Stoichiometry New My

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Question no. 1.2:
Convert μ = 20.0 g/m s to lbf hr/ft2
20 g 1 kg2.2045
lbm1m Lbf s2 1 hr
m s1000 g kg 3.28 ft
32.174 lbm ft
3600 s
=1.160×107 lbf hr/ft2
Question no. 1.3:
K = gcDp2/32(Re/f)
Dp = diameter (m) Dp = ft(American engineering system)
Re/f = unitless
gc = 32.174 lbm ft/lbf s2
(a)
In American engineering system
K = 32.174 lbm.ft . ft2⁄ 32 lbf s2
= 1.005 lbm ft3 ⁄ lbf s2
(b)
In SI system
K = 9.8 m2⁄ 32 s2
= .3062 m2⁄ s2
Question no. 1.4:
Z = 1+ ρB + ρ2C+ ρ3D
Z = Dimensionless
ρ = g mol ⁄ cm3
units of B,C,D = ?
ρ = g mol / cm3 , then B = cm3/g mol
C = (cm3/g mol)2 = cm6 / g mol2
D = ( cm3/ g mol)3 = cm9/g mol3
Z = 1 + ρ*B* + (ρ*)2C = (ρ*)3D
ρ* = lbm /ft3 , then B* = ft3/lbm
C* = (ft3⁄ lbm)2 = ft6/ lbm2
D* = (ft3/lbm)3 = ft9/lbm3
i. B → B*=?Convert cm3/g mol to ft3/lbm
cm3 454 g mol 1 m3 (3.2808 ft)3
g mol lbm mol (1000 cm)3 1 m3
= .01603 ft3 /lbm mol
.01603 ft3 lbm mol
lbm mol lbm mol weight
=.01603 ft3/lbm mol
B* = 62.38 mol weght B
ii. Convert C to C*ft3/lbm → cm6/(g mol)2
cm6 (454 mol)2 1 m6 (3.2808 ft)6
(g mol)2 (lbm mol)2 (100cm)6 1 m6
= .0002570 ft6/(lbm mol)2
.0002570 ft6 (lbm mol)2
(lbm mol)2 (lbm mol weight)2
C = .0002570 C*/(mol weight)2
C* = 3891.050 mol weight C
iii. Convert D → D*cm6/(g mol)3 to ft9/(lbm)3
D* = (B*)3
= (62.3)3 (mol weight)3 D
= 242 × 103 (mole weight)3 D
Question no. 1.6:
Convert:
(a) .04g/min inch3 to lbm hr ft3
.04 g 1 kg 2.2lbm 60 min (12 inch)3
min inch3 1000g 1 kg 1 hr 1 ft3
= 9.134 lbm/hr ft3
(b) 2L/s to ft3/day
2L 1 ft3 3600 s 24 hr
s 28.328 L 1 hr day
= 6101.69 ft3/day
(c) 6 (inch) cm2 / (yr) (s) (lbm) (ft2) to all SI units
6 in cm2 1 ft 1 m2 Lbm
3.280 ft
1000 g
1 yr 3.28 ft
yr s lbm ft2
12 inch
(100cm)2 454 g 1 m 1 kg 3 ft 1 m
= 3.951 × 103 kg1 s1
Question no. 1.7:
CONVERT
K= Btu/hr(ft)2(ºF/ft) to KJ/(day m2 ºC/cm)
Btu 1055J 1 K J24 hr
(3.28 ft)2
1.8 ºF
12 inch
2.54cm
hr(ft)2(ºF/ft) 1 Btu1000
J1
day1 m2 1 ºC 1 ft 1 inch
= 1.49 × 104 KJ/(day) (m2) (ºC/cm)
Question no. 1.8:
(a)
Velocity = 3 ft/s
K.E = (ft) (lbf)/lb
K.E = 1/2mv2
K.E/m = v2/2gc
= (3 ft)2/s2 /2 × 32.174 lbm ft/lbf s2
= 0.139 (lbf) (ft)/lbm
(b)
Diameter (d) = 2 inch
Flow rate = gal/min =?
d = 2/12 ft
Flow area = π/4 d2
= π/4 4/144 ft2
= π / 144 ft2
Flow rate = flow rate × velocity
= π 3 ft3/144 s
= 0.0654 ft3/s
0.0654 ft3 60 s 2832 liter 1 gal
s 1 min 1 ft3 3.786 liter
=29.3600gals /min
FLOW RATE = 29.3600 GAL/MIN
Question no. 1.9:
No, it is not correct to so label a package. Because the contents of packages contain mass, not weight. So 250, is mass which should be
expressed in grams. Weight is not expressed in grams.
Question no. 1.11:
CONVERT :
20 Hp > KW
20 hp 746 Watt 1 KW
1 hp 1000 Watt
= 14.92 KW
It is not possible to drive a 68 KW generator.
Question no. 1.13:
525 mile\hr consumes fuel= 2200gal\hr.
1 mile\hr. consumes=2200gal\525hr
1000 mile\hr consumes= 2200×1000gal\525 hr
=4190.4n gal.
475 mile\hr consumes=2000gal\hr
1 mile\hr consumes=2000gal\475hr
1000 mile\hr consumes=
=4210.5gal
Difference= (4210.44190.4) gal
=20.05 gal
Question no. 1.14:
Given equation:
Cal 454g.mole 4.18J 1BTU 1K g.mole
k.g.mole 1lbm.mole 1cal 1055J 1.8˚F 64g
=0.015BTU\lbm.˚F
×Btu\lbm x F.
Question no. 1.15:
GIVEN:
W=21.3kg
Here weight is given in kg instead of mass.
It means weight is equal mass keeping g constant.
W=mg
=21.3×9.8
=208.7Kg
QUESTION NO. 1.16:
In S.I system:
POWER=F×V
=
In engineering system:
QUESTION NO. 1.17 :
Kg.ft2 2.204lbm lbf.s2 btu
S2 Kg 32.17lbm.ft 778.2ft.lbf
=8.8×105btu
QUESTION NO. 1.18:
180tons.ft2 1000kg 2.2045lbm Lbf.s2
S2 Ton Kg 32.17lbm.ft
K.E=12308.07lbf.ft
QUESTION NO. 1.19:
(1)Conversion of T units :
N lbf M2
M2 4.45 N (3.28ft)2
T=0.0206lbf.\ft2
(2)Conversion of ρ units
Kg 2.204lbm M 3
M 3 Kg (3.28ft) 3
ρ=0.0622lbm\ft3
(3) Conversion of u units:
M 3.28 ft
S m
U=3.28ft\s
QUESTION NO. 1.20:
It can be dimensionally consistent
Question 1.21:
So option © is correct
QUESTION NO. 1.22:
Flow rate=10cm3\s
Distance=125ft to cm
125ft 1m 100cm
3.28ft 1m
=3810.0 cm
FLOW RATE =flow area×velocity
QUESTION NO. 1.23:
Putting units of quantities in expression
It means that h is a dimensionless
QUESTION NO. 1.24:
Therefore v is dimensionless.
QUESTION NO. 1.25:
q = 0.415 (L  0.2 ho) ho1.5
where :
q = volumetric flow rate, ft/s
L = crest height, ft
ho= weir head, ft
g = acceleration of gravity, 32.2 ft/(s)
By putting the values in equation:
ft3/ s = (ft – ft)(ft)1.5
ft3/ s = (ft. ft1.5. ft0.5) / s
ft3/ s = ft3/s
QUESTION NO. 1.26:
kx = Ku0.487
Where :
kx = mol/(cm2.s)
u = cm/s
By putting the values in equation:
mol/(cm2.s) = K (cm/s)0.487
K = mol/(cm2.s) × ( s0.487/cm0.487 )
K = mol/ (cm2.487× s0.0153)
Now:
u = cm/s ft/s
Cm 3.208 ft 1m
S 1m 100cm
u = 0.032808 ft/s
kx = K u0.487
By putting the values in equation:
Mol/ (cm2.s) = K (ft/s)0.487
K = mol/(cm2.s) × (ft/s)0.487
QUESTION NO. 1.27:
A) REYNOLDS NUMBER =
Diameter = 2 inch = ft = 0.166 ft [ : 1 ft = 12inch]
u = 10 ft/s = 10(3600) ft/hr = 36000 ft/hr
= 62.4 lb/ ft3
= 0.3
Re=
Re = 1.243× 106
b) Reynolds number = Diameter = 20 ft
u = 10 mil/hr ft/hr
10 mile 1.6 km 100 m 3.2808 ft
Hr 1 mile 1 km 1 m
u = 5.28 × 104 ft/hr
= 1 lb/ft3
= 0.14
0.14 lbm 3600 s
s. ft 1 hr
= 0.0504
Re =
Re = 2.096 × 107
c) Reynolds number =
Diameter = 1 ft
u = 1 m/s
1 m 3.2808 ft 3600 s
S 1 m 1 hr
u = 11810.88 ft/hr
= 12.5 kg/m3 lbm/ft3
12.5 kg 2.21 lbm 1 m3
m3 1 kg (3.2808 ft)3
= 0.7787 lbm/ft3
= 2 × 10 6 cP
= 2× 109
2× 109 Kg 2.2045 lbm 1 m 3600 s
m.s 1 kg 3.2808 ft 1 hr
= 4.828 × 106
Re =
Re = 1.9049 × 109
d) Reynolds number =
Diameter = 2 mm
2 mm 1 m 3.2808 ft
1000 mm 1 m
Diameter = 0.00656 ft
u = 354.32 cm/s
25 lb/ft3
= 1 × 106 cP
= 1× 103
1× 103 kg 2.2045 lbm 1 m 3600 s
m.s 1 kg 3.2808 ft 1 hr
Re =
Re = 9.6× 104
QUESTION NO. 1.28:
lbm gal 60 min
(3.2808ft)3 1 m3 3.785 lit
1 kg 1 ton
ft3 min 1 hr 1 m3 1000 lit
1 gal 2.204lbm 1000 kg
3.645 × 103
QUESTION NO. 1.30 :
a) What is the molecular weight of CaCO3=?
Molecular weight of CaCO3 = 40 + 12 + 48
= 100 g/mole
b) How many moles are in 10g of CaCO3?
Mole =
Mole = g mole = 0.1 g mole of CaCO3
c) How many lb moles are in 20 lb of CaCO3?
Mole =
Mole = lb mol = 0.2 lb mol of CaCO3
d) How many g are in 2 lb mol of CaCO3?
Mass = mol × molecular mass
Mass = 2 lb mol × 100 lb/(lb mole)
Mass = 200 lb
200 lb 1 kg 100 g
2.2045 lb 1 kg
Mass = 90723 g of CaCO3
QUESTION NO. 1.33 :
Convert the following:
a) 4 g mol of MgCl2 to g=?
4 g mol of MgCl2 95 g of MgCl2
1 g mol of MgCl2
= 4 × 95 g of MgCl2
= 380 g of MgCl2
b) 2 lb mol of C3H3 to g=?
2 lb mol of C3H3
44 lb of C3H3 454 g of C3H3
1 lb mol of C3H3 1 lb of C3H3
= 2 × 44 × 454 g of C3H3
= 39952 g of C3H3
c) 16 g of N2 lb mol=?
= lb mol of N2
= 1.2 × 103 lb mol of N2
d) 3 lb of C2H6O to g mol
3 lb of C2H6O 1 lb mol of C2H6O 454 g mol of C2H6O
46 lb of C2H6O 1 lb mol of C2H6O
= g mol of C2H6O
16 g of N2 1 g mol of N 2 1 lb mol of N2
28 g of N2 454 g mol of N2
= 29.6 g mol of C2H6O
QUESTION NO. 1.36:
% age of C = 42.11%
% age of O = 51.46%
% age of H = 6.43 %
Molecular weight = 341
341 g of compound 42.11 g of carbon 1 g mol of carbon
100 g of compound 12 g of C
= g mol of C
= 11.96 which is approx. 12
341 g of compound 51.46 g of oxygen 1 g mol of oxygen
100 g of compound 16 g of O
= g mol of O
= 10.9 which is approx. 11
341 g of compound 6.43 g of hydrogen 1 g mol of hydrogen
100 g of compound 1 g of H
= g mol of H
= 21.9 which is approx. 22
So compound is C12H22O11
QUESTION NO. 1.38:
The specific gravity of acetic acid is 1.049.what is the density in lbm\ft3?
Sp. Gravity= density of sub/density of water
Density of sub = sp. Gravity x density of water
= 1.049 x62.45 lbm\ ft3
QUESTION NO.1.39:
The specific gravity of a fuel oil is 0.82.what is the density of oil in lbm\ft3?
Sp. Gravity = density of substance/density of H2O
Density of substance = sp. Gravity x density of water
= 0.82 x 62.45 lbm\ ft3
= 51.168 lbm\ ft3
QUESTION NO.1.40:
Mass of the oil is 1000 lbs and specific gravity is 0.926 what is the size of drum in gallons?
Sp. Gravity = density of sub/density of H2O
Density of oil = sp. Gravity x density of water
= 0.926 x 62.45 lbm\ ft3
= 57. 782 lbm/ ft3
Volume = mass/density
=1000 lbm /57. 782 lbm\ ft3
=17.3064 ft3
=17.3064 ft 3 │ 7.485gallons
│ 1 ft3 =129.5 gallons
QUESTION NO.1.41:
The concentration of H2SO4 96% and specific gravity is 1.858 .concentration of the solution to be made is 0.1 molar.
a) the weight of 96 % acid needed per ltr of solution? 0.1molar=0.1moles of
0.1molar = 0.1 moles of H2SO4  98g of H2SO4
1ltr of sol. │1 mole of H2SO4
= 98 g of H2SO4 / ltr
As H2SO4 is 98% pure ; therefore
0.1 M = 9.8
96%
= 10.2 g of 96% H2SO4 \ ltr of sol.
QUESTION NO. 1.43:
Sp. gravity of KOH sol = 1.0824
Mass of KOH = 0.813
Mass fraction of KOH and H2O in soln. =?
Density of KOH solution = 1.0824 ρ of H2O
= 1.0824 62.4 1ft3
ft3 7.485gal
= 9.02 lb/gal
=0.09
=0.91
QUESTION NO. 1.45:
a) Density at 25'C in kg/m3?
Sp.gravity =
Ρ of benzene =0.90 1000kg/ m3
= 900kg/ m3
b) Specific volume at 25’C in ft3/
Sp. volume =
=
= 1.01 103
1.1 103 m3 (3.2808)3 ft3 1kg 454g
kg 1 m3
1000g 1
=
= 0.0178
c) Sp. gravity= 0.90
V = 1.5L
Mass of bottle =232g
Total mass =?
Sp. gravity =ρ of substance/ ρ of H2O
ρ of liquid = 0.90 1000kg/ m3
= 900kg/ m3
ρ of liquid = mass/volume
Mass= ρ of liquid volume
= 900kg/ m3 1.5L
900kg 1.5 1000g 1 m3
m3 1kg 1000
= 1350g
Total mass = 350+232
= 1582g
Total weight = 1582g 9.8
= 15503.6
=
= 15.5036 N
QUESTION NO. 1.46:
a) Weight percent HNO3?
HNO3 wt, percent =
=
= 63٪
b) Pounds HNO3 per cubic foot of sol at 20’C?Sp. gravity = ρ of soln./ ρ of H2O
ρ of soln. = 1.382 1000kg/ m3
= 1382kg/ m3
1382kg 1000g 1 1 m3
m3 1kg 454g (3.2808)3 ft3
=
c) Molarity (gram moles of HNO3 per litre of soln at 20’C)Given mass of HNO3 = 1.704kg
Molar mass of HNO3 = 63g
No of moles =
=
= 27 moles
Molarity = No of moles/ of soln.
= 27
QUESTION NO. 1.47:
Hg = 200mg/analysis
Limit of Hg = 0.0005ppm
Waste water available = 100,000gal/day
Density of water = mass/volume
Mass = ρ volume
= 1000 100,000gal
100,000gal 3.785 1m3 1000kg
1gal 100 m3
Mass = 378.500kg
200mg Hg 1kg H2O
378500kg H2O 106mg
=
= 0.000528 ppm of Hg (b/c )
QUESTION NO. 1.51:
Density (r) of solution=mass/volume
Volume of solution =
=
= 1137.5 gal
Volume =
Volume =
Volume = 151.836 ft
QUESTION NO. 1.53:
Volume = V1 = 5000 bbl Volume = V2 = 20,000 bbl
Density = r1 = 28° API Density = r2 = 15° API
10010 lb gal
8.80 lb
1137.5 gal
3.78 liter 1 m (3.2808 ft)
1 gal 1000 liter
1 m
r1 = r2 =
m1 = r1 m2 = r2
m1 = 28 °API 5000 bbl m2 = 15°API 20,000 bbl
Specific gravity = Specific gravity =
= 0.8871 = 0.965
Specific gravity =
r1 = 0.887 0.999
r1 = 0.886113
m1 = r1 v1
= 0.886113 5000 bbl
= 4430.565
4430.565 g bbl 42 gal 3.78 liter 1000 cm3
Density of substance
Density of H2O
cm3 1 bbl 1 gal 1 liter
= 4430.565 42 3.78 1000 g
= 7.0 108 g
Specific gravity =
r2 = 0.965 0.999
= 0.9640
m2 = r2 v2
= 0.9640 20,000 bbl
= 19280.7
19280.7 g bbl 42 gal 3.78 liter 1000 cm3
cm3 1 bbl 1 gal 1 liter
m2 = 19280.7 42 3.78 1000 g
= 3.06 109 g
m1 + m2 = 7.0 108 + 3.06 109
Density of substance
Density of H2O
= 3.7 109 g
V1 + V2 = 20,000 + 5000
= 25,000 bbl
r of solution =
3.7 109 g 1 lbm
25,000 bbl 454 g
=
= 325.99
3.7 109
g
25,000 bbl
3.7 109 lbm
25,000 454 bbl
=
= 59
1.54:
In a hand book you find that the conversion between ˚API and density is 0.800 and density = 45.28˚API. Is this a misprint?
Solution:
= 0.800
Now we know that
3
3
= 800kg/m3
325.99 lbm
1 bbl 7.485 gal
bbl 42 gal 1 ft3
325.99 7.485 lbm
42 ft3
Yes it is a misprint.
1.55:
Solution:
100 = 19%
1.56:
NIOSH sets standards for CCl4 in air at 12.6mg/m3 of air (a time weighted average over 40hr). The CCl4 found in a sample is 4800ppb. Does the sample exceed the NIOSH standard? Be careful!
Solution:
4800ppbCCl4 103 gmol air Kgmol 154 gCCl4 103mgCCl4109gmol air 1kgmol air 22.8 m3 gmol CCl4 g CCl4 = 32.44mg CCl4/m3
Yes it exceeds the NIOSH standard.
Section. No 1.4
1.59:
You have 130kg of gas of the following composition: 40% N2, 30% CO2 and 30% CH4 in a tank. What is the average molecular weight of the gas?
Solution:
Component Kgmole Molecular weight Kg of gas
N2 40 28 40×28=1128 CO2 30 44 30×44=1320 CH4 30 16 30×16=480
Average molecular weight =
1.60:
You have 25lb of a gas of the following composition: CH4 80%, C2H4 10%, C2H6 10%. What is the average molecular weight of the mixture? What is the weight (mass) fraction of each of the components in the mixture?
Solution:
Components Kgmole Molecular weight Kg of gas CH4 80 16 2400 C2H4 10 28 280 C2H6 10 30 300
1.61:
You analyze the gas in 100kg of gas in a tank at atmospheric pressure, and find the following:CO2: 19.3% N2: 72.1% O2: 6.5% H2O: 2.1%
What is the average molecular weight of the gas?
Solution:
Components Kgmole Molecular weight Kg of gas CO2 19.3 44 849.2 N2 72.1 28 2018.8 O2 6.5 32 208 H2O 2.1 18 37.8
1.62:
Two hundred kg of liquid contains 40% butane, 40% pentane and 20% hexane. Determine the mole fraction composition of the liquid and the mass fraction composition on hexane free bases.
Solution:
Component Mass in kg No of moles Butane (40/100)×200=80kg 1.37 Pentane (40/100)×200=80kg 1.11 Hexane (20/100)×200=40kg 0.46
%
On hexane free basis mass fraction is
1.64:
A fuel gas is reported to analyze, on a mole basis, 20% methane, 5% ethane and the remainder CO2. Calculate the analysis of the fuel gas on the mass percentage basis.
Solution: Base is 100gmol gas Component Kg moles Molecular weight Kg of gas CO2 75 44 3300 CH4 20 16 320 C2H6 05 30 150
Mass percentage of CH4 =
Mass percentage of C2H6=
Mass percentage of CO2=
Self Assessment Test (1.5)
Q.No.1 (a)
The reference point of Celsius is 0˚c freezing point of water and boiling point of water is 100˚c.
(b).
The reference point for ˚F is 32˚F freezing point of water and 212˚F is boiling point of water.
Q.No.3
Yes is higher than 10
Q.No.4
J 1cal K ˚Rgmol. K 4.18J 1.8 R ˚F
Capacity = 92.76+0.198T˚F
Q.No.5
K ˚R
40 40 233 420 25 77 298 537 425 796 698 1256 234 390 38.8 69.8
Q.No.6
Immerse in ice water bath and mark 0 , immerse in boiling water at 1atm
pressure and mark 100 . From 0 to 100 is divided in desired intervals.
Section 1.5
1.67:
Find temperature in Kelvin, Fahrenheit and Rankin.
Also
Now for Rankin
R = 510
1.68:Can negative temperature measurement exist?
Answer:
Yes, if the temperature scale is linear relative or logarithmic scale. No if the scale is absolute.
1.69:
The heat capacity of acetic acid in j/(gmol)(k) can be calculated from the equation
Where T is in K. convert the equation so that T can be introduced into the equation in ˚R instead of K.
Solution:
1.70:
Convert the following temperature to the requested units
10 to
10 to
25 to K
150K to ˚R
Solution:
10 to
10 1.8
1
18 + 32 = 50
10 to
10 1.8
1
18 + 32+460R=510˚R
25 to K
25 1.0
1.8
150K to ˚R
150K 1.8˚R
1.0K
150K = 270 ˚R
1.72:
In a report on the record low temperatures in Antarctica, Chemical and
Engineering news said at one point that the mercury dropped to 76 . In
what sense is that possible?
Answer:
The instrument does not contain mercury but it has to contain a fluid that
responds at 76
1.73:
Further the degree Celsius is exactly the same as Kelvin. The only difference is that zero degrees Celsius are 273.15 Kelvin. Use of Celsius temperature gives us one less digit in most cases. Comment on the quotation. Is it correct? If not, in what way or sense it is wrong?
Answer: The first sentence really mean that the unit interval .
The second sentence is satisfactory when temperature measure in Celsius but only the difference omitted.
1.74:
Calculate all the temperatures from given one temperature.
Solution:
˚R K
A 140 600 333 60 B 77 537 298 25 C 40 500 277.8 4.5 D 40 420 233 40 E 1000 1460 811 538 F 540 1000 555 282 G 1340 1800 1000 727 H 1832 2292 1273 1000
1.75:
The emissive power of a blackbody depends on the fourth power of the temperature and is given by
Where W = emissive power, Btu/(ft)2(hr) A = StefanBoltzmann constant, 0.171×108 Btu/(ft)2(hr)(˚R)4
T = Temperature, ˚RWhat is the value of A in the units J/(m)2(s)(K)4?
Solution:
0.17×108Btu 1.84R4 1hr ft2 Inch2 1002cm2 1055Jft2.hr. R K4 3600s 144inch2 (2.54)2cm2 1m2 1Btu
Self Assessment Test (1.6)(1).
(2).
(3).
Barometric pressure = vacuum pressure = absolute pressure
(4). (a).
80mmHg 14.7Psi
760mmHg = 15.47Psi(b).
80mmHg 1atm760mmHg
= 1.05 atm
(c).
800mmHg 101.325kPa
760mmHg = 106.65kPa
(d).
800mmHg 33.91ft
760mmHg
= 35.69ft
(5).
(a). Gauge Pressure(b). Barometric Pressure(c). Absolute Pressure
(6). Base Atmospheric pressure = 101.325kPa Absolute pressure = barometric pressure = vacuum pressure
Section 1.6
(1.77).
(1.78).
Air in a scuba diver’s tank shows a pressure of 300kPa absolute. What is
the pressure in Psi, atm, inchHg, kg/c , kPa, bar, ft ?
Solution:
(a)
= 43.52Psi(b)
300kPa 1atm
101.325kPa = 2.96atm
(c)
300kPa 1atm 29.92in.Hg101.325kPa 1atm
= 88.58in.Hg
(d)
300kPa 1000Pa 1atm 1.01325BarkPa 101325Pa 1atm
= 3.26 Bar(e)
300kPa 1atm 33.91ft.
101325kPa 1atm
300kPa 1000Pa atm 14.7Psi1kPa 101325Pa atm
= 100.39ft.
(1.79)
(a) Basis = 0.25m of waterAs we know that
(b)
As we know that
25000kg 9.8m sec. N Pa. atm 14.7Psi
100 se Kg.m N 101325Pa 1atm
= 3.55Psi
(1.80)
Side volume =
Volume of edges=
Volume of flour =
Total volume = 281.39
Now
Volume required floating
Now
`
(1.81)
50Psi 101325 N Kg.m
1000kg 14.7Psi Nse
= 35.16m
= 35.16
= 115.3 ft Not sufficient to pump water
(1.82)
Specific gravity of kerosene oil = 0.82
Comparing both sides
0.301in.Hg 25.4mm1 inch
(1.83)
4.2in.Hg 14.7Psi
29.92inHg
P = 2.06Psi
(1.84)
25.78inHg 760mmHg
29.91inHg
P = 655mmHg
(1.85)
According to John
9.75Psi 760mmHg14.7Psi
= 504mmHgYes John is right, but the pressure change at the top of peak.
(1.87)
(1.88)
176.6Pa 760mmHg101325Pa
P = 1.32mmHg
(1.89)
Assume that correction to the gauge reading is directly proportional to gauge reading.The correction is 735/760 to the gauge reading.