What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur
1 answer:
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Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99
Answer:
so the transverse displacement is 0.089963 m
Explanation:
Given data
equation y(x,t)=Acos(kx−ωt)
speed v = 9.00 m/s
amplitude A = 9.00 × 10^−2 m
wavelength λ = 0.480 m
to find out
the transverse displacement
solution
we know
v = angular frequency / wave number
and
wave number = 2 / λ = 2
/ 0.480 = 13.0899
angular frequency = v k
angular frequency = 9.00 × 13.0899
angular frequency = 117.8097 rad/sec = 118 rad/sec
so
equation y(x,t)=Acos(kx−ωt)
y(x,t)=9.00 × 10^−2 cos(13.0899 x−118t)
when x =0 and and t = 0
maximum y(x,t)= 9.00 × 10^−2 cos(13.0899 (0) − 118 (0))
maximum y(x,t)= 9.00 × 10^−2 m
and when x = x = 1.59 m and t = 0.150 s
y(x,t)=9.00 × 10^−2 cos(13.0899 (1.59) −118(0.150) )
y(x,t)=9.00 × 10^−2 × (0.99959)
y(x,t) = 0.089963 m
so the transverse displacement is 0.089963 m