tugas 5f

P3.340 The water jet in fig. P3.40 strikes strikes normal to a fixed plate . neglect gravity gravity and friction friction and compute compute the force F reguired to hold the plate fixed in newtown..

Plate Dj = 10 cm

F Diket :

Dj = 10vjcm ; rj = 5cm = 0,05m = 8 m/s Vj = 8 m/s Ditanya : F = ? Solusi : Akibat tekanan air dengan kecepatan vj pada pipa water jet maka timbul gaya dengan arah sama dengan arah vj yang kita namakan Fj. .

Aliran masa pada pipa water jet kita misalkan : mj .

mj

= ρ.A.vj = 1000kg/m3. π(0,05m)2.8m/s = 62,8 kg/s

Gaya Fj dapat dihitung dengan pers: Fj

.

= mj (vj) = 62,8 kg/s(8m/s) = 502,4 N Gaya F dan Fj bekerja pada satu satu sumbu ( misalkan sumbu x ) karena plate tidak bergerak maka berlaku hukumnewton I: ΣF = 0 ΣFx = 0 Fj – F = 0 F = Fj = 502,4 N

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EXAMPLE 2.12

A drag drag race racerr rest rests s her her coff coffee ee mu mug g on a hori horizo zont ntal al tray tray whil while e she she accelerates at 7 m/s2. The mug is 10 cm deep and 6 cm in diameter and and cont contai ains ns coff coffee ee 7 cm deep eep at rest. st. (a) Assumi Assuming ng rigi rigidb dbod ody y acceleration of the coffee, determine whether it will spill out of the mug. (b) Calculate the gage pressure in the corner at point A if the density of coffee is 1010 kg/m3. Solution

The free surface tilts at the angle θ given by Eq. (2.55) regardless of the shape of the mug. With a z = 0 and standard gravity, -1

θ = tan

a x g

= tan

-1

7,0m / s 2 9,81m / s 2

= 35.5°

If the mug is symmetric about its central axis, the volume of coffee is conserved if the tilted surface intersects the original rest surface exactly at the centerline, as shown in Fig. E2.12.

E2.12

Thus the deflection at the left side of the mug is : z = (3 cm)(tan θ) = 2.14 cm

Ans. (a)

This is less than the 3-cm clearance available, so the coffee will not





pA =ρg( z z surf A) =(1010 kg/m3)(9.81 m/s2)(0.07 m) =694 N/m2 = 694 surf - z Pa

During acceleration, Eq. (2.56) applies, with G = [(7.0) [(7.0)2 + (9.81) (9.81)2]1/2 = 12.05 m/s2. The distance Δs down the normal from the tilted surface to point A is Δs = (7.0 + 2.14)(cos θ) = 7.44 cm

Thus the pressure at point A becomes pA= ρ.G. Δs = 1010(12.05)(0.0744) 1010(12.05)(0.0 744) = 906 Pa (b)

Ans.

which is an increase of 31 percent over the pressure when at rest.

EXAMPLE 2.13

The coffee cup in Example 2.12 is removed from the drag racer, placed on a turntable, and rotated about its central axis until a rigid-body mode occurs. Find (a) the angular velocity which will cause the coffee to just reach the lip of the cup and ( b) the gage pressure at point A for this condition

.

Solution Part a

The cup contains 7 cm of coffee. The remaining distance of 3 cm up to the lip must qual





Solving, we obtain Ω² =1308 (a)

or

Ω = 36.2 rad/s = 345 r/min

Ans.

Part b

To To comp comput ute e the the pres pressu sure re,, it is conv conven enie ient nt to put put the the origi rigin n of coordinates r and z at the bottom of the free-surface depression, as shown in Fig. E2.13. The gage pressure here is p0 = 0, and point A is at (r , z ) = (3 cm, -4 cm). Equation (2.63) can then be evaluated p A = 0 - (1010 kg/m3)(9.81 m/s2)(-0.04 m)+ ½ (1010 kg/m3)(0.03 m)2(1308 rad2/s2) = 396 N/m2 + 594 N/m2 Ans. (b) = 990 Pa

This is about 43 percent greater than the still-water pressure p A = 694 Pa.





P3.12

Water at 20°C flows steadily through a closed tank, as in Fig. P3.12. At section 1, D1= 6 cm and the volume flow is 100 m 3/h. At section 2, D2 =5 cm and the average velocity is 8 m/s. If D3 = 4 cm, what is ( a) Q3 in m3/h and (b) average V 3 in m/s?

Diket: D1 = 6 cm

Ditanya : 3

Q1 = 100 m /h D2 =5 cm = 0,05 m V 2 =8 m/s D3 = 4 cm = 0,04 m

a) Q3 = ? b) V 3 = ?

Solusi : Debit masuk = debit keluar Q in = Qout Q1 = Q2 + Q3 Q1 = ¼ π ( D2)2.V2 + Q3 100 m3/h = ¼ . 3,14 (0,05 m)2. 8 m/s. 3600s/h + Q3 100 m3/h - 56.52 m3/h = Q3 Q3 = 43.48 m3/h Answ (a)





43.48

m3/h

= ¼ . 3,14 (0,04 m)2 .V3 3

.V3

=

43,48 m /h 0,001256m

2

34617,83m

2

V3

=

V3

= 9,62 m/s

/h

3600 s / s Answ (b)

P3.14

being filled through section 1. Assume incompressible incompressible flow. First derive an analytic expression for the water-level change dh/dt in terms of arbitrary volume flows (Q1, Q2, Q3) and tank diameter d . Then, if the water level h is constant, determine the exit velocity V 2 for the given data V1 = 3 m/s and Q3 = 0.01 m3/s.

Diket : V1 = 3 m/s

Q3 = 0.01 m3/s. D1 = 5cm = 0,05 m D2 = 7cm = 0,07 m Ditanya : V2 = ? Solisi : Debit masuk = debit keluar Q in = Qout Q1+ Q3 = Q2 ¼ π ( D1)2.V1 + Q3 = ¼ π ( D 2)2.V2 2 3 ¼ π ( D1 ) .V 1 + Q3 ¼ π ( 0,05 m )2 .3 m/s + 0.01 m /s V







P3.23

The The hypod hypoderm ermic ic needl needle e in Fig. Fig. P3.23 P3.23 conta contains ins a liquid liquid serum serum (SG= (SG= 3 1.05). If the serum is to be injected steadily at 6 cm /s, how fast in in/s should the plunger be advanced ( a) if leakage in the plunger clearance is neglected and (b) if leakage is 10 percent of the needle flow?

Diket :

D1 = 0,75 in D2 = 0,030 in Q1 = 6 cm3/s = 0,366 in3/s

Ditanya :

(a). V2=?; if leakage in the plunger clearance is neglected. (b). V2=?; if leakage is 10 percent of the needle flow.

Solusi

(a). Debit masuk = Debit keluar Q in = Qout Q1 = Q2 0,366 in3/s = ¼ π ( D2)2.V2 0,366 in3/s = ¼ . 3,14 (0,030 in)2. V2





V2 =

Q1 − Q3 ¼ π ( D2 )

2

=

0,366 in 3 /s - 0,0366 in 3 /s. /s. ¼ . 3,14 (0,030 in)

2

V2 = 466,24 in/s

= 466,24 in/s Answ (b)

o

1. Tekana Tekanan n dikam dikampus pus = 87kP 87kPa, a, T=29 T=29 C. Tekanan dipantai = 101,325kPa,T=25oC.Berapa tinggi kampus dari permukaan laut? Jawab : P = Pa(1-B.Z/To)g/R.B (1-B.Z/To) 5,26 = P/Pa {1-(0,0065k/m).Z/288}5,26 = 87/101,325 288-{(0,0065k/m).Z/288} 288-{(0,0065k/m).Z/288} = (0,8586)1/5,26 288-0,065.Z = 279,774 -0,065.Z = -8,226 Z = 1265,539m





P = 96,2584kPa

tugas 5f

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