A.Ensure all circuits are de-energized before beginning work
Answer:
A) 209.12 GPa
B) 105.41 GPa
Explanation:
We are given;
Modulus of elasticity of the metal; E_m = 67 GPa
Modulus of elasticity of the oxide; E_f = 390 GPa
Composition of oxide particles; V_f = 44% = 0.44
A) Formula for upper bound modulus of elasticity is given as;
E = E_m(1 - V_f) + (E_f × V_f)
Plugging in the relevant values gives;
E = (67(1 - 0.44)) + (390 × 0.44)
E = 209.12 GPa
B) Formula for upper bound modulus of elasticity is given as;
E = 1/[(V_f/E_f) + (1 - V_f)/E_m]
Plugging in the relevant values;
E = 1/((0.44/390) + ((1 - 0.44)/67))
E = 105.41 GPa
Answer: The complete part of the question is to find the exit velocity
Explanation:
Given the following parameters
Inlet pressure = 700kpa
outlet pressure = 40kpa
Temperature = 80°C = 353k
mass flow rate = 1 kg/s
The application of the continuity and the bernoulli's equation is employed to solve the problem.
The detailed steps and the appropriate formula is as shown in the attached file.
Answer:
Explanation:
k_max = 26.9 w/mk
k_min = 22.33 w/mk
Explanation:
a) the maximum thermal conductivity is given as
K_MAX = k_m v_m + k_p v_p
where k_m is thermal conductvitiy of metal
k_p is thermal conductvitiy of carbide
v_m = proportion of metal in the cement = 0.15
v_p = proportion of carbide in the cement = 0.85

= 66*0.15 + 20*0.85
k_max = 26.9 w/mk
b) the minimum thermal conductivity is given as

= \frac{20*66}{20*0.15 +66*0.85}
k_min = 22.33 w/mk
the answer is (c)
After the vehicle is involved in a car accident or fire