SOLUTIONS FOR TUTORIAL QUESTIONS COURSE TEP Ideal motor speed = 1450rev/min x 12/120 = 145 rev/min

Transcription

1 SOLUTIONS FOR TUTORIL QUESTIONS COURSE TE 49 ) Maxiu pup diplaceent i c /rev. Motor diplaceent i c /rev. a) t full pup diplaceent: Ideal otor peed 4rev/in x / 4 rev/in The lo in otor peed due to the voluetric (lip) flow lo at N otor torque i: 4-8 rev/in 7 rev/in t half pup diplaceent: Ideal otor peed 4 x 6/ 7. rev/in For a % increae in the otor load torque the preure will have to increae by %. The lo in otor peed due to the lip flow lo at 4 N i: 4 rev/in a the preure ha increaed by % The otor peed will be rev/in b) Load torque 4 N With no loe, otor preure torque diplaceent ( π 4 / rad ) 6 bar For 9% η the otor preure /.9 bar up torque (pup diplaceent of 6c /rev) 6 With no loe, the theoretical pup torque D π.n up echanical efficiency, η, i 9% at full pup diplaceent & 9% at half pup diplaceent. The actual pup torque./.9.4n and Input power pup torque x peed.4 x 4 x π/6.6 kw Chapple aril 6 Solution to Tutorial Quetion TE 49

2 ) a) ( ) iton area π iton preure.96 bar The piton flow, Q, i given by: Q u L / in 8.8L / in The preure drop,, acro the retrictor valve will be - 98bar. The required rated flow, Q R, at a preure drop of bar acro the retrictor valve i: Q R 8. 9L / in 98 b) The circuit in Figure i a eter-in yte, which can only be ued for oppoing load force. For negative (pulling) force the circuit can be odified by placing the retrictor valve in the outlet flow fro the annulu ide of the actuator which i referred to a eterout control. c) preure copenated flow control valve i hown in Figure. ISO Sybol Figure reure copenated flow control valve. The force fro the pring on the piton of the valve in Figure () caue the pool to be fully open. Thi force i oppoed by the preure drop acro the adjutable retrictor downtrea of the pool. Conequently if the force fro the preure drop i greater than that of the pring the valve pool will be oved to the left cauing the inlet retriction created by the pool poition to be increaed. The valve inlet preure,, i kept contant by the relief valve and therefore the flow through the valve will be aintained at a contant preure drop acro the downtrea retrictor. In thi way the flow will be controlled at a contant value hould there be change in the actuator force and hence change in the valve outlet preure. Chapple aril 6 Solution to Tutorial Quetion TE 49

3 ) Valve flow p V Q CQ ; πdx (X valve diplaceent fro null poition) ρ ower, E Q S E ( - ) Q V - load preure drop actuator force / actuator area Total preure in the valve i: ( ) + ( ) ( ) kq V For ax E: ; de E ( - k Q ) Q S kq dq i) Equal rea ctuator V S S S and a kq V S Load reure for axiu power Valve reure Drop ctuator force 4 8 N reure Drop acro each port 6 Flow C q π X ax d ρ 6 Cq 6 and Q U Cqπ Xaxd ctuator velocity U ρ 6π nd the power kW.46/ ii) Flow Q u Q u For zero load force For the extend troke: Q ; Q Chapple aril 6 Solution to Tutorial Quetion TE 49

4 for a valve having yetric etering CQπ dx. ρ Hence Q Q E ue u u E + extending actuator Uing a iilar analytical approach the retract velocity u u R can be obtained: o: Q ; Q ( ) S S Q ( Q ) R and: u ( ) S + u R + Thee velocitie can be rearranged to how that: u u E R i.e. for : area ratio u. 4 u E R iii) Unequal rea ctuator Extend Velocity of.49 / dπcqxax 78 ρ 7 4 Q u 49 Q αq Q 4 Q 4 V ( ) 7 78 v 9 bar v 9 v 48 bar α 4 S v 9 8 bar Chapple aril 6 Solution to Tutorial Quetion TE 49 4

5 ( ) Force N 4) a) The oil volue available for dicharge i V V V () For iotheral copreion of the ga fro to we get V V Thu: V V () λ For adiabatic expanion of the ga fro to the ga law i V con tan t o we get: γ V V () Subtituting equation () and () into () give: γ V V (4) b) The data for the accuulator application are: bar 7bar bar Operating tie/cycle 8 Volue required for the actuator.9l diabatic index, γ for the ga.6 up flow L/in up volue delivered in 8.L and therefore the volue required fro the accuulator i: V L pplying the value to equation 4 give: V ( ) 67. (.6 ) 7.77L.8L Chapple aril 6 Solution to Tutorial Quetion TE 49

6 ) a) Vertical weight coponent of the excavator W in N For tarting the excavator the track force required N Torque required at the track wheel, T T N Taking a axiu preure of bar, the torque fro each otor i given by: T T M ax D η T 49 D. 9 / rad 8. 8c / rev. 8 6 For a peed of 6/ the peed of the track wheel 4.8rev / 88rev / in π.4 The otor type B in the table will provide the required perforance. The otor diplaceent D i 8c /rev. x -6 /rad and ha a axiu peed of rev/in. b) ND 88 8 i) The pup flow required for each otor Q L/ in η 9. Q ii) The diplaceent of each pup D 4. 6c / rev η N. 9 8 iii) Force required to drive the excavator at 6 - F + 6 7N reure required creating thi force The input power to each pup Q v p ηη vp MD v Fr 7. 44bar D η.. 9 p kw c) The total voluetric efficiency for the traniion Thu the leakage lo in the pup otor and to the external drain i % of the total flow at an oil vicoity of cst. For an oil vicoity of cst the leakage lo will increae to 6%. Thi will give a voluetric efficiency of 84%. d) The hydraulic circuit for each pup and otor that provide boot flow input, relief valve and a brake control valve are hown in the Figure. ingle pup could upply the boot flow to both circuit. Chapple aril 6 Solution to Tutorial Quetion TE 49 6

7 e) The value of reduced otor diplaceent that can be ued for operation with axiu flow and preure on level ground i deterined by uing the torque and flow equation. The torque equation give: We have So () give: For the flow we have: ( + U)r D η () M M 6 9 Mη MD. 9 D D D r. 6 + U 9 D () D U D Q U Q r MV ω ηmv r D For Q M L/in, r., η V.9 we get: Equation () and (4) give. η U 4 6 D D D D or 77 D D 79 The olution for thi quadratic i (taking only the poitive root): ± + 44 D. 6 8 / rad 67. 8c / rev 44 () (4) Chapple aril 6 Solution to Tutorial Quetion TE 49 7

8 6) Figure up and otor yte Data Torque required at axiu otor peed 7N Motor diplaceent 8c /rev (D M ) Motor echanical efficiency 9% ( η MM ). Motor voluetric efficiency 9% ( η V ) up echanical efficiency 9% ( η M ) up voluetric efficiency 96% ( η V ) up peed 8rev/in (N ) up diplaceent c /rev (D ) Fluid pecific heat J/kg/ C (C ) Heat diipated in the cooler for a water inlet teperature of TOW C kw 4 ( TOW i the difference between the cooler oil inlet and water inlet teperature). up external drain leakage flow % of the total pup leakage a) 8 Motor diplaceent D M. / rad π ) T Motor inlet preure ηmmd 7 M bar ) up outlet flow Q N D ηv L/ in ) Q. 9 Motor peed N M ηmv 9. 94rev / in DM 8 4) Cooler inlet flow (point in Figure ) QC Q + QD where Q D i the drain flow fro the pup, which i % of the total pup leakage flow. Chapple aril 6 Solution to Tutorial Quetion TE 49 8

9 Thu: D ηv nd Q. ( )N D L/ in Q L/ in C b) ) The total heat generated by the loe i given by W ( ηt )WI where W I i the power input to the pup. The yte efficiency ηt ηmmηmvηmηv ND 8 The input power W I kW 6ηM 6. 9 Total heat generated W 69. ( 78. ). kw ) The teperature increae in the fluid between the pup inlet and the cooler inlet i given by: W. 6 T S 9. C ρqc C Thi aue that there i perfect ixing of the flow at point. ) The teperature that i required at point (cooler inlet) to diipate the heat generated in ) i obtained fro the cooler perforance paraeter: Thu for the cooler to diipate the heat generated of.kw the difference between the fluid teperature at the cooler inlet and the water inlet teperature ut be: 4 T OW. C The fluid teperature at the cooler inlet will be T C C 4) For the value of TOW obtained in ) the reduction in the fluid teperature through the cooler will be the ae a the teperature increae in the yte. Therefore the reervoir teperature will be: T C T Chapple aril 6 Solution to Tutorial Quetion TE 49 9