contestada

Consider water at 27°C in parallel flow over an isothermal, 1‐m‐long flat plate with a velocity of 2 m/s. a) Plot the variation (in MATLAB or Excel) of the local heat transfer coefficient, hx(x), with distance along the plate for three flow conditions corresponding to critical (transition) Reynolds numbers of (i) 5 × 10^5, (ii) 3 × 10^5, and (iii) 0 (the flow is fully turbulent). b) Calculate the average heat transfer coefficients for the entire plate for the three flow conditions of part (a)?

Respuesta :

Answer:

i) h-bar-L = 4110 W/m^2K

ii ) h-bar-L = 4490 W/m^2K

iii) h-bar-L = 5072 W/m^2K

Explanation:

Given:-

- The temperature of water, T = 27°C

- The velocity of fluid flow, U∞ = 2m/s

- The length of the flat place, L = 1 m

Solution:-

- Using table A-6, to determine the properties of water:

                   Density ρ = 997 kg/m^3

                   Dynamic viscosity ν = 0.858*10^-6 m^2/s

                   Pr = 583 , k = 0.613 W/m.K

- The reynold's number for full length (L = 1m):

                   Re = U∞*L / ν

                   Re = (2)*(1) / (0.858*10^-6)

                  Re = 2.33*10^6

- The boundary layer is mixed with Rex,c = 5*10^5. Evaluate the critical length (xc):

                 xc = L* ( Rex,c / Re )

                      = (5*10^5 / 2.33*10^6 )

                      = 0.215 m

a) Using "IHT correlation tool, External Flow, Local coefficients for laminar or Turbulent flows", h (x) was evaluated and plotted with critical Reynolds number for all 3 cases: (i) 5 × 10^5, (ii) 3 × 10^5, and (iii) 0 (the flow is fully turbulent). - (See attachment 1)

b) Using "IHT correlation tool, External Flow, Average coefficients for laminar or Mixed flows", h - bar- (x) was evaluated and plotted with critical Reynolds number for all 3 cases: (i) 5 × 10^5, (ii) 3 × 10^5, and (iii) 0 (the flow is fully turbulent). - (See attachment 2)

c) The average convection coefficient for the plate can be determined from the graphs presented in (Attachments 1 and 2). Since,

                                    h-bar-L = h-bar-x(L)

The values for the flow conditions are:

             ( i) h-bar-L = 4110,  ii ) h-bar-L = 4490 , iii) h-bar-L = 5072 ) W/m^2K

                   

Ver imagen shahnoorazhar3
Ver imagen shahnoorazhar3
lhmarianateixeira

In this exercise we have to use the knowledge of heat transfer in a bar to calculate the power of this, so we will have three cases like this:

A) [tex]4110 W/m^2K[/tex]

B ) [tex]4490 W/m^2K[/tex]

C) [tex]5072 W/m^2K[/tex]

Organizing the information given in the statement, we find that:

  • The temperature of water, T = 27°C
  • The velocity of fluid flow, U∞ = 2m/s
  • The length of the flat place, L = 1 m

Knowing that some properties of water can be characterized as:

  • Density ρ = 997 kg/m^3
  • Dynamic viscosity ν = 0.858*10^-6 m^2/s
  • Pr = 583 , k = 0.613 W/m.K

Calculating the Re, by the formula below we have:

[tex]Re = U*L / v\\ Re = (2)*(1) / (0.858*10^{-6})\\ Re = 2.33*10^6[/tex]

Using the formula that:

[tex]h-bar-L = h-bar-x(L)[/tex]

Putting the conditions given earlier in this exercise, we have that each alternative will be:

A) [tex]4110 W/m^2K[/tex]

B ) [tex]4490 W/m^2K[/tex]

C) [tex]5072 W/m^2K[/tex]

See more about heat transfer at brainly.com/question/12107378

Otras preguntas