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Question 23M.3.HL.TZ2.10

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Date May 2023 Marks available [Maximum mark: 7] Reference code 23M.3.HL.TZ2.10
Level HL Paper 3 Time zone TZ2
Command term Determine, Estimate, State Question number 10 Adapted from N/A
10.
[Maximum mark: 7]
23M.3.HL.TZ2.10

A large tank is used to store oil of density 850 kg m−3 and is filled to a height h1 above the bottom. A valve in the tank wall allows oil to flow out. The centre of the valve is at a height h2 from the bottom of the tank. A circular drainage outlet is at the bottom of the tank.

diagram not to scale

The drainage outlet has a diameter of 100 mm and a metal stopper of mass 2.5 kg is used to plug the outlet.

(a)

Determine the minimum force required to lift the stopper when h1 = 4.0 m.

[3]

Markscheme

calculates Pbottom «= 850 × 9.8 × 4 = Pbottom «= 850 × 9.8 × 4» = 33 320 Pa»

OR

calculates weight of oil «Weightoil = 850 × 0.12π4 × 4 × 9.8» ✓

 

Add the weight of the stopper «Fmin = 33 320 × 0.12×π4 + 2.5 × 9.8» ✓

 

Fmin = 286 N

 

Alternative method may include finding the volume and then weight of oil above the stopper

Award [2 max] if the weight of stopper is ignored to give 262 N

Award [2 max] if plug is taken as a square base to give 358 N

Allow ECF from MP1

With the metal stopper in place, the valve on the side of the tank is opened to let oil flow out.

Using Bernoulli’s equation, it can be shown that the speed v of oil flowing through the valve can be estimated as v = 2g(h1-h2).

(b)

State two assumptions that were used in obtaining the expression for the speed v.

[2]

Markscheme

the pressure «at the valve opening and at the top of the oil in the tank» is constant ✓

the velocity «of oil surface» at the top «of the tank» is zero / negligible ✓

no energy/head losses «as the oil flows through the valve» ✓

no turbulence/ laminar flow occurs «as the oil flows through the valve» ✓

fluid not compressible ✓

ideal fluid ✓

no viscosity ✓

(c)

Estimate the maximum radius of the valve so that turbulent flow does not occur.

The following data are given:

Viscosity of oil = 0.25 Pa s
                 h1 = 4.0 m
                 h2 = 0.5 m

[2]

Markscheme

Use of Re = 1000 OR states Re = 2g(4-0.5)×r×8500.25 ✓

«1000 = 2g(4-0.5)×r×8500.25 so » r = 3.6 «cm» ✓

 

Award [2] if r is taken to be the diameter this gives r = 1.8 cm