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)?
1 answer:
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
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Complete Question
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.
Answer:
The elongation is
Explanation:
In order to gain a good understanding of this solution let define some terms
True Stress
A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as .
True Strain
A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as .
The mathematical relation between stress to strain on the plastic region of deformation is
Where K is a constant
n is known as the strain hardening exponent
This constant K can be obtained as follows
No substituting from the question we have
Making the subject from the equation above
From the definition we mentioned instantaneous length and this can be obtained mathematically as follows
Where
is the instantaneous length
is the original length
We can also obtain the elongated length mathematically as follows
Answer:
Explanation:
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