Fluid Mechanics for MAP/Energy Considerations

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Losses and Energy consideration
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Remember the conservation of energy equation:

$$\dot{Q} + \dot{W}_{schaft} + \dot{W}_{shear} + \dot{W}_{other} = \frac{\partial}{\partial t}\int_{CV}{e\rho dV} + \int_{CS}{\left(e + \frac{P}{\rho}\right)\rho U_{i}n_{i}dA}$$

where

$$e = u + \frac{U^{2}}{2} + gx_{2} = u+\frac{U_{i}U_{i}}{2} + gz$$

The velocity does not have necessarily a plug profile at the inlet and outlet of a control volume and the cross sectional areas might also differ at the inlet and outlet of a CV. The variations in the cross sections of flow conduits and the velocity profiles have to be accounted to calculate the correct pressure drop due to the viscous losses. Therefore, the surface integrals has to be calculated by considering these variations:

$$\dot{Q} = \int_{CS}{\left(e + \frac{P}{\rho}\right)\rho U_{i}n_{i}dA} = \dot{m}(u_{2} - u_{1}) + \dot{m}\left(\frac{P_{2}}{\rho} - \frac{P_{1}}{\rho}\right) + \dot{m}g(z_{2} - z_{1}) +$$

$$\int_{A_{2}}{\frac{U_{2}^{2}}{2}\rho U_{2i}n_{2i}dA} - \int_{A_{2}}{\frac{U_{1}^{2}}{2}\rho U_{1i}n_{1i}dA}$$

For the calculation of the profile effects one can use the kinetic energy coefficient $$\displaystyle \alpha$$ as follows:

$$\int_{A}{\frac{U^{2}}{2}\rho U_{i}n_{i}dA} =\alpha \dot{m}\frac{\bar{U}^{2}}{2}$$

For fully developed laminar velocity profile, where

$$U = 2\bar{U}\left[1 - \frac{r}{R}^{}2\right]$$

$$\displaystyle \alpha = 2$$

For fully developed turbulent flow, due to relatively flat velocity profile, $$ \displaystyle \alpha \approx 1$$

$$\dot{Q} = \dot{m}(u_{2} - u_{1}) + \dot{m}\left(\frac{P_{2}}{\rho} - \frac{P_{1}}{\rho}\right) + \dot{m} \left(\alpha_{2}\frac{\bar{U}^{2}_{2}}{2} - \alpha_{1}\frac{\bar{U}^{2}_{1}}{2}\right)$$

Rearranging the energy equation:

$$\underbrace{\frac{P_{1}}{\rho} + \alpha_{1}\frac{\bar{U}^{2}_{1}}{2} + gz_{1} = \frac{P_{2}}{\rho} + \alpha_{2}\frac{\bar{U}^{2}_{2}}{2} + gz_{2}}_{A} + \underbrace{(u_{2} - u_{1}) - \frac{\dot{Q}}{\dot{m}}}_{B}$$

The conversion of mechanical energy (A) and the irreversible conversion and loss of mechanical energy with viscous dissipation (B) can be better distinguished.

Total pressure drop
Finally the total pressure drop due to viscous effects can be written as

$$\underbrace{P_{1} + \alpha_{1}\rho\frac{\bar{U}^{2}_{1}}{2} + \rho gz_{1} = P_{2} + \alpha_{2}\rho\frac{\bar{U}^{2}_{2}}{2} + \rho gz_{2}}_{Mechanical\ Energy} + \underbrace{\Delta P_{v}}_{Pressure\ drop\ due\ to\ viscous\ losses}$$

$$\Delta P_{v} = (P_{1} - P_{2}) + \rho \left(\alpha_{1}\frac{\bar{U}^{2}_{1}}{2} - \alpha_{2}\frac{\bar{U}^{2}_{2}}{2}\right) + \rho g(z_{1} - z_{2})$$

Major and minor pressure losses
The total pressure drop consist of major losses, which are due to frictional effects in fully developed flow in constant area tubes, and minor losses, which occur at entrances, sudden area changes, bends, elbows, fittings, valves, contraction and diffusers, etc..

Major losses in pipe flows


The pressure drop along a pipe, which has no elevation change, can be written as:

$$\displaystyle \Delta P_{v} = (P_{1} - P_{2})$$

Nondimensionalizing with the kinetic energy per unit mass of the flow:

$$\zeta = \frac{\Delta P_{v}}{0.5 \rho\bar{U}^{2}} = \frac{L}{D}\ f\left(Re, \frac{e}{D}\right)$$

where $$\zeta$$ is the loss coefficient and f is the friction factor.

Non-circular ducts
For non circular ducts, which appears in air conditioning, heating and ventilating applications, an equivalent diameter is calculated so that correlations for circular pipes can be utilized.

This equivalent diameter is called as hydraulic diameter:

$$\displaystyle D_{h} = \frac{4 \times Area}{Perimeter} = \frac{4A}{P}$$

Thus for a rectangular duct with a width of b and height of h:

$$D_{h} = \frac{4bh}{2(b + h)}$$

And for a square duct

$$\displaystyle D_{h} = h$$

Minor losses
For the other components which might cause pressure drop, the loss coefficient (resistance coefficient) can similarly be defined:

$$\zeta = \frac{\Delta P_{v}}{0.5\rho \bar{U}^{2}}$$

However, for the vast variety of geometries, it is not possible to obtain relations as for the pipe flows. Thus, $$\zeta$$ for each component is measured and documented.

Calculation of total loss in a single pipe system
A single pipe system may have many minor losses. Since all are correlated with $$0.5\rho\bar{U}^{2}$$, they can be summed into a single total system loss if the pipe has constant diameter:

$$\Delta P_{total} = \Delta P_{pipe} + \sum \Delta P_{components} =$$

$$0.5\rho\bar{U}^{2}\left[\frac{L}{D}f + \sum_{i}\zeta_{i}\right]$$

However, we must sum the losses separately, if the pipe size changes, i.e. mean velocity changes.

Multiple path system


The solution for pipe network problems is often carried out by use of node and loop equations similar in many ways to that done in electrical circuits.


 * 1) The net flow into any junction must be zero.
 * 2) The net head loss around any closed loop must be zero. At each junction there is one single pressure.
 * 3) All head losses must satisfy the major and minor-loss friction correlations.

In such problems the pipes and components might have different areas, thus pressure loss at each pipe and component should be calculated separately. Moreover, there can be components with unknown properties.

For such cases it is found useful to formulate the pressure loss as a function of the flow rate:

$$\Delta P = \zeta 0.5 \rho \bar{U}^{2} = \underbrace{\zeta \frac{0.5\rho}{{A}^2}}_{K}Q^{2} = KQ^{2}$$

Resultant loss coefficients, Kr, for serial and parallel connections
Serial connection: Flow rate is the same for all components

$$\displaystyle K_{r} = K_{1} + K_{2}$$

$$\displaystyle \Delta P_{r} = K_{r}Q^{2} = \Delta P_{1} + \Delta P_{2} = (K_{1} + K_{2})Q^{2}$$

Parallel connection: Pressure drop is the same inlet and outlet junctions



$$\displaystyle K_{r} = \frac{K_{1}K_{2}}{K_{1} + K_{2} + 2\sqrt{K_{1}K_{2}}}$$

$$\Delta P_{r} = K_{r}Q^{2}_{total} = K_{1}Q_{1}^{2} = K_{2}Q^{2}_{2}$$

$$\displaystyle Q_{total} = Q_{1} + Q_{2}$$

Design of a dry powder inhaler
The utilization of the explained concepts will be demonstrated in the class by developing an inhaler from scratch. The basic steps of the design are as follows:
 * 1) Make a conceptual design
 * 2) Decide pressure drop-flow rate characteristic of the inhaler (Inhaler patient matching)
 * 3) Dimension each component of the inhaler to ensure
 * 4) *Pressure drop-flow rate characteristic
 * 5) *Full emptying of the blister
 * 6) *Sufficient dispersion of the powder composition

An overview of dry powder inhaler design could be found in in the following publication

Reference

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