Step 2 df A
FTESELH'B under 1B GONE i5 QWEI'I BE:
P=Z
Herer p is the pressure applied under the writ. and F; is the terse applied an the park.
The area Aelthe force is applied an the circular area] is calculated as,
i
4
Here Ania the Area, at: is the diam
Step 1 of2 A
From Bernoulli's equation for laminar flow,
1 l
Pl +3pr +psh = P; +5.01%? + sts
Herer hub: is the height above and below a thin roof respectively. vl is the velocity of air inside
the house, v1 is the velocity of air outside the house. pl, F1
(11)
Consider the equation (7)
V =( 2(9.81mfs1)(1.03 1a)]
V1 2 4.49 nifs
Hence the numerical value of velocity is 4.49 this .
('3)
Consider the equation of flowsir rate in terms of V and:
JV
Flow(r) = m3!s
d:
. JV
Hence, the orair rate is n13 Is .
if:
(
EXPEI l HIISWEI
Increase in heighi = initial height I coefcient of thermal expansion 1 temperature rise
= 324x 1.2 x 105 x 19.5 = T5315 mm = 0.075316 rn
Comment )
Recall tltat weight is mass times acceleration due to gtatdty. 5o. weight of the duck will be:
W = my
Since, density is mass perunit vohune, we can write mass asm = pix. Hence. the aboveequation will
become:
W = cfw_Midst
cfw_b
Snbstintte the values in ab
LengthatDDC isL= 1m
Coefficient of 1ll'olume expansion of steel :5 = 3.6 x10 '5 f C
Coefficient of linear expansion of steel Cr : 1"31'3 : cfw_3.5 xiii]I '5 I 0C I 3 = 12 xl '5 I C
Coefficient of 1ll'olume expansion of invar :6 ' = 2.?xl '5 I C
Coefficien
0)
Consider the equation of Bernoullis equation at the opening of the
pool
1
P2 + pghz + E pi? = Constant . (4)
1
Pr + ,0th + E pig: 2 Constant . (5)
Where, P2 = Pifatnlospherio pressure) and 112 = II] III
1
P0 + E 19V: 2 Constant . (6)
Hence, the Bernoul
FWVEF I'l. IS an HUI. a UIIEIEFIIIHL HHHEUIE. UHCHUHE IE IIDW I5 CFHHQIHQ II'I li'lE way IE FJUUL I5 uralnlng.
the solution of this diferential equation is not the object of this problem.
Fiowlit] =
f. Express the cross sectional area in the openng A in t
Step 3 df A
This pressure is transmitted td the ddttam df the jug as per Pascal's law.
The extra terse applied at the hattem df thejug is calculated as,
Here, F is the extra terse exerted at the bdttdm dijug and Al is the area dtjug and di is the
diameter
in
It the ether tactcrs are kept ccnstant cther than the "ISM! rate Q and the radius r then the flew
rate is:
The radius is increases by 5% therelcre,
sQ =cfw_i .cslg Q
= Q[1.cs 1]
games
9
=21.55%
Hence the increase in flew rate is obtained trcm a 5.00% i
A host pours the remnants of several bottles ct wine into a jug after a party. The host then inserts
a cart: with a EDD-cm diameter mm the bottle, placing it in direct contact with the wine. The hdst
is amazed when the host pounds the cork into place and
If the ether feature are kept eenatant ether than the flaw rate Q and the radius r then the flaw
rate is:
The radius is deereaaea by 5% therefere,
a9 = Q [0.95) Q
= Q[1cfw_u.95)]
E I cfw_1.135
*2
= 13.5%
Henee, this is 1ireriiieizl that an 13.5% or nearly
There are two cans of food are sitting on a kitchen counter. The same pressure is exerted on the table by
each If the weights of the cans: however the cans are not identical and the densities of the oontents are
not identical. The radius of can 1 is r1 =
(51)
Consider the equation of Bernoullis equation
Pl +pghl +pl = P1 +33th +1.9V1 = Constant
. (1)
Consider the equation of Bernoullis equation at the top of the pool
P + pghl + % pl]: 2 Constant . (2)
Where, P1 = P[atn1ospherio pressure)
At the top of the
More Fluids
3-31-2017
Buoyancy
. We talked before about the force a fluid exerts on an object in it
Buoyancy
. We talked before about the force a fluid exerts on an object in it
. Always points in toward the object
Buoyancy
. We talked before about the fo
Thermal Physics
4-10-2017
Units of Heat
. A few units have traditionally been used to describe heat
. The calorie is defined as the energy needed to increase the temperature of 1
g of water from 14.5 C to 15.5 C
. This is different than the Calorie used i
More on Pressure
3-29-2017
Pascals Principle
. Since the pressure in a fluid depends on both the depth below the surface of the
fluid and the pressure at the surface, any change in that surface pressure must be
felt by each point in the fluid
Pascals Prin
Fluids in Motion, Thermal
4-7-2017
Finally, Bernoullis Equation
1
1
P1 V P2 V = mv22 mv12 + mgy2 mgy1
2
2
. Dividing both sides by the volume V (and noting that = m/V )
1
1
P1 P2 = v22 v12 + gy2 gy1
2
2
. Gathering terms from the lower end of the pipe and
Solids and Fluids
3-22-2017
States of Matter
. All matter is normally classified as one of four types
. Solid - definite volume and shape
. Liquid - definite volume but shape changes to conform to vessel
. Gas - volume and shape change to conform to vesse
Fluids in Motion
4-3-2017
Submerged Object
. Lets look at Newtons second law for a submerged object
X
Fy = B mg = ma
. Using the same substitutions as before (m = object Vobject ; B = fluid Vdisp g ),
Newtons second law can be written
(fluid object )Vobje