Đề thi môn Công Nghệ thủy lực và Khí nén

lOMoARcPSD|25311344

PNHY230529 Cơng nghệ thủy lực và khí nén 9 6 2023 0001
thủy lực khí nén (Trường Đại học Sư phạm Kỹ thuật Thành phố Hồ Chí Minh)

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lOMoARcPSD|25311344

Downloaded by Nguyen Vu ()


lOMoARcPSD|25311344

Downloaded by Nguyen Vu ()


lOMoARcPSD|25311344

Downloaded by Nguyen Vu ()


lOMoARcPSD|25311344

Downloaded by Nguyen Vu ()


lOMoARcPSD|25311344

Downloaded by Nguyen Vu ()


lOMoARcPSD|25311344

Downloaded by Nguyen Vu ()


lOMoARcPSD|25311344

Downloaded by Nguyen Vu ()


lOMoARcPSD|25311344

Downloaded by Nguyen Vu ()


TRl/dNG DH SIT PHAM KY THUAT TPHCM
KHOA CO KHt CHE TAO MAY
BQ MON CO DIEN Tty
Chu ky giam thi 2
Chu ky giam thi 1

Diem \;i chfr k>
CB cham thi thu nhat

CB cham thi thu hai

DAP AN CUOI KY HOC KY 1 NAM HOC 19-20

Mon: Cong nghe thuy lyc va khi nen
Ma mon hoc: PNHY330529
Deso/Made:01
De thi co 4 trang.
Thai gian: 60 phut.
Buac nheo su duns tai lieu siav dehorn duac sir
duns laptop vd dien thoai).
SV lam b&i true tiep tren de thi va n6p l?i de
Ho va te n :..................................................................
Ma so SV :..................................................................
So T T :...................... Phong th i:...............................

Cau 1 (5 diem) Thiet ke mach dieu khien dien - khi nen theo phwmg phap ket hop

Chia ding cho phmmg phap thiet kk mach KET HOP:
B+A+C-

/

A-

TANG 1

/

TANG 2

/ lOgiayA+B/

TANG 3

/
/

A-C+
TANG 4

Tin hi§u dieu khiin:
El = START AS6
E2 = S5
E3 = S1ATimer
E4 = S3
XylanhA:

A+ = Y1 = (LI AS4) v L3

XylanhB:

B+ = Y2 = L1

Xy lanh C:

C- = Y4 = (L1 AS2) v L2 v L3 v (L4AS1)

^

B- = Y 3= L 3AS2

SI, S2 su dung 2 l&n do do can dung Ro le trung gian KS1, KS2
n

5

0
3

+9

S3

rHMl

ik iv j

V 2

S6

S4

!

9

S2

O

A+

SS
$1

ft

Chia tang va xac dinh tin hieu dieu khien: Id
Phin thiet ke mach 4d, mot loi sai tru 0.5d
Phirtmg phap thiet ke mach TANG:
Chia tang cho phirtmg phap thiet ke mach KET HQfP:
B+A+C-

/

A-

TANG1

/

TANG 2

/ lOgiayA+B/

TANG 3

/
/

A-C+

TANG 4

Tin hieu dilu khiln:
El = START AS6
E2 = S5
E3 = S1ATimer
E4 = S3 A S2
XylanhA:

A+ = Y1 = (LI AS4) v L3

XylanhB:

B+ = Y2 = L1

Xy lanh C:

C- = Y4 = (LI AS2) v L2 v L3 v (L4AS1)

B- = Y3 = L3AS2

S1, S2 su dung 2 lan do do can dung Ra le trung gian KS1, KS2

O


+

2

4

V

1

2

3

f l

1

0

1 1

1 ?

1 5

1 6

THIET k! mach theo Phmmg phap NHIP:
K1 = START AS6 AK8
K2 = K1 AS4
K3 = K2 AS2
K4 = K3 AS5
K5 = K 4 AS1 ATimer

K6 = K5 A S2
K7 = K6 AS3
K8 = K7 A S1
B+ = K1
B- = K6
A+ = K2 v K3 v K5 v K6
C- = K3 v K4 v K5 v K6 v K7

Q


LiAJ lu'p'ng ke
(Flowmeter)

i 14 i

V a n Ipc (F ilter)

j 12 i

O png

ca

th u y ly c (H y d ra u lic m o to r )

V an k h o a /v a n c fo n g m cr (S h u t-o ff v a lv e )

!3

Opng

ca

d ip n (E le c tric m o to r )

e .................. ...

6

i

;

j

...........................

T h i e t bi l£ m m a t ( c o o le r )

1 15 1 V a n t r a n ( P r e s s u r e r e l i e f v a l v e )
............................

T h i e t bj b a o m i r e d a u (fillin g l e v e l i n d i c a t o r )

1 11 i V a n c a n ( C o u n t e r - b a l a n c e v a lv e )

i

f ....... ]

!7!

A p ke (P ressure gau g e)

i 13 i

L tr u I t r y n g k e ( F l o w m e t e r )

V an a p s u a t d ie u k h ien t u a n t i/

.........1 (P re ss ure s e q u e n c e valve)

:

.. ................ "

i

j

i

i

j V an d ieu a p (P re ss u re re g u la to r)

.. ................... ...

j 4

;

•

j V a n m p t c h i e u d i e u k h i e n d i r p c h u ’d n g c h p n

g

V an m p t c h ie u (C h eck v a lv e )

N h ift ke (T h e rm o m e te r)

............................

(Lockable piloted n o n - r e t u r n valve)

[

V a n t a e d p n g k h 6 a Ian

;

[

C u m b o rn (H y d ra u lic p o w e r p ack )

i

............................

(Delockable p ilo te d d o u b l e n o n - r e tu r n valve)

f

]

V a n d i p n tu* t h u y l y e 4 / 2 (4 /2 w a y e le c tr o -

i

hydraulically o p e r a t e d dir ec tiona l

:

2

i

B o m d a u (H y d ra u lic p u m p )

i

B o n / b e c h tra d a u ( T a n k /R e s e rv o ir)

i

B inh tr f c h c h u 'a t h u y ly c / A c q u y t h u y ly c

i * .................. ..

......................

| 1Q ]
• .....................

i

c o n tr o l valve)

V a n d i p n t u 1 t h u y l y e 4 / 3 (4 /3 w a y e le c tr o hydraulically o p e r a t e d dir ec tiona l

c o n tr o l valve)

i

5

! 8

(H y d ra u lic A c c u m u la to r)

a. Goi ten cac thiet bi thuy lire tir so 1 den 15 trong mach thuy lye phia tren (ghi so thu tu cua thiet
bi ngay canh ten thiet bi trong Bang danh muc thiet bi thuy lire)? (l,5d)
b. Trinh bay cong dung cua thiet bi so 11? (0,5d)
c. Tinh cong suat dong ca dien biet hieu suat bam 84,5%, ap suat bam 250 bar, bam banh rang
thuy lye co the tich bom 7.1 cm3/vong, dong ca dien quay 4000 vong/phut? (Id)
d. Tinh lye can thiet de xy lanh co the keo vat m = 2000 kg co lai theo phirang ngang tu vi tri dung
yen vdi gia toe = 5m/s2 ; g = 9,81 m/s2 ; he so ma sat tren mat phang ngang p= 0,15; luc ma sat
trong xy lanh Fmst = 4500 N. (Id)

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r

?

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in


e. Tinh luu lirgng cap vao xy lanh khi xy lanh co lai biet van toe xy lanh di chuyen la 70mm/s,
duong kinh xy lanh 250mm , duong klnh true piston 160 mm, hieu suat hoat dong cua van va xy
lanh la 90%? (Id)
DAP AN;
Cau a: Doc t$n thiet bi: xem tren hinh (doc dung ten moi thiet bi 0,ld).
Cau b: Trinh bay c6ng dung cua van so 11; Khi cuon day Y4 cua van dien tir thuy luc 4/3 dugc kich
hoat, van se hoat dong o 6 phia ben phai va cap dau vao cua ben phai cua dong co thuy luc, van can so 11
se mo ra khi ap suat dau xa ra tu dong co thuy luc Ion hon muc ap suat cai dat tren than van la 30 bar, khi
do dong co se quay theo chieu kim dong ho va dau se dugc xa ve bon thong qua van loc so 14.
Cau c; Tinh cong suat bum
Qbom = n x V = 4.000 vong/phut x 7.1 cm3/vong = 28.400 cm3/phut = 28,4 dm3/phut = 28,4 lit/phut (0.3d)
Cong suit bom = P xQ/600 = 250 x 28,4/600 = 11,83 KW (0.3d)
Cong suat dong co dien = Cong suat bom/Hieu suat bom = 11.83/0,845 = 14 KW (0.4d)
Cau d: Tinh luc keo vat m co lai
Fco - Fms - Fmst = ma (0.2d)
Fms = jLi.mg = 0,15 x 2.000 x 9,81 = 2943 N (0,2d)
Fco = Fms + Fmst + ma = (i.mg + 4500 + 2000x5m/s2= 0,15x2000x9,81 +4500 + 2000x5 =
17443 (N) (0.6d)
Cau e: Tinh luu luung cap vao xy lanh khi xy lanh co lai

Luu luong cap vao xy lanh = tiet dien x van t6c dich chuyen -> qR =Hieu suit x (A-a) x v (0.3d).
-» qR=0,9 x3,14 x (D2 - d2)/4 x v = 0,9 x 3,14 x (0,252 - 0,162)/4 x 0,07 = 0,00183 m3/s = 109,512 lit/phut
(0.7d).

Ghi chu.’Can bo coi thi khdng du gc g id i thick tie thi.

Chuan diu ra cua hgc phin (ve kien thuc)
[G 1.2]: Ve dugc ky hieu, trinh bay dugc chuc nang va nguyen ly hoat dong cua
cac thiet bi dugc sir dung trong he thing truyen dgng thuy luc va khi nen.
[G 1.3]: Co kha nang thiet ke cac mach dieu khien khi nen - dien khi nen
fG 4.11: Tinh toan dugc thong si co ban cac thiit bi khi nen - thuy luc

N$i dung kiim tra
Cau 2
Cau 1
Cau 2

Ngay 10 thang 12 nam 2019
Thong qua Truong nganh

C

A

1 . ;

n i/fl/A 'T

KẾT HỢP

E1=START^S3---E2=S4--E3=S2^S5--A+=Y1=K2-A-=Y2=K3--B+=Y3=(K1^S6)ORK2OR
(K3^S1-)--C+=Y5=K1






TRÜÖNG DH SU-PHAM KY THUÄT TPHCM
KHOA DAO TAO CHAT Ll/giNG CAO

Mön: Cong nghe Thüy lue vä Khi nen
Mä mön hoc: PNHY330529 - PNHY230529

Chü ky giäm thi 2

Chü ky giäm thi 1

DE THI CI HQC KY 2 NÄM HOC 2022/2023

D6 sư/Mä d€: 2

D6 thi co 4 trang.

Thai gian: 75 phüt.
Dugc phep sü düng täi lieu, khong duge düng laptop.
SV läm bäi tryc tiep tren de thi vä nop lai de

Diem vä chü1ky
CB chäm thi thu nhät

CB chäm thi thu hai

Ho vä ten:................................................................
Mä so SV :......................................................................
So TT:......................Phöng thi:..............................

Cäu 1 (1.5 diöm): Ve ky hieu hoäc dien ten cäc thiet bi khi nen theo bang ducri
Relay thdi gian khi nen
ngät chäm

~s

1
:

_x:

■ r ~ 11
;l —
12

1

Van tiet liru 1 chieu
1

3

2

Van Logic OR

1

3

Cum bäo duang khi
nen

Van däo chieu 5/2 täc
döng häng dien vä nüt
an 2 phia
Xv lanh khi nen täc
döng ken cö giäm chän
cö the dieu chinh, cö tu

>

Cäu 2 (3.5 diem): Thiet ke mach dieu khien dien - khi nen theo phucmg phäp ket hop cho
so dö hänh trinh birdc hinh 1:
1

2

3

4

Delay 3s

5

6

7
52

S1
54
53

S6
55

Sö hieu: BM1/QT-PBT-RBTV/02

Län soät xet: 02

Ngäy hieu lue: 15/5/2020

Trang: 1/4


Sur düng 3 xylanh khi nen täc döng kep, van dien tvr täc döng 2 phia, dieu chinh duac van
töc chieu di ra cüa xy lanh B.

Sư hieu: BM1/QT-PDT-RBTV/02

Län sột xet: 02

Ngäy hieu Iure: 15/5/2020

Trang: 2/4


Van 4/3

tiết lưu 1 chiều

Van giảm áp

xylanh tác động kép

Cäu 3: Cho so dö mach thuy luc bö phän dieu chinh vi tri trong mäy cuön thep täm nhu hinh 2
a. Ghi ten cäc thiet bi trirc tiep väo nhän tuong üng tren hinh (1 diem)
b. Tinh luc cüa xylanh chieu di ra Foirrvä chieu di ve Fin, biet äp suät läm viec P = 2070 PSI, äp
suät buöng xä tuong ung = 300 PSI, hieu suät he thong lä 95%. (1 diem)
c. Tinh cöng suät thüy luc cho he thüy luc tren. (1 diem)
d. Tinh van töc cüa 2 xy lanh khi täc dong coil a-RAISE - V raise (mm/s) vä khi täc dong coil bV lower (mm/s), biet duong kinh cän piston lä d = 60mm. (1 diem)
e. Tinh thoi gian hoän thänh hänh trinh cüa 2 xy lanh khi täc dong coil a-RAISE - tRAisE (s) vä
khi täc dong coil b-LOWER tLowER(s). (1 diem)

So hieu: BM1/QT-PDT-RDTV/02

Län soät xet: 02

Ngäy hieu l\rc: 15/5/2020

Trang: 3/4


Ghi chü: Cän bö coi thi khöng duoc giäi thich de thi.
Chuän däu ra cüa hoc phän (ve kien thirc)
[CLOl]: Co khä näng ve dugc ky hieu, doc hieu vä phän tich mach dieu
khien, trinh bäy dugc chüc näng vä nguyen ly hoat dong cüa cäc thiet bi
dugc sü düng trong he thong thüy luc vä khi nen
[CL05]: Cö khä näng thiet ke mach dieu khien, mö phöng bäng cäc phän
mem ho trg he thong khi nen, dien - khi nen, thüy luc, dien - thüy luc theo
yeu cäu dät ra
[CL06]: Cư khä näng tinh tộn, phän tich cäc ket quä tu quä trinh mö phöng
he thong truyen döng thüy luc - khi nen

Nöi düng kiem tra
Cäu 1
Cäu 2, Cäu 3
Cäu 3

Ngäy 29 thäng 5 näm 2023

p, Triröng ngänh

So hieu: BM1/QT-PDT-RDTV/02