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Calculate the coefficient of discharge (Cd) for an orifice with a diameter of 50 mm in the side of a tank containing water at a constant head of 3 m. The actual discharge of water from orifice is 45 000 litres per hour (l/h).

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Answers

3 parts
5.1

-106.97 W\text{-106.97 W}

5.2

0.83\text{0.83}

6.4

8900.0 N\text{8900.0 N}

Step-by-step solution

10 steps

Step 1: Calculate the cross-sectional areas and velocities.
The lower diameter D1=170D_1 = 170 mm =0.17= 0.17 m.
The upper diameter D2=290D_2 = 290 mm =0.29= 0.29 m.
The quantity of water flowing Q=50Q = 50 L/s =0.05= 0.05 m3^3/s.
The cross-sectional area at the lower end A1A_1:

A1=π4D12=π4(0.17 m)2≈0.022698 m2A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (0.17 \text{ m})^2 \approx 0.022698 \text{ m}^2

The velocity at the lower end v1v_1:

v1=QA1=0.05 m3/s0.022698 m2≈2.2027 m/sv_1 = \frac{Q}{A_1} = \frac{0.05 \text{ m}^3/\text{s}}{0.022698 \text{ m}^2} \approx 2.2027 \text{ m/s}

The cross-sectional area at the upper end A2A_2:

A2=π4D22=π4(0.29 m)2≈0.066052 m2A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (0.29 \text{ m})^2 \approx 0.066052 \text{ m}^2

The velocity at the upper end v2v_2:

v2=QA2=0.05 m3/s0.066052 m2≈0.7569 m/sv_2 = \frac{Q}{A_2} = \frac{0.05 \text{ m}^3/\text{s}}{0.066052 \text{ m}^2} \approx 0.7569 \text{ m/s}

Step 2: Calculate the change in height.
The length of the pipe L=12L = 12 m.
The inclination θ=25∘\theta = 25^\circ.
The change in height z2z_2 (assuming z1=0z_1 = 0 at the lower end):

z2=Lsin⁡(θ)=12 m×sin⁡(25∘)≈12 m×0.4226≈5.0712 mz_2 = L \sin(\theta) = 12 \text{ m} \times \sin(25^\circ) \approx 12 \text{ m} \times 0.4226 \approx 5.0712 \text{ m}

Step 3: Apply Bernoulli's equation to find the pressure at the upper end.
Bernoulli's equation (neglecting head losses):

P1ρg+v122g+z1=P2ρg+v222g+z2\frac{P_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} + z_2

Given P1=180P_1 = 180 kPa =180×103= 180 \times 10^3 Pa, ρ=1000\rho = 1000 kg/m3^3, g=9.81g = 9.81 m/s2^2.
Rearranging for P2P_2:

P2=P1+12ρ(v12−v22)−ρg(z2−z1)P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) - \rho g (z_2 - z_1) P2=180×103 Pa+12(1000 kg/m3)((2.2027 m/s)2−(0.7569 m/s)2)−(1000 kg/m3)(9.81 m/s2)(5.0712 m)P_2 = 180 \times 10^3 \text{ Pa} + \frac{1}{2}(1000 \text{ kg/m}^3)((2.2027 \text{ m/s})^2 - (0.7569 \text{ m/s})^2) - (1000 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(5.0712 \text{ m}) P2=180000+500(4.8518−0.5729)−49748.472P_2 = 180000 + 500(4.8518 - 0.5729) - 49748.472 P2=180000+500(4.2789)−49748.472P_2 = 180000 + 500(4.2789) - 49748.472 P2=180000+2139.45−49748.472P_2 = 180000 + 2139.45 - 49748.472 P2≈132390.978 Pa≈132.39 kPaP_2 \approx 132390.978 \text{ Pa} \approx 132.39 \text{ kPa}

The pressure at the upper end of the pipe is 132.39 kPa\boxed{\text{132.39 kPa}}.

Part 5.1· 1 step

2 The change in kinetic energy of the liquid over the length of the pipe

  1. Calculate the rate of change of kinetic energy.

    The change in kinetic energy of the liquid over the length of the pipe refers to the rate of change of kinetic energy (power) of the fluid as it flows from the lower to the upper end.
    The mass flow rate m˙=ρQ=1000 kg/m3×0.05 m3/s=50 kg/s\dot{m} = \rho Q = 1000 \text{ kg/m}^3 \times 0.05 \text{ m}^3/\text{s} = 50 \text{ kg/s}.
    The change in kinetic energy rate ΔKE\Delta KE:

    ΔKE=12m˙(v22−v12)\Delta KE = \frac{1}{2} \dot{m} (v_2^2 - v_1^2) ΔKE=12(50 kg/s)((0.7569 m/s)2−(2.2027 m/s)2)\Delta KE = \frac{1}{2} (50 \text{ kg/s}) ((0.7569 \text{ m/s})^2 - (2.2027 \text{ m/s})^2) ΔKE=25(0.5729−4.8518)\Delta KE = 25 (0.5729 - 4.8518) ΔKE=25(−4.2789)\Delta KE = 25 (-4.2789) ΔKE≈−106.97 W\Delta KE \approx -106.97 \text{ W}

    The change in kinetic energy of the liquid over the length of the pipe is -106.97 W\boxed{\text{-106.97 W}}.

Part 5.2

0.83\text{0.83}

Calculate the coefficient of discharge (CdC_d) for an orifice

  1. Convert the actual discharge to m3^3/s.

    The actual discharge Qactual=45000Q_{actual} = 45000 litres/hour.

    Qactual=45000 L/h×1 m31000 L×1 h3600 s=453600 m3/s=0.0125 m3/sQ_{actual} = 45000 \text{ L/h} \times \frac{1 \text{ m}^3}{1000 \text{ L}} \times \frac{1 \text{ h}}{3600 \text{ s}} = \frac{45}{3600} \text{ m}^3/\text{s} = 0.0125 \text{ m}^3/\text{s}

  2. Calculate the theoretical discharge.

    The orifice diameter d=50d = 50 mm =0.05= 0.05 m.
    The area of the orifice AA:

    A=π4d2=π4(0.05 m)2≈0.0019635 m2A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.05 \text{ m})^2 \approx 0.0019635 \text{ m}^2

    The constant head H=3H = 3 m.
    The theoretical velocity vtheoreticalv_{theoretical}:

    vtheoretical=2gH=2×9.81 m/s2×3 m=58.86 m/s≈7.672 m/sv_{theoretical} = \sqrt{2gH} = \sqrt{2 \times 9.81 \text{ m/s}^2 \times 3 \text{ m}} = \sqrt{58.86} \text{ m/s} \approx 7.672 \text{ m/s}

    The theoretical discharge QtheoreticalQ_{theoretical}:

    Qtheoretical=A×vtheoretical=0.0019635 m2×7.672 m/s≈0.01507 m3/sQ_{theoretical} = A \times v_{theoretical} = 0.0019635 \text{ m}^2 \times 7.672 \text{ m/s} \approx 0.01507 \text{ m}^3/\text{s}

  3. Calculate the coefficient of discharge (CdC_d).

    Cd=QactualQtheoretical=0.0125 m3/s0.01507 m3/s≈0.82946C_d = \frac{Q_{actual}}{Q_{theoretical}} = \frac{0.0125 \text{ m}^3/\text{s}}{0.01507 \text{ m}^3/\text{s}} \approx 0.82946

    The coefficient of discharge (CdC_d) is 0.83\boxed{\text{0.83}}.

Part 6.1

Name the TWO main types of static loads on bearings.

The two main types of static loads on bearings are:

  • Radial load
  • Axial load (or thrust load)

Part 6.2

Name FOUR of these factors to be considered when selecting the correct type and size of bearing.

Four factors to consider when selecting the correct type and size bearing are:

  • Type of load (radial, axial, or combined)
  • Magnitude of load (static and dynamic)
  • Operating speed
  • Required life

Part 6.3· 2 steps

Calculate the expected service life of the bearing in hours.

  1. Identify given values and the life formula.

    Equivalent dynamic bearing load P=4100P = 4100 N.
    Basic dynamic load rating C=44.3C = 44.3 kN =44300= 44300 N.
    The bearing is a roller bearing, so the life exponent a=10/3a = 10/3.
    The life formula in millions of revolutions (L10L_{10}):

    L10=(CP)aL_{10} = \left(\frac{C}{P}\right)^a

    The life in hours (L10hL_{10h}) is given by:

    L10h=10660n(CP)aL_{10h} = \frac{10^6}{60n} \left(\frac{C}{P}\right)^a

    where nn is the operating speed in revolutions per minute (rpm).

  2. Calculate the life in millions of revolutions.

    L10=(44300 N4100 N)10/3=(10.804878)10/3≈1800.7 million revolutionsL_{10} = \left(\frac{44300 \text{ N}}{4100 \text{ N}}\right)^{10/3} = (10.804878)^{10/3} \approx 1800.7 \text{ million revolutions}

    The problem asks for the service life in hours but does not provide the operating speed (nn) of the bearing. Without the operating speed, it is not possible to convert the life from revolutions to hours.

    The expected service life of the bearing in hours cannot be calculated without the operating speed (rpm).

Part 6.4

8900.0 N\text{8900.0 N}

Calculate the radial force between the 20° involute gears.

  1. Calculate the angular velocity.

    The speed N=1100N = 1100 r/min.

    ω=2πN60=2π×110060 rad/s=110π3 rad/s≈115.19 rad/s\omega = \frac{2 \pi N}{60} = \frac{2 \pi \times 1100}{60} \text{ rad/s} = \frac{110 \pi}{3} \text{ rad/s} \approx 115.19 \text{ rad/s}

  2. Calculate the pitch line velocity.

    The pitch circle diameter dp=142d_p = 142 mm =0.142= 0.142 m.

    v=ωdp2=115.19 rad/s×0.142 m2≈8.1785 m/sv = \frac{\omega d_p}{2} = \frac{115.19 \text{ rad/s} \times 0.142 \text{ m}}{2} \approx 8.1785 \text{ m/s}

  3. Calculate the tangential force.

    The power transmitted Ppower=200P_{power} = 200 kW =200×103= 200 \times 10^3 W.

    Ft=Ppowerv=200×103 W8.1785 m/s≈24455.7 NF_t = \frac{P_{power}}{v} = \frac{200 \times 10^3 \text{ W}}{8.1785 \text{ m/s}} \approx 24455.7 \text{ N}

  4. Calculate the radial force.

    The pressure angle ϕ=20∘\phi = 20^\circ.
    The radial force FrF_r is related to the tangential force FtF_t by:

    Fr=Fttan⁡(ϕ)F_r = F_t \tan(\phi) Fr=24455.7 N×tan⁡(20∘)F_r = 24455.7 \text{ N} \times \tan(20^\circ) Fr≈24455.7 N×0.36397F_r \approx 24455.7 \text{ N} \times 0.36397 Fr≈8900.0 NF_r \approx 8900.0 \text{ N}

    The radial force between the gears is 8900.0 N\boxed{\text{8900.0 N}}.

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