Answer:
a) 0.01
b) 150 cm^3/s
c) 300 cm^3/s
d) 25 cm^3/s
Explanation:
a) We know that :
Q=ΔP/R
R=8ηl/π*r^4
Givens:
r^2 = 0.1 r_1
Plugging known information to get :
Q=ΔP/R
=ΔP*π*r^4/8*η*l
Q_2/r_2^4 =Q_1/r_1^4
Q_2=Q_1/r_1^4*r_2^4
=Q_1/r_1^4*r*0.0001*r_1^4
Q_2 = 0.01
b) From the rate flow of the fluid we know that :
Q=ΔP/R (1)
F=η*Av/l (2)
R=8*ηl/π*r^4 (3)
<em>Where: </em>
ΔP is the change in the pressure .
r is the raduis of the tube .
l is the length of the tube .
η is the coefficient of the vescosity of the fluid .
R is the resistance of the fluid .
Givens: Q1 = 100 cm^3/s , ΔP= 1.5
Plugging known information into EQ.1 :
Q=ΔP/R
Q_2/ΔP2=Q_1/ΔP
Q_2=150 cm^3/s
c) we know that :
F = η*Av/l
can be written as :
ΔP = F/A = η*v/l
Givens: η_2 = 3η_1
Q=ΔP/R
Q=η*v/l*R
Q_2/η_2=Q_1/η_1
Q_2=300 cm^3/s
d) We know that :
Q=ΔP/R
R=8*ηl/π*r^4
Givens: l_2 = 4*l_1
Plugging known information to get :
Q=ΔP/R
Q=ΔP*π*r^4/8*ηl
Q_2/l_2=Q_1/l_1
Q_2 = 25 cm^3/s
Answer:
1211 N.
Explanation:
Okay, we are given the following data or parameters or information in the question above;
=> "aluminum (E = 70 GPa) tube of length 8-m.
=> "The aluminum is used as a simply supported column carrying a 1.2 kN axial load"
The axial load to provide a factor of safety of 2 against buckling =[ (22/7)^2 × 224348 × 1000 × 70] ÷ (8 × 1000)^2 × 2.
The axial load to provide a factor of safety of 2 against buckling = 1211 N.
Answer:
a true b false thos is the solution
Answer:
distance = 22.57 ft
superelevation rate = 2%
Explanation:
given data
radius = 2,300-ft
lanes width = 12-ft
no of lane = 2
design speed = 65-mph
solution
we get here sufficient sight distance SSD that is express as
SSD = 1.47 ut + ..............1
here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here
so put here value and we get
SSD = 1.47 × 65 ×2.5 +
solve it we get
SSD = 644 ft
so here minimum distance clear from the inside edge of the inside lane is
Ms = Rv ( 1 - ) .....................2
here Rv is = R - one lane width
Rv = 2300 - 6 = 2294 ft
put value in equation 2 we get
Ms = 2294 ( 1 - )
solve it we get
Ms = 22.57 ft
and
superelevation rate for the curve will be here as
R = ..................3
here f is coefficient of friction that is 0.10
put here value and we get e
2300 =
solve it we get
e = 2%