A long cylindrical black surface fuel rod of diameter 25 mm is shielded by a surface concentric to the rod. The shield has diame
1 answer:
Answer:
surface temp of fuel rod = 678.85 K
Explanation:
Given data :
D1 = 25 mm
D2 = 50 mm
T2 = 335 k
T∞ = 300 k
hconv = 0.15 w/m^2.k
ε2 = 0.05
ε1 = 1
Determine energy at Q23
Q23 = Qconv + Qrad
attached below is the detailed solution
Insert given values into equation 1 attached below to obtain the surface temperature of the fuel rod ( T1 )
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Answer:
ratio = 412
speed of wheel = 13.63
Explanation:
given data
head h = 91.5 m
discharge Q = 0.04 m³/s
wheel rotates N = 720 rpm
velocity coefficient Cv = 0.98
efficiency of the wheel η = 80 %
ratio of bucket speed to jet speed = 0.46
to find out
wheel to jet diameter ratio and power specific speed of the wheel
solution
we know that discharge = area × velocity .............1
here velocity = Cv√(2gh)
velocity = 0.98√(2×9.81×91.5) = 41.52 m/s
and area =
so from equation 1
0.04 = × 41.52
d = 0.00123 m = 1.23 mm
we know bucket speed to jet speed = 0.46
so bucket speed u = 0.46 v
bucket speed u = 0.46 × 41.52 = 19.1 m/s
and bucket speed =
so 19.1 =
so D = 0.506 m = 506.62 mm
so ratio is
ratio = 412
and
we know efficiency =
0.80 × ρQgh = output power
power output = 0.80 ×1000×0.04×9.81×91.5
power output P = 28.72 kW
so
speed of wheel is
speed of wheel =
speed of wheel =
speed of wheel = 13.63
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
