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Answer:
The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.
Explanation:
Diameter of pipe = 2 in = 0.0508 m
Steam temperature = 300 F = 422.04 K
Duct temperature = 70 F = 294.26 K
Emmisivity of surface 1 = 0.79
Emmisivity of surface 2 = 0.276
Net emmisivity of both surfaces ∈ = 0.25
Stefan volazman constant = 5.67 ×
Heat transfer per foot length is given by
Q = ∈ A (
) ------ (1)
Put all the values in equation (1) , we get
Q = 0.25 × 5.67 × × 3.14 × 0.0508 × 1 × (
)
Q = 54.78 Watt per foot.
This is the value of heat transferred watt per foot length.
Answer:
(a) The magnitude of force is 116.6 lb, as exerted by the rod CD
(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.
Explanation:
Step by step working is shown in the images attached herewith.
For this given system, the coordinates are the following:
A(0, 0, 0)
B(26, 0, 0)
And the value of angle alpha is 20.95°
Hope that answers the question, have a great day!
Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
