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Flow with pressure losses

The expression for the mass flow rate (4.46) is appropriate regardless the flow is isentropic or adiabatic. That expression was derived based on the theoretical total pressure and temperature (Mach number) which does not based on the considerations whether the flow is isentropic or adiabatic. In the same manner the definition of $A^{*}$ referred to the theoretical minimum area (''throat area'') if the flow continues to flow in an isentropic manner. Clearly, in a case where the flow isn't isentropic or adiabatic the total pressure and the total temperature will change (due to friction, and heat transfer). A constant flow rate requires that ${\dot{m}}_{A}= {\dot{m}}_{B}$ . Denoting subscript A for one point and subscript B for another point mass equation (4.47) can be equated as

$\displaystyle \left( {k P_0 A^{*}\over R T_{0}} \right) \left( 1 + {k -1 \over ... ... k- 1 \over 2(k -1)}} = constant %{\left( {k P_0 A^{*} \over R T_{0}} \right)$

(4.71)

From equation (4.71), it is clear that the function $f(P_{0}, T_{0},A^{*}) = constant$ . There are two possible models that can be used to simplify the calculations. The first model for neglected heat transfer (adiabatic) flow and in which the total temperature remained constant (Fanno flow like). The second model which there is significant heat transfer but insignificant pressure loss (Rayleigh flow like).

If the mass flow rate is constant at any point on the tube (no mass loss occur) then

$\displaystyle \dot{m} = A ^{*} \sqrt{{k \over R T_{0} } \left( 2 \over k +1 \right)^{k+1 \over k -1}} P_0$

(4.72)

For adiabatic flow, comparison of mass flow rate at point A and point B leads to

$\displaystyle \left. {P_{0}} {A^{*}}\right\vert _{A}$

$\displaystyle = \left. {P_{0}} {A^{*}}\right\vert _{B}$

$\displaystyle \leadsto {\left.{P_0} \right\vert _{A} \over \left.{P_0} \right\vert _{B}}$

$\displaystyle = {\left.{A^*} \right\vert _{A} \over \left.{A^*} \right\vert _{B}}$

(4.73)

And utilizing the equality of $A^{*} = {A^{*} \over A} A$ leads to

$\displaystyle {\left.{P_{0}}\right\vert _{A} \over \left.{P_{0}}\right\vert _{B... ...}}}\strut} {{\left. A \right\vert _{A} \strut} \over \left. A \right\vert _{B}}$

(4.74)

For a flow with a constant stagnation pressure (frictionless flow) and non adiabatic flow reads

$\displaystyle {\left.{T_{0}}\right\vert _{A} \over \left.{T_{0}}\right\vert _{B... ...{\left. A \right\vert _{B} \strut} \over \left. A \right\vert _{A}} \right]^{2}$

(4.75)

$\begin{examl} At point A of the tube the pressure is $3[Bar]$, Mach number is 2.... ...and adiabatic steady state flow, calculated the total pressure lost. \end{examl}$

Solution

Both Mach numbers are known, thus the area ratios can be calculated. The total pressure can be calculated because the Mach number and static pressure are known. With these information, and utilizing equation (4.74) the stagnation pressure at point B can be obtained.

Isentropic Flow		Input: M			k = 1.4
M	T/T0	ρ/ρ0	A/A*	P/P0	PAR	F/F*
1.5	0.689655	0.394984	1.17617	0.272403	0.320392	0.55401
2.5	0.444444	0.131687	2.63672	0.0585277	0.154321	0.626929

First, the stagnation at point A is obtained from Table (4.2) as

$\displaystyle \left. {P_{0}}\right\vert _{A} = \left. P \over {\underbrace{\lef... ...right)}_{M=2.5} } \right\vert _{A} = { 3 \over 0.058527663 } = 51.25781291[Bar]$

Utilizing equation (4.74) provides

$\displaystyle \left.{P_{0}}\right\vert _{B} = 51.25781291 \times { 1.1761671 \over 2.6367187 } \times {0.01 \over 0.015} \approx 15.243 [Bar]$

Hence

$\displaystyle \left.{P_{0}}\right\vert _{A} - \left.{P_{0}}\right\vert _{B} = 51.257 - 15.243 = 36.013[Bar]$

Note that the large total pressure loss is much larger than the static pressure loss (Pressure point B the pressure is $0.27240307 \times 15.243= 4.146$ [Bar]).

Next: Isentropic Tables Up: Mass Flow Rate (Number) Previous: ``Naughty Professor'' Problems in Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21