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3.2: Mass Flow Rate

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    81484
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    Before proceeding, we need to develop a better understanding of how to calculate the mass flow rate. Recall that we earlier defined the mass flow rate as the time rate at which mass crosses the boundary of a system. Note that a mass flow rate really only has meaning as it relates to the boundary for which it is defined.

    A differential area dA of a curved surface. A unit outwards normal vector n points outwards from dA, perpendicular to the surface at that point, and a vector V points out of the surface at an acute angle to the vector n.

    Figure \(\PageIndex{1}\): Geometry for flow across a system boundary.

    To calculate the mass flow rate, consider the rate at which mass flows across the system boundary with area \(A\) in Figure \(\PageIndex{1}\). By definition the mass flow rate out of a system is defined by the equation

    \[ \dot{m}_{out} = \int\limits_{A_{sys}} \rho \left( \mathbf{V}_{rel} \cdot \mathbf{n} \right) \, dA \nonumber \]

    where

    \[ \begin{align*} \rho &= \text{the fluid density,} \\ \mathbf{V}_{rel} &= \text{velocity (a vector) of the mass crossing the boundary measured with respect to the system boundary, and} \\ \mathbf{n} &= \text{is the unit vector normal to the differential area } dA \text{ and pointing out of the system.} \end{align*} \nonumber \]

    The key to understanding the meaning of Eq. \(\PageIndex{1}\) is to recall the meaning of a simple scalar or dot product of two vectors. First recall that the scalar or dot product of two vectors produces a scalar. Now applying the definition of a this operation to the unit normal vector and the velocity vector gives:

    The vector V_rel points up and to the right. The unit vector n, which originates at the same point as V_rel, points up and to the right at a steeper angle. The angle between the two vectors is theta. V_rel,n is a vector that lies along n, whose magnitude equals the product of the magnitude of V_rel, the magnitude of n, and the cosine of theta; it represents the normal velocity of the mass relative to the boundary surface.

    Figure \(\PageIndex{2}\): Definition of the dot product of \(\vec{V}_{rel}\) and \(\hat{n}\).

    \[ \begin{align*} V_{rel, n} &= |\mathbf{V}_{rel, n}| \\ &= \mathbf{V} \cdot \mathbf{n} = |\mathbf{V}| |\mathbf{n}| \cos \theta \\ &= \text{normal velocity of the mass relative to the boundary surface} \end{align*} \nonumber \]

    Using this result, the mass flow rate expression can be written as

    \[ \dot{m} = \int\limits_{A_{sys}} \rho V_{rel, n} \, dA \nonumber \]

    In general, the density \(\rho\) and the normal velocity \(V_{rel,n}\) may change with position on the boundary surface. Also note that we have dropped the "out" subscript because our observation of the direction of \(V_{rel,n}\) with respect to the system determines whether the mass flow rate is into or out of the system.

    There are numerous modeling assumptions that are used to describe the behavior of real systems in the construction of mathematical models. If the flow has uniform density at the flow boundary, then the density is spatially uniform at the flow boundary. Under these conditions,

    \[ \dot{m} = \int\limits_{A_c} \rho V_{rel, n} \, dA = \rho \underbrace{ \int\limits_{A_c} V_{rel, n} \, dA } _{= \dot{ V\kern-0.8em\raise0.3ex- }} = \rho \dot{ V\kern-0.8em\raise0.3ex- }, \nonumber \]

    more simply expressed as

    \[ \dot{m} = \rho \dot{ V\kern-0.8em\raise0.3ex- } \quad \text{(uniform density)} \nonumber \]

    where the integral of relative normal velocity over the flow cross-sectional area is the volumetric flow rate, \(\dot{ V\kern-0.8em\raise0.3ex- }\). The dimensions of volumetric flow rate are \([\text{L}]^3/[\text{T}]\) and typical units are \(\text{m}^3/\text{s}\) in SI and \(\text{ft}^3/\text{s}\) in AES. There are many other commonly used units for volumetric flow rate, including gallons per min (\(\text{gpm}\)) and cubic feet per min (\(\text{cfm}\)).

    If we restrict ourselves to planar (flat) flow boundaries, then several additional simplifications are possible. If the flow at a boundary has uniform velocity and uniform density, both the density and relative velocity come outside the integral and the mass flow rate becomes

    \[ \dot{m} = \rho A_c V_n \nonumber \]

    This is one of the most commonly used forms for calculating the mass flow rate. Notice that we have dropped the "rel" subscript to simplify the notation; however, you are warned that all mass and volumetric flow rates must of necessity be calculated with respect to a flow boundary. For flow conditions where the density and normal velocity vary across the flow cross section, it is common practice to break the flow cross section into many small elements where Eq. \(\PageIndex{4}\) applies and then sum the results, e.g. \(\dot{m} = \sum_{j} \left( \rho A_c V_{rel, n} \right) _j \). In many cases, the velocity will vary across the cross section but the density will be relatively uniform. Under these conditions it is frequently useful to write the mass flow rate in terms of an average velocity \(V_{n, avg}\).

    \[ \dot{m} = \rho A_c V_{n, avg} \quad \text{where } V_{n, avg} = \frac{ \dot{m} }{\rho A_c} = \frac{ \dot{ V\kern-0.8em\raise0.3ex- } }{A_c} \nonumber \]

    Example — Flow in a rectangular duct

    Air flows steadily through a heating duct with a square cross section \(2H \times 2H\). The measured velocity profile over the cross-sectional area can be described mathematically in terms of the position in the duct \((x, \, y)\) and the centerline velocity \(V_o\) (velocity at (0,0)).

    A standard-orientation 2D Cartesian coordinate system is centered on a square with sides of length 2H.

    \[ V_n = V_o \left[ 1 - \left( \frac{x}{H} \right)^2 \right] \left[ 1 - \left( \frac{y}{H} \right)^2 \right] \nonumber \]

    Determine:

    (a) the volumetric flow rate, in \(\text{m}^3/\text{s}\), if \(V_o = 10 \text{m/s}\) and \(H = 0.3 \text{m}\), and

    (b) the ratio of \(V_{avg}\) to \(V_o\).

    Solution

    Known: Air flows in a square duct with a specified velocity profile.

    Find: (a) Volumetric flow rate if \(V_o = 10 \text{m/s}\) and \(H = 0.3 \text{m}\). (b) Ratio of \(V_{avg}\) to \(V_o\).

    Given: Sketch and velocity profile shown above.

    Analysis: Strategy \(\rightarrow\) Since Vn and the area are given, use the definition of volumetric flow rate. We start with the definition for volumetric flow rate

    \[ \begin{align*} \dot{ V\kern-0.8em\raise0.3ex- } &= \int\limits_{A_c} V_n \, dA \quad \text{where } dA = dx \ dy \\ &= \int\limits_{-H}^{H} \int\limits_{-H}^{H} \underbrace{ V_o \left[ 1 - \left( \frac{x}{H} \right)^2 \right] \left[ 1 - \left( \frac{y}{H} \right)^2 \right] }_{V_n} \, \underbrace{ dx \ dy }_{dA} \end{align*} \nonumber \]

    Bringing \(V_o\) outside the integral and using symmetry to change the limits of integration, we have

    \[ \dot{V} = 4 V_o \int\limits_{0}^{H} \int\limits_{0}^{H} \underbrace{ \left[ 1 - \left( \frac{x}{H} \right)^2 \right] \left[ 1 - \left( \frac{y}{H} \right)^2 \right] }_{V_n} \underbrace{ dx \ dy }_{dA} \nonumber \]

    Integrating first with respect to the \(x\)-axis gives

    \[ \int\limits_{0}^{H} \left[ 1 - \left( \frac{x}{H} \right)^2 \right] \ dx = \left[ x - \frac{1}{3} \frac{x^3}{H^2} \right]_0^H = \frac{2}{3} H \nonumber \]

    Substituting back into the full integral and integrating with respect to \(y\) gives

    \[ \begin{align*} \dot{ V\kern-0.8em\raise0.3ex- } &= 4 V_o \left( \frac{2}{3} H \right) \int\limits_{0}^{H} \left[ 1 - \left( \frac{y}{H} \right)^2 \right] \ dy = 4 V_o \left( \frac{2}{3} H \right) \left[ y - \frac{1}{3} \frac{y^3}{H^2} \right]_0^H = 4 V_o \left[ \frac{2}{3} H \right]^2 \\ &= \frac{16}{9} H^2 V_o \end{align*} \nonumber \]

    Substituting the numbers into this expression gives the volumetric flow rate as

    \[ \dot{ V\kern-0.8em\raise0.3ex- } = \frac{16}{9} \left( 0.3 \ \text{m} \right)^2 \left( 10 \ \frac{ \text{m} }{ \text{s} } \right) = 1.6 \ \text{m}^3 / \text{s} \nonumber \]

    Now to find the ratio of the average to the centerline velocity, use the definition of the average velocity:

    \[ V_{avg} = \frac{ \dot{m} }{\rho A_c} = \frac{\dot{ V\kern-0.8em\raise0.3ex- }}{A_c} = \frac{ \frac{16}{9} H^2 V_o }{ \left[ 2H \right]^2 } = \frac{4}{9} V_o \nonumber \]

    Thus the average velocity is \(\dfrac{4}{9}\) of the centerline velocity.

    Example — River flow rate

    Water flows steadily in a river. The river cross-section is shown in the figure below. In addition, surface velocities measured by watching pieces of wood float downstream are also shown.

    Cutaway view of a river, mapped to a coordinate plane. The horizontal axis shows distance from the left side of the river in feet; the lower section of the vertical axis shows depth of the river in feet and the higher section shows surface velocity of the river at the corresponding point in feet per second.

    Compute the (a) volumetric flow rate in \(\text{ft}^3/ \text{s}\) and (b) mass flow rate if the density of water is \(62.4 \ \text{lbm/ft}^3\).

    Solution

    Known: Velocities on the surface of a river and the depth of the river at each location.

    Find: Volumetric flow rate in \(\text{ft}^3/\text{s}\) and the mass flow rate if \(\rho = 62.4 \text{lbm/ft}^3\).

    Given: Channel depth and surface velocity given in figure above.

    Analysis:

    Strategy \(\rightarrow\) Should be able to apply defining equation for volumetric and mass flow rate.

    Assume \(\rightarrow \) Velocity is uniform from top to bottom of channel. The defining equation for volumetric flow rate is

    \[ \begin{align*} \dot{ V\kern-0.8em\raise0.3ex- } &= \int\limits_{A_{surface}} V_n \, dA \\ &\simeq \sum_{i=1}^{N} A_{i} V_{avg, \ i} \simeq \sum_{i=1}^{N}\left(h_{i} \Delta x_{i}\right) V_{avg, \ i} \end{align*} \nonumber \]

    To perform the necessary integration numerically, we can set up a table:

    \(i\) \(x \,\, (\text{ft})\) \(V_n \,\, (\text{ft/s})\) \(\Delta x\,\, (\text{ft})\) \(h_i\,\, (\text{ft})\) \(V_n (h \Delta x)\,\,(\text{ft}^3/\text{s})\)
    \(1\) \(2.5\) \(2.5\) \(5\) \(2.5\) \(31.25\)
    \(2\) \(7.5\) \(7.5\) \(5\) \(8.0\) \(300.00\)
    \(3\) \(12.5\) \(11.25\) \(5\) \(12.5\) \(703.10\)
    \(4\) \(17.5\) \(13.75\) \(5\) \(16.0\) \(1100.00\)
    \(5\) \(22.5\) \(15.0\) \(5\) \(18.0\) \(1350.00\)
    \(6\) \(27.5\) \(15.0\) \(5\) \(26.0\) \(1950.00\)
    \(7\) \(32.5\) \(15.0\) \(5\) \(37.0\) \(2775.00\)
    \(8\) \(37.5\) \(15.0\) \(5\) \(42.0\) \(3150.00\)
    \(9\) \(42.5\) \(15.0\) \(5\) \(36.0\) \(2700.00\)
    \(10\) \(47.5\) \(12.5\) \(5\) \(25.0\) \(1562.50\)
    \(11\) \(52.5\) \(7.5\) \(5\) \(15.0\) \(562.50\)
    \(12\) \(57.5\) \(2.5\) \(5\) \(5.0\) \(62.50\)
    \(.\)         \(16246.85\)

    Thus the average volumetric flow rate is \[ \dot{ V\kern-0.8em\raise0.3ex- } = 16.2 \times 10^3 \, \text{ft}^3/\text{s} \nonumber \]

    The mass flow rate can be found by multiplying the water density times the volumetric flow rate:

    \[\dot{m} =\rho \dot{ V\kern-0.8em\raise0.3ex- } = \left(62.4 \ \frac{\mathrm{lb}_{\mathrm{m}}}{\mathrm{ft}^{3}}\right) \left(16200 \ \frac{\mathrm{ft}^{3}}{\mathrm{~s}} \right) = 1.01 \times 10^{6} \ \frac{\mathrm{lb}_{\mathrm{m}}}{\mathrm{s}} \nonumber \]

    Comment: If we recognize that the velocity is at a maximum at the free surface and zero at the bottom, a better estimate might be obtained by assuming that the average velocity is some fraction of the amount at the free surface. For example, if we assume that the average velocity is \(4 / 5\) of the maximum velocity, then the actual volumetric flow rate is \(13.0 \times 10^{3} \, \mathrm{ft}^{3} / \mathrm{s}\).


    This page titled 3.2: Mass Flow Rate is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Donald E. Richards (Rose-Hulman Scholar) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.