(20pts) Air T[infinity] = 10 °C and u[infinity] = 100 m/s flows over a flat plate. Assume that the density of air is 1.0 kg/m3 a
nd the dynamics viscosity is 1.8 ×10-6 kg/m∙s. (a) Determine the velocity boundary layer thickness at, δ at x = 2 m (x indicates the distance from the leading edge of the plate). (b) Determine local friction coefficient at x = 2 m. (c) Determine the Nusselt number at x = 2 m. Assume that the velocity boundary layer thickness is the same as the thermal boundary layer thickness (δ = δT).
1 answer:
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Answer:
A) EXIT TEMPERATURE = 14⁰C
b) rate of heat transfer of air = - 13475.78 = - 13.5 kw
Explanation:
Given data :
diameter of duct = 20-cm = 0.2 m
length of duct = 12-m
temperature of air at inlet= 50⁰c
pressure = 1 atm
mean velocity = 7 m/s
average heat transfer coefficient = 85 w/m^2⁰c
water temperature = 5⁰c
surface temperature ( Ts) = 5⁰c
properties of air at 50⁰c and at 1 atm
= 1.092 kg/m^3
Cp = 1007 j/kg⁰c
k = 0.02735 W/m⁰c
Pr = 0.7228
v = 1.798 * 10^-5 m^2/s
determine the exit temperature of air and the rate of heat transfer
attached below is the detailed solution
Calculate the mass flow rate
= p*Ac*Vmean
= 1.092 * 0.0314 * 7 = 0.24 kg/s
Answer:
r=0.31
Ф=18.03°
Explanation:
Given that
Diameter of bar before cutting = 75 mm
Diameter of bar after cutting = 73 mm
Mean diameter of bar d= (75+73)/2=74 mm
Mean length of uncut chip = πd
Mean length of uncut chip = π x 74 =232.45 mm
So cutting ratio r
r=0.31
So the cutting ratio is 0.31.
As we know that shear angle given as
Now by putting the values
\
Ф=18.03°
So the shear angle is 18.03°.