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help me with 2 related questions

3. Shown is the hydraulic circuit of a load-lifting hydraulic system. The lowering speed is controlled by means of a throttle-check valve. Discuss the construction and operation of this system. Redraw the hydraulic circuit in the load-lowering mode, then calculate the pressure in the cylinder rod side, P the inlet flow rate, Qu outlet flow rate, Q pump flow rateé, Q pump output power, N, and the area of the throttle valve, A, Neglect the hydraulic losses in the system elements, except the throttle valve. The flow rate through the throttling element is given by: Q = CA,/2aP/p, where Q= Flow rate, m2/s A,-Throttle area, m2 p Oil density, kg/m2 Pe Given Qout Pump exit pressure 30 bar Piston area A,-78.5 cm2 Oil density 870 kg/m2 Safety valve is pre-set at 350 bar W C,-Discharge coefficient AP Pressure difference, Pa PQP Piston speed -0.07 m/s Piston rod side area A, = 40 cm2 Discharge coefficient= 0.611 Weight of the body 30 kN Page 4 of 5 4. Redraw the circuit of problem 3 in lifting mode. For the same pump flow rate, safety valve setting, and dimensions, calculate the maximum load that the system can lift. Calculate all of the system operating parameters at this mode. Neglect the hydraulic losses in the system elements, except for the throttle valve.

Answer 1

SOLUTION:

CIRCUIT DIAGRAM OF LOAD LIFTING MODE:

Pe P Qp

Calculation of maximum lifting load

Maximum pump discharge pressure P= safety valve set pressure= 350 bar

Hence P_C= 350 bar

P_A=0

350 x 10 x 40 x 10-4- 0 140 kN Wmar PcAr PAAp

Other system parameters during lifting mode.

P_C= 350 bar

P=pump pressure=350 bar

P_A= 0 (bar gage)

Qin VA, (0.07) (40 x 10-4)0.00028 m23/s

Qout VA, (0.07) (78.5 x 10) = 0.00055 m2/s

P= Qin = 0.00028 m3/

Pump power is

$\small N_h=PQ_p=350\times10^5(0.00028)=9800\: W=9.8\: kW$

Problem 8

The throttle check valve combines one way flow and flow control features.

The throttling is achieved typically through a series of radial holes and passing through a restriction area defined by adjustment sleeve and inner housing.

By rotating the adjusting sleeve, the restriction area between adjustment sleeve and inner housing can be varied which will allow more or less flow for any particular pressure drop.

The flow in the reverse direction is blocked typically by piston of a check valve (spring loaded) located concentrically inside the inner housing.

In the forward direction if the pressure force is more than spring force flow happens.

In the reverse direction the piston of check valve is pressed against the seat by pressure and spring forces and flow is shut-off.

The scheme with load lowering configuration is shown below.

in Pe Cout

Calculation of P_C

By force balance

$\small W=P_CA_r-PA_p$

Or

30(103) x 78.5(10-)30(10) PAp W PC 133.9 x 10 Pa 40(10-4) Ar

The inlet flow rate is given by

$\small Q_i_n=VA_p=0.07\times78.5(10^{-4})=0.00055\: m^3/s$

Outlet flow rate is given by

$\small Q_{out}=VA_r=0.07\times40(10^{-4})=0.00028\: m^3/s$

Pump flow rate is

$\small Q_p=Q_i_n=0.00055\: m^3/s$

Pump power is given by

$\small N_h=Q_PP=0.00055\times30(10^5)=1648.5\: W=1.6485\: kW$

Calculation of A_t

_{$\small Q=Q_{out}=C_dA_t\sqrt{\frac{2\Delta P}{\rho}}$}

_{$\small \Delta P=P_C-0=133.9\: bar$}

Hence

$\small 0.00028=0.611A_t\sqrt{\frac{2\times133.9\times10^5}{870}}$

Or

A_t= 2.61x10^-6 m²= 0.0261 cm²