Answer:
In the steel: 815 kPa
In the aluminum: 270 kPa
Explanation:
The steel pipe will have a section of:
A1 = π/4 * (D^2 - d^2)
A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2
The aluminum core:
A2 = π/4 * d^2
A2 = π/4 * 0.7^2 = 0.3848 m^2
The parts will have a certain stiffness:
k = E * A/l
We don't know their length, so we can consider this as stiffness per unit of length
k = E * A
For the steel pipe:
E = 210 GPa (for steel)
k1 = 210*10^9 * 0.1178 = 2.47*10^10 N
For the aluminum:
E = 70 GPa
k2 = 70*10^9 * 0.3848 = 2.69*10^10 N
Hooke's law:
Δd = f / k
Since we are using stiffness per unit of length we use stretching per unit of length:
ε = f / k
When the force is distributed between both materials will stretch the same length:
f = f1 + f2
f1 / k1 = f2/ k2
Replacing:
f1 = f - f2
(f - f2) / k1 = f2 / k2
f/k1 - f2/k1 = f2/k2
f/k1 = f2 * (1/k2 + 1/k1)
f2 = (f/k1) / (1/k2 + 1/k1)
f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN
f1 = 200 - 104 = 96 kN
Then we calculate the stresses:
σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa
σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa
Answer:
(b) Constant (minimum) volume
Explanation:
In the idealized Otto cycle there are 4 process that are
- Reversible adiabatic compression
- Addition of heat at constant volume
- Reversible adiabatic expansion
- Rejection of constant volume
So from above discussion we can see that heat is added when there is constant (minimum) volume which is given in option (b) so option (b) will be the correct answer
Answer:
Space mean speed = 44 mi/h
Explanation:
Using Greenshield's linear model
q = Uf ( D -
/Dj )
qcap = capacity flow that gives Dcap
Dcap = Dj/2
qcap = Uf. Dj/4
Where
U = space mean speed
Uf = free flow speed
D = density
Dj = jam density
now,
Dj = 4 × 3300/55
= 240v/h
q = Dj ( U -
/Uf)
2100 = 240 ( U -
/55)
Solve for U
U = 44m/h
True
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