# How to calculate kinetic energy of gas flow in pipe given tempearture, pressure, pipe diameter and velocity

Ive been given a question and asked to calculate the kinetic energy of a gas given the temperature, pressure, diameter of pipe and velocity. I've been looking and have been unable to find any equations that can use this information. The only hint Ive been given is "STP conditions" but am unsure how that is relevant to solving the question. I cannot figure out how to solve this question. Below is the question given.

Air at 300 °C and 130 kPa flows through a horizontal 7-cm ID pipe at a velocity of 42.0 m/s. a) Calculate the kinetic energy (in J/s), assuming ideal gas behaviour.

Any formulas or advice about how to approach this question would be greatly appreciated.

• What happens when you try to use the general formula for kinetic energy? – J. Ari May 21 '20 at 13:08
• Your KE will likely just need to be per kg since a continuous flow doesn't model well into KE. But if you need something per time or similar, turn temp and pressure into mass flow rate. – Tiger Guy May 21 '20 at 15:06

I worked out an answer below. If you want to work it out for yourself just look at the equations that don't have any units in them.

# Quick and dirty guess:

$$\text{K.E.}=(\text{mass flow rate}) \cdot (\text{KE term from Bernoulli equation})$$

$$\text{K.E.}=(\bar{V} \cdot A \cdot \rho) \cdot (\frac{\bar{V}^2}{2})$$

Use the temperature, pressure, and composition of Earth air to calculate density $$\rho$$. Calculate $$A$$ as cross sectional flow area of the pipe. $$\bar{V}$$ is average velocity.

# Verbose exploration:

When asked about kinetic energy or potential energy you should immediate think "energy balances".

Bernoulli equation

The "Bernoulli equation without friction" (as derived and shown in Chapter 4, page 87 of Unit Operations of Chemical Engineering by Warren L. McCabe 7th Edition) is possibly a useful energy balance equation for a unit mass flowing along a streamline through a tube. The energy of fluid at point $$a$$ along the tube is balanced with energy at point $$b$$. Here is the equation and some accompanying explanations from McCabe:

$$\frac{p_a}{\rho} + gZ_a + \frac{u^2_a}{2} = \frac{p_b}{\rho} + gZ_b + \frac{u^2_b}{2}$$

Where:

$$p_a$$ : absolute pressure of fluid in the tube at point $$a$$

$$p_b$$ : absolute pressure of fluid in the tube at point $$b$$

$$\rho$$ : fluid density (assumed constant in this example)

$$g$$ : gravitational constant

$$Z_a$$ : elevation of the tube at point $$a$$

$$Z_b$$ : elevation of the tube at point $$b$$

$$u_a$$ : velocity of a unit mass of fluid in the tube at point $$a$$ along a streamline

$$u_b$$ : velocity of the same unit mass of fluid in the tube at point $$a$$ along the streamline

Each term in [the equation] is a scalar and has the dimensions of energy per unit mass, representing a mechanical energy effect based on a unit mass of the flowing fluid. Terms $$gZ$$ and $$u^2/2$$ are the potential and kinetic energy, respectively, of a unit mass of fluid; and $$p/\rho$$ represents the mechanical work done by forces, external to the stream, on the the fluid in pushing it into the tube or the work recovered from the fluid leaving the tube.

...

The term $$u^2/2$$ in [the equation] is the kinetic energy of a unit mass of fluid all of which is flowing at the same velocity $$u$$.

A later version of this equation takes friction and the fact that $$u$$ isn't average velocity:

$$\frac{p_a}{\rho} + gZ_a + \frac{\alpha \bar{V}^2_a}{2} = \frac{p_b}{\rho} + gZ_b + \frac{\alpha \bar{V}^2_b}{2} + h_f$$

where:

$$\alpha$$ : kinetic energy correction factor ("$$\alpha$$ is $$2.0$$ for laminar flow and is about $$1.05$$ for highly turbulent flow.")

$$\bar{V}^2_a$$ : average velocity of the fluid at point $$a$$

$$\bar{V}^2_b$$ : average velocity of the fluid at point $$a$$

$$h_f$$ : mechanical energy lost to friction

## Applicability to compressible flow

Although the equation is first used in McCabe's chapter on incompressible flow, it is used in the McCabe's treatment of compressible flow in a subsequent chapter as a way to show how the energy balance between kinetic energy, pressure, and friction loss. In other words, that over a short enough length of pipe, all three quantities must sum to zero:

$$\frac{dp}{\rho} + d \left( \frac{\alpha \bar{V}^2}{2} \right) + gdZ + dh_f = 0$$

where $$d$$ carries the same meaning as the $$d$$ in a derivative of differential calculus. The equation's applicability to compressible flow depends upon whether density changes are important. The air in our example is compressible but we are not having to account for density changes anywhere in the example as stated. Therefore, using the kinetic energy term $$\frac{\alpha \bar{V}^2}{2}$$ to answer the question should be appropriate.

## Strategy

You were given $$\bar{V}=42.0 \space \frac{\text{m}}{\text{s}}$$ and you don't care about what happens at two points. Also, since you were not provided material of construction information about the pipe you cannot calculate $$\alpha$$ (which is a function of pipe smoothness) so therefore the easiest way to move forward is to assume $$\alpha=1$$ and $$h_f=0$$ (flow is turbulent and pipe wall is frictionless). Regarding $$\alpha$$ on page 113, McCabe indicates "In most practical situations [$$\alpha$$ is] taken as unity in turbulent flow". From personal experience, my guess is the inner pipe diameter of $$D=7 \space \text{cm}$$ coupled with air velocity of $$\bar{V}=42 \space \frac{\text{m}}{\text{s}}$$ is turbulent, but it may be useful to check that.

Once we have a density we can use the cross-sectional area of the pipe plus the average air velocity to calculate a mass flow rate. Then, we just have to multiply this mass flow rate with the kinetic energy term of the Bernoulli equation to calculate the kinetic energy component of the stream.

## Check turbulence state via Reynolds number

If the Reynolds number is greater than $$4,000$$ then flow in the pipe is definitely turbulent. (from McCabe pg. 54: "Under ordinary conditions, the flow in a pipe or tube is turbulent at Reynolds numbers above about $$4,000$$").

The formula for Reynolds number is:

$$\text{Re} = \frac{D \bar{V} \rho}{\mu}$$

where:

$$D$$ : inner diamter of tube

$$\bar{V}$$ : average velocity of fluid

$$\rho$$ : density of fluid

$$\mu$$ : [dynamic] viscosity of liquid (not kinematic viscosity)

From the information provided I believe it should be possible to calculate $$\rho$$ since you are permitted to treat air as an ideal gas. $$\mu$$ can be looked up from a table.

## Calculate density

From the ideal gas law:

$$PV = \frac{m}{M} RT$$

density can be calculated

$$\rho = \frac{m}{V} = \frac{PM}{RT}$$

where:

$$\rho$$ : gas density

$$m$$ : unit mass of gas

$$V$$ : unit volume of gas

$$P$$ : absolute pressure of gas

$$M$$ : molecular weight of gas

$$R$$ : ideal gas constant

$$T$$ : absolute temperature of the gas

Given $$P=130 \space \text{kPa}$$ (which I am assuming is absolute pressure, not gauge pressure), $$M_\text{air}=28.8 \space \text{mol}$$, $$R=8.314462618 \space \frac{\text{J}}{\text{mol} \cdot \text{K}}$$, and $$T=(300+273.15)=573.15 \space \text{K}$$, then $$\rho$$ can be calculated.

$$\rho = \frac{(130 \space \text{kPa}) \cdot ({28.8 \space \frac{\text{g}}{\text{mol}}})}{(8.314462618 \space \frac{\text{J}}{{\text{mol}} \cdot \text{K}}) \cdot (573.15 \space \text{K})} \cdot (\frac{\text{J}}{\text{kg} \cdot \frac{\text{m}^2}{\text{s}^2}}) \cdot (\frac{1000 \space \text{Pa}}{\text{kPa}}) \left( \frac{\left( \frac{\text{kg}}{\text{s}^2 \cdot \text{m}} \right)}{\text{Pa}} \right) \cdot (\frac{\text{kg}}{1000 \space \text{g}})$$

$$\rho = 0.786 \space \frac{\text{kg}}{\text{m}^3}$$

This value seems somewhat reasonable since it is close to the air density calculated by this real air density calculator which I'm guessing takes compressibility into account.

## Look up air viscosity

Using this webpage for calculating air viscosity as a function of temperature, we get:

$$\mu = 29.28 \cdot 10^{-6} \space \text{Pa} \cdot \text{s}$$

## Calculate Reynolds number to determine if turbulent state present

Plugging our values into the Reynolds equation, we get:

$$\text{Re} = \frac{\left( 7 \text{cm} \right) \left( 42.0 \frac{\text{m}}{\text{s}} \right) \left( 0.786 \frac{\text{kg}}{\text{m}^3} \right)}{\left( 29.28 \cdot 10^{- 6} \text{Pa} \cdot \text{s} \right)} \cdot \left( \frac{\text{m}}{100 \space \text{cm}} \right) \cdot \left( \frac{\text{Pa}}{\left( \frac{\text{kg}}{\text{s}^2 \text{m}} \right)} \right)$$

$$\text{Re} = 80,000$$

This is well above $$4,000$$ and therefore the flow is turbulent.

## Calculate mass flow rate

Since we're given average velocity $$\bar{V}$$, pipe diameter $$D=(7 \space \text{cm})$$, and density $$\rho$$, we can calculate mass flow rate $$\dot{m}$$ like so:

$$\dot{m} = (\text{volumetric flow rate}) \cdot (\text{density})$$

$$\dot{m} = (\bar{V} \cdot A) \cdot (\rho)$$

$$\dot{m} = (42.0 \frac{\text{m}}{\text{s}}) \cdot (\pi \cdot \left( \frac{7 \space \text{cm}}{2} \right)^2 ) \cdot (0.786 \frac{kg}{\text{m}^3}) \cdot \left( \frac{\text{m}}{100 \space \text{cm}} \right)^2$$

$$\dot{m} = 0.127 \frac{\text{kg}}{\text{s}}$$

## Apply Bernoulli equation

Since each term in the Bernoulli equation is in units of energy per unit mass (but with the actual mass factor divided out), all you need to do to calculate kinetic energy is to multiply the mass flow rate by the kinetic energy term, $$\alpha \frac{\bar{V}}{2}$$,from one side of the Bernoulli equation.

$$\text{K.E.} = \dot{m} \cdot \frac{\alpha \bar{V}^2}{2}$$

This makes sense from a physics perspective because kinetic energy of any object is proportional to its mass multiplied by its velocity squared. Here, the constant of proportionality is $$\frac{\alpha}{2}$$.

Because flow is turbulent ($$\text{Re}>4000$$) we can set $$\alpha=1.0$$, simplifying our calculation.

$$\text{K.E.} = \dot{m} \cdot \frac{\bar{V}^2}{2}$$

Plugging in our values we get:

$$\text{K.E.} = \left( 0.127 \frac{\text{kg}}{\text{s}} \right) \cdot \left( \frac{(42.0 \space \frac{\text{m}}{\text{s}})^2}{2} \right) \cdot \left( \frac{\text{J}}{\text{kg} \cdot \frac{\text{m}^2}{\text{s}^2}} \right)$$

$$\text{K.E.} = 112 \space \frac{\text{J}}{\text{s}}$$

## Summary

When sizing a gas compressor to move air at an average velocity of $$42.0 \space \frac{\text{m}}{\text{s}}$$ at this point through the pipe at temperature $$300^{\circ}\text{K}$$, and absolute pressure $$130 \space \text{kPa}$$, $$112 \space \frac{\text{J}}{\text{s}}$$ of energy is needed in the form of kinetic energy. Other calculations would need to be performed to:

• Determine the energy required to bring air starting (presumably) from STP conditions ($$0^{\circ}\text{C}$$, $$100 \space \text{kPa}$$) to $$130 \space \text{kPa}$$

• Determine the energy required to overcome and elevation increase (or energy available from elevation decreases).

• Determine the energy required to overcome friction losses throughout the system.

• Very thorough explanation – SimpleJack May 22 '20 at 11:02