Answer:
3150 L/min
Explanation:
The volume of the room is the product of the height of the room and its area. It is given as:
Volume (V) = height (h) × Area (A)
V = h × A
Height = 2.7 m, Area = 200 m². Hence:
V = h × A = 2.7 × 200 = 540 m³
The flow capacity () is given by
ACH = air changes per hour = 0.35
But 1 m³ = 1000 L, 1 hr = 60 min
Answer:4050 W
Explanation:
Given
Heat transfer Coefficient(h)=
Air temperature =75 F
surface area(A)=
Temperature of hot tube is 102 F
We know heat transfer due to convection is given by
Answer:
a) I=0 b) 4.17V c) 0.354 A d) 14.5s
Explanation:
a) consider circuit in the attachment
i(t)= E/R (1- e^(-t/RL))
i(0)= 12.5/3×(1-e^(0/RL))
i(0)=0
b) at t⇒∞
i(∞)= 12.5/3× (1- e^(-∞/RL))
= 4.17V
c) 1/RL= 1/(6.95×3)= 0.0479616
i(1.85) = 12.5/3 × (1- e^(-1.85×0.0479616)
= 0.354A
d) I/2= I (1- e^(-t/RL))
t= - RL ln0.5
t= - 3×6.95 × (-0.693)
t= 14.5 s
Answer:
the critical flaw length is 10.06 mm
Explanation:
Given the data in the question;
plane strain fracture toughness = 92 Mpa√m
yield strength σ = 900 Mpa
design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa
Y = 1.15
we know that;
Critical crack length = 1/π(
/ Yσ )²
we substitute
= 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²
= 1/π( 92 Mpa√m / (517.5 Mpa )²
= 1/π( 0.177777 )²
= 1/π( 0.03160466 )
= 0.01006 m = 10.06 mm
Therefore, the critical flaw length is 10.06 mm
{ = ( 10.06 mm ) > 3 mm
The critical flow is subject to detection