Answer:
Time taken by airplane to cover given distance = 2 hour
Explanation:
Given:
Distance cover by airplane = 1,000 miles
Velocity of airplane = 500 miles per hour
Find:
Time taken by airplane to cover given distance
Computation:
Time taken = Distance / Velocity(Speed)
Time taken by airplane to cover given distance = Distance cover by airplane / Velocity of airplane
Time taken by airplane to cover given distance = 1,000 / 500
Time taken by airplane to cover given distance = 2 hour
Answer:
0.0297M^3/s
W=68.48kW
Explanation:
Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
state 1
X=quality=1
T=-26C
density 1=α1=5.27kg/m^3
entalpy1=h1=234.7KJ/kg
state 2
T2=70
P2=8bar=800kPa
density 2=α2=31.91kg/m^3
entalpy2=h2=306.9KJ/kg
Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.
m1=m2
(Q1)(α1)=(Q2)(α2)

the volumetric flow rate at the exit is 0.0297M^3/s
To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation
W=m(h2-h1)
m=Qα
W=(0.18)(5.27)(306.9-234.7)
W=68.48kW
the compressor power is 68.48kW
Answer:

Explanation:
For close gas turbine:
Gas turbine works on Brayton cycle.Gas turbine have lots of applications like ,it is use in aircraft,in land applications etc.
Reheating is the method to improve the efficiency of the gas turbine.In reheating gas is expanding in two turbine instead of one turbine alone.Two turbine like high pressure turbine and low pressure turbine are used for expansion.
In the above diagram 1-2 is a compressor,2-3 heat addition,3-4 high pressure turbine,4-5 reheating of cycle 5-6 low pressure turbine,6-1 heat rejection,
We know that 
Now take
represent the enthalpy of point 1,2,3,4,5,6 in the cycle respectively.
So total heat supplied
=

Net work out put
=
So efficiency 

Answer:
a) P = 86720 N
b) L = 131.2983 mm
Explanation:
σ = 271 MPa = 271*10⁶ Pa
E = 119 GPa = 119*10⁹ Pa
A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²
a) P = ?
We can apply the equation
σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N
b) L₀ = 131 mm = 0.131 m
We can get ΔL applying the following formula (Hooke's Law):
ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)
⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm
Finally we obtain
L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm
Answer: 1.38g
Explanation:
Width of planting area = 100m
Field size = 6-hectares(14.425 acres)
Weight of 1000 black eye seed = 230g
1 lb = 453.4g
1 black eye seed = 230g/1000 = 0.23g = 0.00023kg
1 hectare = 10,000sq metre
6 hectare = 60,000sq metre
(Weight/Area) kg/m2
0.00023kg / 10,000 = 2.3×10^-8kg/m^2
And field size = 6hectares = 60,000m^2
(2.3×10^-8) × 60,000 = 0.00138kg of black eye seed
0.00138kg × 1000 = 1.38g