A light aircraft with a wing area of 200 ft^2 and a weight of 2000 lb has a lift coefficient of 0.39 and a drag coefficient of 0
1 answer:
Answer: power required to maintain level flight=82.20hp
Explanation:
Given
Area = 200 ft^2
Weight = 2000 lb
Cl( Lift coefficient)= 0.39
Cd( Drag coefficient) = 0.06
The density ρ of air at standard atmospheric pressure = 2.38 X 10^-3 slugs/ft^3
For Equilibrium to be maintained during flight conditions, the lift force must be balanced by the weight of the aircraft such that
Lift force = Weight of aircraft
(1/2)ρAU²Cl= W
1/2X 2.38 X 10^-3 X 200 X U² X 0.39 = 2000
U²= 2000 X 2 / 2.38 X 10^-3 X 200 X 0.39
U=
Velocity, U= 146.7892ft/s
Drag force of the velocity can be deduced from the formulae
Cd= Drag force(D) /1/2 ρU²A
Drag force=1/2 ρU²ACd
D=1/2 x (2.38 X 10^-3 slugs/ft^3) x (146.7892ft/s)² x 200 ft^2 x 0.06
D=307.69
Drag force= 308lb
power required to maintain level flight is given as
P = Drag force x Velocity = D x U
=308lb X 146.7892ft/s
=45,211.0736lb.ft/s
Changing to hp we have that
1 Horsepower, hp = 550 ft lbf/s
??=45,211.0736lb.ft/s
45,211.0736lb.ft/s/ 550 ft lbf/s= 82.20hp
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Answer:
a) 40.6%
b)72.19kg/s
Explanation:
The Rankine cycle with two reheat stages has 9 stages in total.
The maximum pressure will be at the first inlet stage of the HP turbine which is stage 3. The minimum pressure will be the exit stage of the condenser because the condenser operates under vacuum pressure which is stage 1.
The following assumptions can be made:
1 - Each component in the cycle is analyzed as an open system operating at steady-state.
2 - All of the processes are internally reversible.
3 - The turbine and pump operate adiabatically and are internally reversible, so they are also isentropic.
4 - Condensate exits the condenser as saturated liquid.
5 - The effluent from the HP turbine is a saturated vapor.
6 - No shaft work crosses the system boundary of the boiler or condenser.
7 - Changes in kinetic and potential energies are negligible
a) The thermal efficiency of the cycle is defined as the work of the cycle divided by the total heat input to the system. The stages that have heat input is stages 2-3, 4-5, 6-7.
For stage 2:
s₁=s₂ assuming isentropic
enthalpy will be a compressed liquid so after interpolation
For stage 3:
Superheated steam @ T=500⁰C and P=15MPa
Stage 4:
superheated vapor
P=5MPa
s₃=s₄=6.3480 kJ/kg, we must use interpolation to find h₄
Stage 5:
Superheated steam @ T=500⁰C and P₄=P₅=5 MPa
Stage 6:
Superheated steam at P₆= 1MPa
s₅=s₆
We find h₆ using interpolation from the steam tables:
Stage 7:
P₇=P₆=1MPa
T=500⁰C superheated steam
The heat into the cycle is:
We can determine the work out by the condenser from stage 9 to stage 1:
Stage 1:
saturated liquid P=5kPa
Stage 9:
We assume that its a saturated liquid with quality of 1 at 5kPa and
s₇=s₉ and after interpolation
Qout = [/tex]2568.53-137.75=2430.79kJ/kg[/tex]
The thermal efficiency can be written in terms of qin and qout:
Efficiency of 40.61%
b)
The mass flow rate can be calculated from the Wnet:
Work of the turbines minus the work of the pumps:
mass flow rate of steam is 72.19 kg/s
