Free jet

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a jet of water issues from a 2in diameter orifice and, after some contraction, reaches a speed of 9 m/s under a head of 5m. What is the velocity coefficient of the jet? Also, if the measured volumetric flowrate is .016 m3/s, what are the contraction coefficient and discharge coefficient

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This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:
This problem is discussing a free jet, comparing experimentally observed values to those predicted by ideal behavior, and describing the difference through coefficients. The contraction coefficient describes how much the jet contracts to a diameter smaller than the orifice size, the velocity coefficient describes how much the jet’s speed is lower than that predicted by Bernoulli, and the discharge coefficient describes how the flow rate is lower than than predicted by ideal discharge i.e., lossless discharge in a free jet with diameter equal to that of the orifice.

We start with the ideal speed predicted by Torricelli’s relation:

U_\text{ideal} = \sqrt{2gh}

Taking g=9.81~\mathrm{m/s^2} and h=5~\mathrm{m},

U_\text{ideal} = \sqrt{2 \times 9.81 \times 5} \approx \sqrt{98.1} \approx 9.91~\mathrm{m/s}\,.

The problem states that the free jet reaches a speed of 9~\mathrm{m/s}. Thus the velocity coefficient is

C_v=\frac{U_\text{actual}}{U_\text{ideal}}=\frac{9}{9.91}\approx0.91\,.

Next, the orifice is specified as 2 in in diameter. Converting to meters,

2~\mathrm{in} = 0.0508~\mathrm{m}\,,

so the area is

A_\text{orifice}=\frac{\pi D^2}{4}=\frac{\pi (0.0508)^2}{4}\approx0.00203~\mathrm{m^2}\,.

The ideal volumetric flow rate predicted by using the full orifice area and U_\text{ideal} is

Q_\text{ideal}=A_\text{orifice}\, U_\text{ideal}=0.00203 \times 9.91\approx 0.0201~\mathrm{m^3/s}\,.

Given a measured flow rate of

Q_\text{measured}=0.016~\mathrm{m^3/s}\,,

the discharge coefficient is

C_d=\frac{Q_\text{measured}}{Q_\text{ideal}}=\frac{0.016}{0.0201}\approx 0.80\,.

Since the discharge coefficient is the product of the contraction coefficient and the velocity coefficient,

C_d=C_c\, C_v\,,

we can solve for the contraction coefficient,

C_c=\frac{C_d}{C_v}=\frac{0.80}{0.91}\approx 0.88\,.

Thus, the answers are:

  • Velocity coefficient, C_v \approx 0.91
  • Contraction coefficient, C_c \approx 0.88
  • Discharge coefficient, C_d \approx 0.80

Assumptions made:

  • Standard gravity g=9.81~\mathrm{m/s^2} was used.
  • The full orifice area is employed to calculate the ideal volumetric flow rate.
  • The ideal exit speed from the reservoir is given by Torricelli’s relation without viscous or entrance effects other than those captured by the coefficients.

To formulate this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked their significance below. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 4 Part 1 (HIGHLY RELEVANT)
  2. Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 2 (HIGHLY RELEVANT)
  3. Chapter 5 Fundamentals: Forming a free jet (HIGHLY RELEVANT)
  4. Chapter 1 Fundamentals: The area–velocity relation is derived by combining conservation of mass, co… (HIGHLY RELEVANT)
  5. Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 3 Part 1 (HIGHLY RELEVANT)

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