Solve for me branch d and e (d) The force, F, of a turbine generator is a function of density p, area A and velocity v. ByassumingF = αρ Αθνθ,and dimensional homogeneity, find a, b and c and express F in terms of p, A and v. (a, a, b and care real numbers). Make the following assumptions to determine the dimensionless parameter:F = 1 k N if the scalar values of pAv= 1milli.(e) The dynamic coefficient of viscosityµ (viscosity of a fluid) is found from the formula:µAvF =Fis the force exerted on the liquid, A is the cross sectional area of the path, v is the fluid velocityand l is the distance travelled by the fluid. Using dimensional analysis techniques, determine theequation that governs u and its dimensions using the results of (b) and the equation in c, clearlyshowing all steps in the dimensional analysis. Task 1(a) A low voltage transformer manufacturing line has a purchase order of 1700 piece. If themanufacturing line production rate is 22 piece per minute use dimensional analysistechniques to determine the time taken to produce the requested transformers to thenearest minute in hours and minutes.(b) Assume that the production time of a transformer can be given by the following formula:1t = Gmav-312Where m is the mass of the transformer in kg, v is the velocity of the manufacturing linein m/s and l is the length of the production line in meters. Find the dimensions of G.(c) In b, if m = 220 grams, 1= 10.2 m, v=2.3 m/s and time is 30.1 ms. Find G?

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Solve for me branch d and e

(d) The force, F, of a turbine generator is a function of density p, area A and velocity v. By
assuming
F = αρ Αθνθ,
and dimensional homogeneity, find a, b and c and express F in terms of p, A and v. (a, a, b and c
are real numbers). Make the following assumptions to determine the dimensionless parameter:
F = 1 k N if the scalar values of pAv= 1milli.
(e) The dynamic coefficient of viscosityµ (viscosity of a fluid) is found from the formula:
µAv
F =
Fis the force exerted on the liquid, A is the cross sectional area of the path, v is the fluid velocity
and l is the distance travelled by the fluid. Using dimensional analysis techniques, determine the
equation that governs u and its dimensions using the results of (b) and the equation in c, clearly
showing all steps in the dimensional analysis.

Task 1
(a) A low voltage transformer manufacturing line has a purchase order of 1700 piece. If the
manufacturing line production rate is 22 piece per minute use dimensional analysis
techniques to determine the time taken to produce the requested transformers to the
nearest minute in hours and minutes.
(b) Assume that the production time of a transformer can be given by the following formula:
1
t = Gmav-312
Where m is the mass of the transformer in kg, v is the velocity of the manufacturing line
in m/s and l is the length of the production line in meters. Find the dimensions of G.
(c) In b, if m = 220 grams, 1= 10.2 m, v=2.3 m/s and time is 30.1 ms. Find G?

Expert Solution

Step 1

Dimensional Analysis

Dimensional Analysis is the study of dimensions of various complex quantities and represents them in terms of simpler dimensions.

There are two types of dimensions in Physics:

Primary Dimension: A primary dimension is a dimension that is independent of all other dimensions. Large dimensions are always represented in terms of primary dimensions. Mass, length, time, current, luminous intensity and temperature are primary dimensions.
Secondary Dimension: A secondary dimension is a dimension that is represented in terms of primary dimensions. Example: Force, work, electric field, charge, etc.

Step 2

d) In the given question the force is represented as

$F = α ρ^{a} A^{b} v^{c}$

$α$ is a dimensionless constant, $ρ$ is the density, $A$ is the area and $v$ is the velocity.

The dimension of density $ρ = [ρ] = M L^{- 3}$

The dimension of the area $A = [A] = L^{2}$

The dimension of the velocity $v = [v] = L T^{- 1}$

The dimension of force $F = [F] = M L T^{- 2}$

Thus we have

$[F] = [α] [ρ^{a}] [A^{b}] [v^{c}] \Rightarrow M L T^{- 2} = {(M L^{- 3})}^{a} {(L^{2})}^{b} {(L T^{- 1})}^{c} \Rightarrow M L T^{- 2} = M^{a} L^{- 3 a} L^{2 b} L^{c} T^{- c} \Rightarrow M L T^{- 2} = M^{a} L^{- 3 a + 2 b + c} T^{- c}$

Comparing both the sides we get three equations

$a = 1 c = 2 - 3 a + 2 b + c = 1$

Substituting the value of $a$ and $c$ in the third equation

$- 3 + 2 b + 2 = 1 \Rightarrow b = \frac{2}{2} = 1$

Thus we get the force

$F = α ρ^{a} A^{b} v^{c} = α ρ A v^{2}$

Now to find the value of $α$ we use the condition given in the question that $F = 1 k N$ when all the values of $ρ A v = 1 m i l l i$ . This gives

$1 \times 10^{3} N = α \times 1 \times 10^{- 3} k g m^{- 3} \times 1 \times 10^{- 6} m^{2} \times 1 \times 10^{- 6} m^{2} s^{- 2} \Rightarrow α = \frac{10^{3}}{10^{- 3} \times 10^{- 6} \times 10^{- 6}} = 10^{18}$

Thus the force is given as

$F = 10^{18} ρ A v^{2}$

e) The force exerted on the fluid

$F = \frac{μ A v}{l}$

Thus from the above expression, the coefficient of viscosity is

$μ = \frac{F l}{A v}$

Taking dimensions on both the sides

$[μ] = \frac{[F] [l]}{[A] [v]}$

This is the dimensional equation for the coefficient of viscosity. From the dimensional equation

$[μ] = \frac{M L T^{- 2} \times L}{L^{2} \times L T^{- 1}} = M L^{1 + 1 - 2 - 1} T^{- 2 + 1} = M L^{- 1} T^{- 1}$

This is the dimension of the coefficient of viscosity.

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