Answer:
P= 5.5 bar
Explanation:
Given that
L= 4000 m
d= 0.2 m
Friction factor(F) = 0.01
speed V= 2 m/s
Head = 5 m
Head loss due to friction
So the total head(H) = 5 + 40.77 + 10.3 =56.07
Where 10.3 m is the atmospheric head.
We know that
P=ρ g H
So total Pressure
P= 1000 x 9.81 x 56.07 Pa
P= 5.5 bar
Answer:
More Drag on the down going wing and More Lift on the up going wing
Explanation:
The autorotation spins of blades used in airborne wind energy technology sectors help drive and move the winds and water propeller-type turbines or shafts of generators to produce electricity at altitude and transmit the electricity to earth through conductive tethers.
Sometimes autorotation takes place in rotating parachutes, kite tails. Etc.
As a result, more Drag usually induces the autorotation spin characteristics of a straight-wing aircraft on the downgoing wing and More Lift on the up-going wing.
Answer:
Two stroke cycle Four stroke cycle
1.Have on power stroke in one revolution. 1.have one power
stroke in two revolution
2.Complete the cycle in 2 stroke 2.Complete the cycle in 4 stroke
3.It have ports 3.It have vales
4.Greater requirement of cooling 4.Lesser requirement of cooling
5.Less thermal efficiency 5.High thermal efficiency
6.Less volumetric efficiency 6.High volumetric efficiency
7.Size of flywheel is less. 7.Size of flywheel is more.
Answer:
(a) T = W/2(1-tanθ) (b) 39.81°
Explanation:
(a) The equation for tension (T) can be derived by considering the summation of moment in the clockwise direction. Thus:
Summation of moment in clockwise direction is equivalent to zero. Therefore,
T*l*(sinθ) + W*(l/2)*cosθ - T*l*cosθ = 0
T*l*(cosθ - sinθ) = W*(l/2)*cosθ
T = W*cosθ/2(cosθ - sinθ)
Dividing both the numerator and denominator by cosθ, we have:
T = [W*cosθ/cosθ]/2[(cosθ - sinθ)/cosθ] = W/2(1-tanθ)
(b) If T = 3W, then:
3W = W/2(1-tanθ),
Further simplification and rearrangement lead to:
1 - tanθ = 1/6
tanθ = 1 - (1/6) = 5/6
θ = tan^(-1) 5/6 = 39.81°
