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:
The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3
Heat flux is 9.67×10^7 Btu/hrft^2
Explanation:
Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr
Area (A) = πD^2/4
Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft
A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2
Volume (V) = A × Length
L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft
V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3
Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3
Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2
To get rockets into orbit, they need much more thrust than the amount that will get them up to the required altitude. They also need sufficient thrust to allow them to travel with very high orbital speed. ... If speed is less than this, an object will fall back to the Earth
Answer:
Explanation:
Reward him if he does his homework/work
If he doesnt do his homework, take things that he loves off of him. and tell him if he does his homework/work he will get them things back
