4.2 One-Dimensional FlowNiklas Andersson
Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics
Most of the problems addressed in this book can be analyzed using steady state equations in one dimension. For control volume problems, we can derive flow equations that relates the flow state at the upstream end of the control volume to the the flow state at the downstream end by applying the integral form of the governing equations (Eqns. (3.5) (3.24) A 1 A_1 A 1 A 2 A_2 A 2 A 1 = A 2 = A A_1=A_2=A A 1 = A 2 = A (4.1) (4.4)
ρ 1 u 1 A 1 = ρ 2 u 2 A 2 ⇒ ρ 1 u 1 = ρ 2 u 2 \rho_1 u_1 A_1 = \rho_2 u_2 A_2 \Rightarrow \rho_1 u_1 = \rho_2 u_2 ρ 1 u 1 A 1 = ρ 2 u 2 A 2 ⇒ ρ 1 u 1 = ρ 2 u 2 ( p 1 + ρ u 1 2 ) A 1 = ( p 2 + ρ u 2 2 ) A 2 ⇒ p 1 ρ u 1 2 = p 2 ρ u 2 2 \left(p_1 + \rho u_1^2\right) A_1 = \left(p_2 + \rho u_2^2\right) A_2 \Rightarrow p_1 \rho u_1^2 = p_2 \rho u_2^2 ( p 1 + ρ u 1 2 ) A 1 = ( p 2 + ρ u 2 2 ) A 2 ⇒ p 1 ρ u 1 2 = p 2 ρ u 2 2 ( h 1 + 1 2 u 1 2 ) A 1 = ( h 2 + 1 2 u 2 2 ) A 1 ⇒ h 1 + 1 2 u 1 2 = h 2 + 1 2 u 2 2 \left(h_1+\dfrac{1}{2}u_1^2\right)A_1=\left(h_2+\dfrac{1}{2}u_2^2\right)A_1 \Rightarrow h_1+\dfrac{1}{2}u_1^2 = h_2+\dfrac{1}{2}u_2^2 ( h 1 + 2 1 u 1 2 ) A 1 = ( h 2 + 2 1 u 2 2 ) A 1 ⇒ h 1 + 2 1 u 1 2 = h 2 + 2 1 u 2 2 Since h o = h + u 2 / 2 h_o=h+u^2/2 h o = h + u 2 /2 (4.3)
h o 1 = h o 2 h_{o_1}=h_{o_2} h o 1 = h o 2 In many of the applications that we will analyse later, we will need to investigate the continuous change of flow quantities through a specific flow section, which means that we need to have the differential form of Eqns. (4.1) (4.4)
From the continuity equation (Eqn. (4.1)
ρ 1 u 1 = ρ 2 u 2 = c o n s t \rho_1 u_1 = \rho_2 u_2 = const ρ 1 u 1 = ρ 2 u 2 = co n s t and thus
d ( ρ u ) = 0 d(\rho u) = 0 d ( ρ u ) = 0 Expanding the derivative, we get
d ( ρ u ) = 0 ⇒ ρ d u + u d ρ = 0 d(\rho u) = 0 \Rightarrow \rho du + ud\rho =0 d ( ρ u ) = 0 ⇒ ρ d u + u d ρ = 0 If we now divide by ρ u \rho u ρ u (4.6)
d ρ ρ = − d u u \dfrac{d\rho}{\rho}=-\dfrac{du}{u} ρ d ρ = − u d u Now, let’s have a look at the momentum equation. From Eqn. (4.1)
p 1 ρ u 1 2 = p 2 ρ u 2 2 = c o n s t p_1 \rho u_1^2 = p_2 \rho u_2^2 = const p 1 ρ u 1 2 = p 2 ρ u 2 2 = co n s t and thus the differential form of the equation becomes
d ( p + ρ u 2 ) = 0 d(p+\rho u^2) = 0 d ( p + ρ u 2 ) = 0 If the derivative of momentum is expanded we get
d ( p + ρ u 2 ) = d p + ρ u d u + u d ( ρ u ) ⏟ = 0 = 0 d(p+\rho u^2)=dp+\rho udu + \underbrace{ud(\rho u)}_{=0}=0 d ( p + ρ u 2 ) = d p + ρ u d u + = 0 u d ( ρ u ) = 0 where the differential form of the momentum equation is identified and can be removed.
d p + ρ u d u = 0 ⇒ d p = − u 2 d u u dp+\rho udu=0 \Rightarrow dp=-u^2\dfrac{du}{u} d p + ρ u d u = 0 ⇒ d p = − u 2 u d u Assuming calorically perfect gas, u = M γ R T u=M\sqrt{\gamma RT} u = M γ RT p = ρ R T p=\rho RT p = ρRT
d p p = − M 2 γ R T R T d u u ⇒ \dfrac{dp}{p}=-\dfrac{M^2\gamma RT}{RT}\dfrac{du}{u}\Rightarrow p d p = − RT M 2 γ RT u d u ⇒ d p p = − γ M 2 d u u \dfrac{dp}{p}=-\gamma M^2 \dfrac{du}{u} p d p = − γ M 2 u d u The energy equation (Eqn. (4.4)
h o 1 = h o 2 = c o n s t h_{o_1}=h_{o_2}=const h o 1 = h o 2 = co n s t and thus the corresponding differential equation becomes
or, if the total enthalpy h o h_o h o h + u 2 / 2 h+u^2/2 h + u 2 /2
By differentiating the equation of state for perfect gases (p = ρ R T p=\rho RT p = ρRT
d p = R ( ρ d T + T d ρ ) dp=R(\rho dT + Td\rho) d p = R ( ρ d T + T d ρ ) divide by ρ R T \rho RT ρRT
d p ρ R T = d T T + d ρ ρ ⇒ d T T = d p p − d ρ ρ \dfrac{dp}{\rho RT}=\dfrac{dT}{T}+\dfrac{d\rho}{\rho}\Rightarrow \dfrac{dT}{T}=\dfrac{dp}{p}-\dfrac{d\rho}{\rho} ρRT d p = T d T + ρ d ρ ⇒ T d T = p d p − ρ d ρ We can substitute d p / p dp/p d p / p d ρ / ρ d\rho/\rho d ρ / ρ (4.8) (4.14)
d T T = ( 1 − γ M 2 ) d u u \dfrac{dT}{T}=(1-\gamma M^2)\dfrac{du}{u} T d T = ( 1 − γ M 2 ) u d u From the definition of total enthalpy we get
d h o = d h + u d u dh_o=dh+udu d h o = d h + u d u Assuming the gas to be calorically perfect, we can rewrite this expression as
d T o = d T + 1 C p u d u = d T + γ − 1 γ R T T u 2 d u u = d T + ( γ − 1 ) M 2 T d u u dT_o=dT+\dfrac{1}{C_p}udu=dT+\dfrac{\gamma-1}{\gamma RT}Tu^2\dfrac{du}{u}=dT+(\gamma-1)M^2T\dfrac{du}{u} d T o = d T + C p 1 u d u = d T + γ RT γ − 1 T u 2 u d u = d T + ( γ − 1 ) M 2 T u d u Substituting d T dT d T (4.20)
d T o T = ( 1 − M 2 ) d u u \dfrac{dT_o}{T}=(1-M^2)\dfrac{du}{u} T d T o = ( 1 − M 2 ) u d u Multiply and divide by T o T_o T o
d T o T o T o T = ( 1 − M 2 ) d u u \dfrac{dT_o}{T_o}\dfrac{T_o}{T}=(1-M^2)\dfrac{du}{u} T o d T o T T o = ( 1 − M 2 ) u d u from before (Eqn. XXX) we know that
T o T = 1 + γ − 1 2 M 2 \dfrac{T_o}{T}=1+\dfrac{\gamma-1}{2}M^2 T T o = 1 + 2 γ − 1 M 2 and thus we get
d T o T o = 1 − M 2 1 + γ − 1 2 M 2 d u u \dfrac{dT_o}{T_o}=\dfrac{1-M^2}{1+\dfrac{\gamma-1}{2}M^2}\dfrac{du}{u} T o d T o = 1 + 2 γ − 1 M 2 1 − M 2 u d u For a calorically perfect gas, the local Mach number can be calculated as
M = u γ R T M=\dfrac{u}{\sqrt{\gamma RT}} M = γ RT u Differentiating the expression above gives
d M = ( γ R ) − 1 / 2 [ T − 1 / 2 d u + u d ( T − 1 / 2 ) ] = d u γ R T − u 2 1 γ R T − 3 / 2 d T = dM=(\gamma R)^{-1/2}\left[T^{-1/2}du+ud(T^{-1/2})\right]=\dfrac{du}{\sqrt{\gamma RT}}-\dfrac{u}{2}\dfrac{1}{\sqrt{\gamma R}}T^{-3/2}dT= d M = ( γ R ) − 1/2 [ T − 1/2 d u + u d ( T − 1/2 ) ] = γ RT d u − 2 u γ R 1 T − 3/2 d T = = u γ R T d u u − 1 2 u γ R T d T T = M d u u − M 2 d T T =\dfrac{u}{\sqrt{\gamma RT}}\dfrac{du}{u}-\dfrac{1}{2}\dfrac{u}{\sqrt{\gamma RT}}\dfrac{dT}{T}=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T} = γ RT u u d u − 2 1 γ RT u T d T = M u d u − 2 M T d T With d T / T dT/T d T / T (4.20)
d M M = 1 + γ M 2 2 d u u \dfrac{dM}{M}=\dfrac{1+\gamma M^2}{2}\dfrac{du}{u} M d M = 2 1 + γ M 2 u d u Finally, the entropy equation gives
d s = C v d p p − C p d ρ ρ ds=C_v\dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho} d s = C v p d p − C p ρ d ρ d p / p dp/p d p / p d ρ / ρ d\rho/\rho d ρ / ρ (4.14) (4.8)
d s = C v γ ( 1 − M 2 ) d u u ds=C_v\gamma(1-M^2)\dfrac{du}{u} d s = C v γ ( 1 − M 2 ) u d u