Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Answer: Option D, piezoelectric pressure guage
Explanation: Quartz crystal possess a very useful quality in science as they can generate small charges when pressure is applied to them or when they are hit. This property can be harnessed to construct a piezoelectric pressure gauge which would be used to measure and indicate changes in pressure, the quartz crystal releases little voltage each time there is an applied pressure . This device would be able to sense changes in pressure as there would voltage proportional to the applied pressure.
Answer:
thickness1=1.4m
thickness2=2.2m
convection coefficient=0.33W/m^2K
Explanation:
you must use this equation to calculate the thickness:
L=K(T2-T1)/Q
L=thickness
T=temperature
Q=heat
L1=0.04*(0--350)/10=1.4m
L2=0.1(220-0)/10=2.2m
Then use this equation to calculate the convective coefficient
H=Q/(T2-T1)
H=10/(250-220)=0.33W/m^2K
Answer:
critical stress = 595 MPa
Explanation:
given data
fracture toughness = 74.6 MPa-
crack length = 10 mm
f = 1
solution
we know crack length = 10 mm
and crack length = 2a as given in figure attach
so 2a = 10
a = 5 mm
and now we get here with the help of plane strain condition , critical stress is express as
critical stress = 
......................1
put here value and we get
critical stress = 
critical stress = 595 MPa
so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.
plain stress condition occur in thin body where stress through thickness not vary by the thinner section.
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