Answer:
minimum electric power consumption of the fan motor is 0.1437 Btu/s
Explanation:
given data
area = 3 ft by 3 ft
air density = 0.075 lbm/ft³
to find out
minimum electric power consumption of the fan motor
solution
we know that energy balance equation that is express as
E in - E out =
......................1
and at steady state
= 0
so we can say from equation 1
E in = E out
so
minimum power required is
E in = W = m
=
put here value
E in =
E in =
E in = 0.1437 Btu/s
minimum electric power consumption of the fan motor is 0.1437 Btu/s
Answer:
The power developed in HP is 2702.7hp
Explanation:
Given details.
P1 = 150 lbf/in^2,
T1 = 1400°R
P2 = 14.8 lbf/in^2,
T2 = 700°R
Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h
Using air table to obtain the values for h1 and h2 at T1 and T2
h1 at T1 = 1400°R = 342.9 Btu/h
h2 at T2 = 700°R = 167.6 Btu/h
Using;
Q - W + m(h1) - m(h2) = 0
W = Q - m (h2 -h1)
W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h
W = (-65000 Btu/h ) - (-1928.3) Btu/s
W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s
W = -18.06Btu/s + 1928.3 Btu/s
W = 1910.24Btu/s
Note; Btu/s = 1.4148532hp
W = 2702.7hp
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