Answer:
a) 0.35 W
b) 5.25 W
Explanation:
Given that
Width of the chip, W = 5 mm
The environment temperature, T(∞) = 15° C
Surface temperature, T(s) = 85° C
The initial convection heat coefficient, h1 = 200 W/m².K
The final convection heat coefficient, h2 = 3000 W/m².K
See attachment for solution
Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
Answer:
Speed of aircraft ; (V_1) = 83.9 m/s
Explanation:
The height at which aircraft is flying = 3000 m
The differential pressure = 3200 N/m²
From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3
Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.
Thus, let's apply the Bernoulli equation :
P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2
Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.
We'll obtain ;
P1/ρg + (V_1)²/2g = P2/ρg
Let's make V_1 the subject;
(V_1)² = 2(P1 - P2)/ρ
(V_1) = √(2(P1 - P2)/ρ)
P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question
Thus,
(V_1) = √(2 x 3200)/0.909)
(V_1) = 83.9 m/s
