1. A 55 gal. drum of organic solvent, full and sealed, weighs 400 lbs. The empty drum weights 40 lbs. What is the density of the solvent, in kg/L? What is its specific gravity?
Variables:
Vdrum = 55 gal. = 0.208 m3
mempty = 40 lbs = 18.14 Kg
mfull = 400 lbs = 181.44 Kg
msolvent = ?
ρsolvent = ?
Relationships:
msolvent = mfull – mempty
ρ = m/V
Computation:
msolvent = 163.29 Kg
ρsolvent = msolvent/Vdrum = 163.29Kg/0.208m3 = 785.05 Kg/m3
2. A cylindrical object 0.5m in diameter and 2m long weights has a mass of 200kg. The object is floating end-down in water. How many cm of the object are above the waterline?
Variables:
d= 0.5m
h = 2m
m= 200Kg
ρl = 1000Kg/m3 (density of pure water)
Wcyl = ?
Total volume of object, V = ?
Volume of object submerged, Vsub = ?
Volume of liquid displaced, V1 = ?
Fb = ?
Submerged height of object, hsub = ? (Final answer)
Relationships:
V = πd2h/4
Volume of liquid displaced = Volume of object submerged, or, V1 = Vsub
Buoyant force, Fb = Weight of liquid displaced = ρlVsubg (Archimedes’ Principle)
Buoyant force, Fb = Total weight of object, Wcyl (Condition for floating)
Wcyl = mcylg = ρcylVg
Hence, ρcylVg = ρlVsubg, or Vsub/V = ρcyl / ρl = mcyl/(Vρl)
Vsub/V = hsub/h (constant cross-section of cylinder)
Hence, hsub = (hmcyl)/(Vρl)
Computation:
V = πdcyl2hcyl/4 = (3.14 x (0.5m)2 x 2m)/4 = 0.3925 m3
hsub = (hmcyl)/(Vρl) = (2m x 200Kg)/(0.3925m3 x 1000Kg/m3) = 1.02 m
3. A steel pipe runs vertically from a wellhead on the floor of the ocean to a point 10m above the surface. The pipe is filled with drilling mud having a density of 2.5 kg/L . What is the pressure on the pipe 200m below the surface of the water? At that point, what is the force on 1 cm2 of the pipe wall? (Density of seawater = 1.03 kg/L)
Variables:
mud density, ρ = 2.5 kg/L = 2500 kg/m3
Density of seawater, ρsw = 1.03 kg/L = 1030 kg/m3
Depth of P from top of the pipe, h = 200m(below surface) + 10m(above surface) = 210m
Area on which force is to be calculated, A = 1cm2 = 10-4 m2
Pressure exerted at point P, P = ?
Force on this area, F = ?
Relationships:
Pressure on surface of the pipe = Atmospheric pressure, Patm= 101325 Pa
Pressure exerted at a point due to material above it = ρgh
Total pressure, P = Patm + ρgh
P = F/A
Computation:
P = 101325 Pa + 1030Kg/m3 x 9.8m/s2 x 210m = 22.21x105 Pa = 2221 KPa
F = 22.21x105 Pa x 10-4 m2 = 222.1 N
4. Water is flowing through a circular pipe with an internal diameter of 400 cm, at a rate of 1.00 m3/s. What is the flow velocity? What is the dynamic pressure?
Variables:
d = 400cm = 4m
Volumetric flow rate, V = 1m3/s
Density of pure water, ρ = 1000 kg/m3
Flow velocity, v = ?
PD = ?
Relationships:
A = πd2/4
V = Av
PD = ½ ρv2
Computation:
A = 3.14 x (4m)2/4 = 12.56 m2
v = (1m3/s) / 12.56 m2 = 0.08 m/s
PD = ½x 1000 kg/m3 x (0.08 m/s) = 3.2 Pa
5. The static water pressure at point 1 in a pipe is 100 kPa, and the fluid velocity is 2 m/s. At point 2 downstream, the pipe narrows and the velocity increases to 3 m/s. What is the static pressure at that point? The usual assumptions concerning laminar flow, no friction losses, incompressibility, etc. all apply.
Variables:
Density of pure water, ρ = 1000 kg/m3
Ps1 = 100kPa
Ps2 = ?
v1 = 2m/s
v2 = 3m/s
PD1 = ?
PD2 = ?
Relationships:
PD = ½ ρv2
Ps1 + PD1 = Ps2 + PD2 (Bernoulli’s principle)
Computation:
PD1 = ½x 1000 kg/m3 x (2 m/s) = 1000 Pa = 1kPa
PD2 = ½x 1000 kg/m3 x (3 m/s) = 1500 Pa = 1.5kPa
Ps2 = 100 kPa + 1 kPa – 1.5 kPa = 99.5 kPa
References
UnitConversion.org. (2009). Free unit converter. Retrieved 07 09, 2012, from Unitconversion.org: http://www.unitconversion.org/unit_converter
