Answer:
, 
Explanation:
The drag force is equal to:
Where 
is the drag coefficient and 
is the frontal area, respectively. The work loss due to drag forces is:
The reduction on amount of fuel is associated with the reduction in work loss:
Where 
and 
are the original and the reduced frontal areas, respectively.
The change is work loss in a year is:
![\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)](https://tex.z-dn.net/?f=%5CDelta%20W%20%3D%20%280.3%29%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Ccdot%20%281.20%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%29%5Ccdot%20%2827.778%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%29%5E%7B2%7D%5Ccdot%20%5B%281.85%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%20-%20%281.50%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%5D%5Ccdot%20%2825%5Ctimes%2010%5E%7B6%7D%5C%2Cm%29)
The change in chemical energy from gasoline is:
The changes in gasoline consumption is:
Lastly, the money saved is:
Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
Answer:
COP of the heat pump is 3.013
OP of the cycle is 1.124
Explanation:
W = Q₂ - Q₁
Given
a)
Q₂ = Q₁ + W
= 15 + 7.45
= 22.45 kw
COP = Q₂ / W = 22.45 / 7.45 = 3.013
b)
Q₂ = 15 x 1.055 = 15.825 kw
therefore,
Q₁ = Q₂ - W
Q₁ = 15.825 - 7.45 = 8.375
∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124
Answer:
Temperature
Explanation:
In an ideal gas the specific enthalpy is exclusively a function of Temperature only this can be also written as h = h(T)
A gas is said be ideal gas if obeys PV= nRT law
And in a ideal gas both internal energy and specific enthalpy are a function of Temperature only. Therefore the constant volume and constant pressure specific heats Cv and Cp are also function of temperature only.
