Question

Water circulates through a hot-water heating system in a house. The water leaves the basement with a speed of 0.5 m/s through a 4-cm-diameter pipe, der a total pressure of 3Parm. (Assume that all the pipes have circular cross-sectional areas and that the pipes don't "branch off at any point.)

1 If the empty steel container takes up a total volume of 1.1 x 102 m, determine the buoyant force acting on a container that's completely filled with gasoline, if 98% of the container (with gasoline) is under the ocean surface. 2) Determine a numerical value for the density of steel (in kg/ms)

Answer 1

1). Total Volume of the container, $V= 1.1 \times 10^2 m^3$ (Assuming that you provided wrong units for volume)

Volume submerged in the ocean,

Then Buoyancy force on the container= Mass of displaced water by the submerged volume

using Density= Mass/ Volume

thus Buoyancy Force on the container, $F_b= M_{displaced}g= V_s \rho_{water}g$

Knowing that density of water, $\rho_{water}= 1000 kg/m^3$

Buoyancy force will be $F_b= 0.98 \times 1.1 \times 10^2 \times 1000 \times 9.8= 10.5644 \times 10^{5} N$ (ANS)

2). For determining the density of steel, we need the thickness and other dimensions of the container as for calculating the mass of steel contained in the container, we need to check for the actual volume in which steel is contained. I am solving this part using the variable. So please follow the procedure and then put the values after crosscheckng the correct question to get the final value.

   Now when the container is in equilibrium in the ocean water that means

Weight of the container= Buoyancy Force on the container

or Weight of steel content +Weight of gasoline= F_{b....................................(1)}

Now for the weight of steel,

Mass= Density of steel * volume in which steel is contained

where volume in which steel is contained= Area of container_{(including the area of base)} * thickness ( as container is only made up of the steel but contains gasoline)

then Weight of steel= Mass of Steel * g

or $W_{steel}= A_{tank} \times t \times \rho_{steel} \times g$ ( where t= thickness of tank)

Using above value in equation 1,

   $(A_{tank} \times t \times \rho_{steel} \times g)+ (V \times \rho_{gasoline})g= F_b$

where density of gasoline= 719.7 kg/m³

then by crosschecking for the values and using Buoyancy force from 1st part, you can solve above equation for density of steel.