable displacement pump is used to control fixed displacement motor
Properties of the pump Vp 164 10 6 m 3 /rev n 25 rev/s (constant, fixed to electric motor) λ p 0.9 10 6 m 3 /(s bar) mec 85%
Properties of the motor Vm 65 10 6 m 3 /rev λ m 0.9 10 6 m 3 /(s bar) η mec 85% J m 1.0 kg m 2 IHA‐2307
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Variable displacement pump is used to control fixed displacement motor Compressibility of the fluid and pressure losses in the pipe line are neglected. Calculate: a) acceleration of the hydraulic motor, when it’s rotational speed is 33 rev/s and pump displacement is 60 %. b) required power of electric motor when system state is same than in question a).
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Variable disp..: a) acceleration of the hydraulic motor, when it’s rotational speed is 33 rev/s and pump displacement is 60 %. Q n p Vp λ p Δp n m Vm λ m Δp Δp
Mm
n p Vp n mVm (λ p λ m )
25 0.6 164 10 6 33 65 10 6 (0.9 0.9) 10
6
178bar
dω m J dt
2π M m Vm Δp 2π η m_mec Δp Vm
η m_mec dω m dt
J 2π
0.85 178 105 65 10 6 rad 156 2 2π 1.0 s IHA‐2307
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Variable disp..: b) required power of electric motor when system state is same than in question a).
ΔpQ Pem ηp_tot Δp (n p Vp ηp_vol ) ηp_tot Δpn p Vp ηp_mec 178 105 25 0.6 164 10 6 0.85 51.4kW
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One former exam question: There is below the hydraulic diagram of the hydrostatic power transmission (a closed loop hydraulic system). Name the numbered components of the hydraulic system and explain their operation in the system briefly.
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