7. The "3 second rule" is the time you should pause at an intersection marked with a stop sign.
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Answer:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
Explanation:
Given:
- The diameter of the pipe d = 5-in
- The pipe is supported every L = 9 ft of pipe in length
- The weight if the pipe + contents W = 10 lb/ft
Find:
determine the magnitude and location of the maximum bending moment in members AC and BC.
Solution:
- The figure (missing) is given in the attachment.
- We will first determine the external forces acting on each member:
Section: 9-ft section of pipe.
Sum of forces perpendicular to member AC = 0
F_d - 0.8*W*L = 0
F_d = 0.8*10*9 = 72 lb
Sum of forces perpendicular to member BC = 0
F_e - 0.6*W*L = 0
F_e = 0.6*10*9 = 54 lb
F_d = 72 lb , F_e = 54 lb
- Then we will determine the support reactions for each member AC point A and BC point B.
Section: Entire Frame.
Sum of moments about point B = 0
-A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0
-A_y*(1.5625) + 15 + 39.375 = 0
A_y = 34.8 lb
Sum of forces in vertical direction = 0
A_y + B_y - 0.8*F_d - 0.6*F_e = 0
B_y = 0.8*(72) + 0.6*(54) - 34.8
B_y = 55.2 lb
Sum of forces in horizontal direction = 0
A_x + B_x - 0.6*F_d + 0.8*F_e = 0
A_x + B_x = 0
Section: Member AC
Sum of moments about point C = 0
F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0
72*2.5 - 34.8*12 - 9*A_x = 0
A_x = -237.6 / 9 = - 26.4 lb
B_x = - A_x = 26.4 lb
A_x = -26.4 lb , B_x = 26.4 lb
- Now we will calculate bending moment for each member at different sections.
Member AC:
From point A till just before point D
-0.6*A_x*x - A_y*0.8*x + M = 0
15.84*x - 27.84*x + M = 0
M = 12*x ..... max value at D, x = 12.25 in
M_max = 12*12.25/12 = 12.25 lb-ft
Member BC:
From point B till just before point E
-0.8*B_x*x + B_y*0.6*x + M = 0
-21.12*x + 33.12*x + M = 0
M = -12*x ..... max value at E, x = 11.25 - 2.5 = 8.75 in
M_max = -12*8.75/12 = -8.75 lb-ft
- The maximum bending moments and their locations are:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
Answer:
work=281.4KJ/kg
Power=4Kw
Explanation:
Hi!
To solve follow the steps below!
1. Find the density of the air at the entrance using the equation for ideal gases
where
P=pressure=120kPa
T=20C=293k
R= 0.287 kJ/(kg*K)= gas constant ideal for air
2.find the mass flow by finding the product between the flow rate and the density
m=(density)(flow rate)
flow rate=10L/s=0.01m^3/s
m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s
3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow
Work
w=Cp(T1-T2)
Where
Cp= specific heat for air=1.005KJ/kgK
w=work
T1=inlet temperature=20C
T2=outlet temperature=300C
w=1.005(300-20)=281.4KJ/kg
Power
W=mw
W=(0.0143)(281.4KJ/kg)=4Kw