calculate the magnitude of the force acting on the pin at D. Pin C is fixed in DE and bears against the smooth slot in the trian
gular plate.
1 answer:
You might be interested in
Answer:
(b). T = 22.55 ⁰C
(c). q = 557.8 W
Explanation:
we take follow a step by step process to solving this problem.
from the question, we have that
The two glass pieces is separated by a 1.8 cm distance layer of air.
the thickness of glass piece is 1 cm
width = 4 m
the height = 3 m
(a). the sketch of the thermal circuit is uploaded in the picture below.
(b). the thermal resistance due to the conduction in the first glass plane is given thus;
R₁ = Lg / Kg A ................(1)
given that Kg rep. the thermal conductivity of the glass plane
A = conduction surface area
Lg = Thickness of glass plane4
taking the thermal conductivity of glass plane as Kg = 0.78 w/mk
inputting values into equation (1) we have,
R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]
R₁ = 1.068 ˣ 10 ⁻³ k/w
Being that we have same thermal resistance in the first and second plane,
therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w
⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus
R₂ = La/KaA .....................(2)
given Ka = thermal conductivity of air
A = surface area
La = thickness of air
substituting values into the equation we have
R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]
R₂ = 5.73 ˣ 10⁻² k/w
Given the thermal resistance on the outer surface due to convection, we have
R₄ = 1/hA
inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w
R₄ = 6.94 ˣ 10⁻³k/w
Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄
R-total = 0.0663 kw
From this we can calculate the rate of heat loss
using q = Ti - To / R-total ..............(3)
given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C
from equation (3),
q = 27- (-10) / 0.0063 = 557.8 W
q = 557.8 W
⇒ Applying the heat transfer formula for inside surface glass temperature gives;
q = Ti - T₂ / R₃ + R₄
T₂ = Ti - q (R₃ + R₄)
T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C
T₂ = 22.55°C
cheers i hope this helps
Answer: hello the complete question is attached below
answer:
A) Group symbol = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Explanation:
<u>A) Classifying the soil according to USCS system</u>
( using 2nd image attached below )
<em>description of sand</em> :
The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also
The soil is a sand given that more than 50% particles passed from No 4 sieve
The soil can be a clean sand given that fines ≤ 12%
The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time
Group symbol as per the 2nd image attached below = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
