Mass Flow Rate Per Unit Area

University:
Duke University
Course:
PHYSICS 814 | Introduction to Fluid Mechanics
Academic year:

2023
Views:

51

Pages:

2
Author:

Pemuda H

Blue sweater, red plaid shirt: 111 Mass Flow Rate Per Unit Area (3.3.1.1) m=pvAA= Want to write this ratio as a function of M and stagnation states Po = v Ma=MNORT = MNOR (To/(1+5M))" =(1/2)(M) → Po RTo = A = NRTO NTM (1+M²) (81) moved these three dudes to LHS to make non-dimensional NRT A Po = f(M, 8) Plotting this in Figure 3.5. See that there is a maximum at M=1 and air is higher than the combustion products +1 Y+1 M +1 2(8-1) (1+ M2) M² = 0 7-1 1+2M²- 2+1 2 M²=0 ← 1-M²=0 M=1 is a maximum When M=1, then we have a critical mass flow rate ++ ・(+1)=(Ⅱ) M -+ (1+5 M²). 7+1 2(8-1) (2)M} Area Ratio (3.3.1.2) Can find A(x) A Problem: Given Acr, find M Can use Newton's Method to linearize it x²= M F(x) = 0 (to find roots) F(x+1)= F(xn) + (x²+1-x") F'(xn) + ((x+x)²) = 0 1+14 x+1= x- F(x) →x" is an initial guess n Non-Isentropic Flow (3.3.2) How does Po and To vary? isn't this eq. only for isentropic flows? To T 8-1 1+2M² →To Tz In To InT+ Inz (1) To = ++ dz Po →P=Pz In P = InP+ji Inz (2)=+ P Tds=du+pdr→h=u+pr =d(h-pr) + pdr = dhd (pr) + pdr = dh-pdr-rdp + pdr (Tas=dh-rdp) RT ds R YRAT 8-1 RT = P Tdp = - (3) To =(-)-(-) R To ds R rd To If adiabatic and no work: Pods Po R 89-8w=dh+vdv = dho 0=dh+vdv = dho