how to find p , power required by motor ?

Question

if t= 7.3 sec , Gv= 1125 kg , δa= 1.088, g= 9.8 m/s , vb = 14.78 m/s , Coefficient of Friction= 1 , Density of Air= 1.2kg/m3, Frontal Area (Ac)= 2.42 , Aerodynamic Drag (Cd)= 0.36 , vf= 27.7 m/s , find p , power required by motor ? ​

Answer 1

The first thing to note is that the variable v does not appear in either of the integrands, so integration is simple---it's just the difference between the limits.

t=k₁V_b-k₂(V_f-V_b), where k₁ is the first fraction (containing power P as a constant) and k₂ is the second fraction (also containing P as a constant).

We should be able to calculate the quantity in the large fractions (effectively k₁ and k₂) by substituting the given values (apart from P which is to be found).

Start with the numerators: let A=G_vδ_a/g=1125×1.088/9.8=124.898 approx.

I assume that P in the denominator is the required pressure to be found.

G_vμ_f=1125 because μ_f is given as 1.

Let B=½ρC_dA_c=½×1.2×0.36×2.42=0.52272.

V_b²=218.4484; V_f²=767.29.

The constants can be taken outside the integrals:

7.3=A/(P/14.78-1125-218.4484B)∫[0,14.78]dv-

       A/(P/27.7-1125-767.29B)∫[14.78,27.7]dv.

(The limits are parenthesised in square brackets [low,high].)

After integration:

7.3/A=1/(P/14.78-1125-218.4484B)(14.78)-1/(P/27.7-1125-767.29B)(27.7-14.78),

7.3/A=14.78/(P/14.78-1125-218.4484B)-12.92/(P/27.7-1125-767.29B).

Let C=1125+218.4484B=1239.19 approx and D=1125+767.29B=1526.08 approx.

Let a=7.3/A=14.78/(P/14.78-C)-12.92/(P/27.7-D),

a=218.4484/(P-14.78C)-357.884/(P-27.7D),

a(P-14.78C)(P-27.7D)=218.4484(P-27.7D)-357.884(P-14.78C),

aP²-aP(14.78C+27.7D)+409.406aCD=-139.4356P-6051.02068D+5289.52552C,

aP²-P(14.78aC+27.7aD-139.4356)+409.406aCD+6051.02068D-5289.52552C=0.

In the quadratic aP²+bP+c=0 (we can expect up to two solutions):

a=0.05845 approx;

b=-(14.78aC+27.7aD-139.4356)=-3401.77 approx;

c=409.406aCD+6051.02068D-5289.52552C=47931365 approx, and:

P=(-b±√(b²-4ac)/(2a).

P=34,277 or 23,925 approx (calculations should be checked).