What forced induction device is more efficient?
2 answers:
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Answer:
i) SF = 0.83
ii) 10 psi
iii) 3.16 psi
iv) 0.0083
Explanation:
Constant load = 1000 Psi
mean resistance = 1200 psi
Probability of failure = 1.9 * 10^-3
<u>Determine </u>
<u>i) Safety factor </u>
S.F = 1 / (mean load / load design ) = 1 / ( 1200 / 1000 ) = 0.83
<u>ii) Standard deviation </u>
1.9 X 10^-3 = e ( -1/2 ( μ / б) ^3 / (2.5 * 6 )
0.004756 = e^- 20000 / б^3
hence std = 10 psi
<u>iii) Variance </u>
= = 3.16 psi
<u>iv) coefficient of variation </u>
Cv = std / mean resistance
= 10 / 1200 = 0.0083
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Answer:
net boiler heat = 301.94 kW
Explanation:
given data
saturated steam = 6.0 bars
temperature = 18°C
flow rate = 115 m³/h = 0.03194 m³/s
heat use by boiler = 90 %
to find out
rate of heat does the boiler output
solution
we can say saturated steam is produce at 6 bar from liquid water 18°C
we know at 6 bar from steam table
hg = 2756 kJ/kg
and
enthalpy of water at 18°C
hf = 75.64 kJ/kg
so heat required for 1 kg is
=hg - hf
= 2680.36 kJ/kg
and
from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg
so here mass flow rate is
mass flow rate =
mass flow rate m = 0.10139 kg/s
so heat required is
H = h × m
here h is heat required and m is mass flow rate
H = 2680.36 × 0.10139
H = 271.75 kJ/s = 271.75 kW
now 90 % of boiler heat is used for generate saturated stream
so net boiler heat =
net boiler heat =
net boiler heat = 301.94 kW