How many ase certifications are there for automotive technicians?
1 answer:
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Answer:
The percentage ductility is 35.5%.
Explanation:
Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.
Step1
Given:
Diameter of shaft is 10.2 mm.
Final area of the shaft is 52.7 mm².
Calculation:
Step2
Initial area is calculated as follows:
A = 81.713 mm².
Step3
Percentage ductility is calculated as follows:
D = 35.5%.
Thus, the percentage ductility is 35.5%.
Answer:
the highest rate of heat transfer allowed is 0.9306 kW
Explanation:
Given the data in the question;
Volume = 4L = 0.004 m³
V = V
= 0.002 m³
Using Table ( saturated water - pressure table);
at pressure p = 175 kPa;
v = 0.001057 m³/kg
v = 1.0037 m³/kg
u = 486.82 kJ/kg
u 2524.5 kJ/kg
h = 2700.2 kJ/kg
So the initial mass of the water;
m₁ = V/v
+ V
/v
we substitute
m₁ = 0.002/0.001057 + 0.002/1.0037
m₁ = 1.89414 kg
Now, the final mass will be;
m₂ = V/v
m₂ = 0.004 / 1.0037
m₂ = 0.003985 kg
Now, mass leaving the pressure cooker is;
m = m₁ - m₂
m = 1.89414 - 0.003985
m = 1.890155 kg
so, Initial internal energy will be;
U₁ = mu
+ m
u
U₁ = (V/v
)u
+ (V
/v
)u
we substitute
U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)
U₁ = 921.135288 + 5.030387
U₁ = 926.165675 kJ
Now, using Energy balance;
E - E
= ΔE
QΔt - mh
= m₂u₂ - U₁
QΔt - mh
= m₂u
- U₁
given that time = 75 min = 75 × 60s = 4500 sec
so we substitute
Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675
Q(4500) - 5103.7965 = 10.06013 - 926.165675
Q(4500) = 10.06013 - 926.165675 + 5103.7965
Q(4500) = 4187.690955
Q = 4187.690955 / 4500
Q = 0.9306 kW
Therefore, the highest rate of heat transfer allowed is 0.9306 kW