
(a)
The position of the maximum fluid accelerating along

Answer to Problem 4.7P
The position of the maximum fluid accelerating along
Explanation of Solution
Given information:
The fluid velocity along the streamline
Write the expression for the acceleration
Here, the acceleration is
Substitute
Differentiate the above expression with respect to
Calculation:
Equate
Substitute
Further simplify the above expression.
Conclusion:
The position of the maximum fluid accelerating along
(b)
The time required to travel for a fluid particle from

Answer to Problem 4.7P
The time required to travel for a fluid particle from
Explanation of Solution
Given information:
The fluid velocity along the streamline
Write the expression for the velocity in terms of displacement.
Here, displacement is
Substitute
Further solve the above expression.
Integrate the above expression.
Here, the initial limit for time is
Substitute
Further solve the above expression.
Thus, for a fixed velocity
Conclusion:
The time required to travel for a fluid particle from
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Chapter 4 Solutions
Fluid Mechanics
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You should measure these anticlockwise from the y-axis. 20 [5 marks] (iii) maximum shear strain in the plane (ymax)? Ex = Ea Ey = εc [5 marks] (epol) (apob) é Ea = A = -210 2 B=E₁ = -50 E₁ = C = 340 D = 45° bril elled ✓A bedivordan nemigas olloho shot on no eonsoup Imeneo alubom shine sail-no viss ieqse sidetiva bnat sabied 2arrow_forward1) Solve and show which is converage or diyverage a = 2+(0.1)" 3 16) a = n 1-2n 2) a = In n 1+2n 17) a = n 1-5n4 3) an = n* +8n³ 18) a =√4"n n² -2n+1 n! 20) a = 4) a₁ = 10 n-1 (Ina) 5) a=1+(-1)" 21) a= 6) a 7) an = * = (12+) (1-1) 2n (-1)+1 2n-1 3n+1 22) a= 3n-1 x" 23) a= .x>0 2n+1 2n 3"x6" 8) a = 24) a = n+1 π 9) a = sin 2 sin n 10) an = n + 2 x n! 25) a = tanh(n) n² 1 26) a = -sin- 2n-1 27) a = tan(n) n n 11) a = 2" 12) a = n 13) a = 8/ +=(1+2)" 14) a = 15) a = √10n In(n+1) 29) a = n 30) an-√n²-1 1 28) a = + √2" (In n)200 n 31) a=- = 1 dx nixarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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