A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas
1 answer:
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Answer:
The answer is 0.4490Mpa
Explanation:
Given :
P in direction of 001 , P = 1.1MPa
slip plane = 111, slip plane normal direction of 111,
slip direction = 101
= δcosФcosλ
= (001 * 101) = 1 = cosλ
(111 * 001) 1 cosФ
= 1.1 MPa * *
1.1 Mpa * 1/1.4142 * 1/1.7320
1.1 Mpa * 0.7072 * 0.5773
= 0.4490 MPa
therefore the resolved shear stress = 0.4490MPa
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
Answer:
cutting speed is 365.71 m/min
Explanation:
given data
diameter D = 250 mm
length L = 625 mm
Feed f = 0.30 mm/rev
depth of cut = 2.5 mm
n = 0.25
C = 700
to find out
the cutting speed that will allow the tool life to be just equal to the cutting time for the three parts
solution
we will apply here cutting time formula that is express as
Tc = .......................1
here D is diameter and L is length and f is feed and V is speed
so we get
Tc =
Tc =
and we know tool life is
T = 3 × Tc ................................2
here T is tool life and Tc is cutting time
so find here tool life by put value in equation 2
T = 3 ×
by taylor tool formula cutting speed is
V ×
× 8.37 = 700
V = 365.71
so cutting speed is 365.71 m/min