101. Consider fully developed flow in a circular pipe with negligible entrance length effects. Assuming the mass flow rate, density and friction factor to be constant, if the length of the pipe is doubled and the diameter is halved, the head loss due to friction will increase by a factor of

1. 1. 4

   
   
   

   2. 16

   
   
   

   3. 32

   
   
   

   4. 64

   
   
   

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1. View Hint View Answer Discuss in Forum

   Head loss =	fLv2	=	fl(Q/A)2
   	2gd		2gd

   
   

   ⇒  Head loss ∝	L
   	d5

   
   

   h1* ∝	L
   	d5

   
   

   h2* ∝	2.L	=	L	× 64
   	(d/2)5		d5

   
   

   h1	= 1/64 ⇒	h2	= 64
   h2		h1

   Correct Option: D

   Head loss =	fLv2	=	fl(Q/A)2
   	2gd		2gd

   
   

   ⇒  Head loss ∝	L
   	d5

   
   

   h1* ∝	L
   	d5

   
   

   h2* ∝	2.L	=	L	× 64
   	(d/2)5		d5

   
   

   h1	= 1/64 ⇒	h2	= 64
   h2		h1

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102. Water is flowing through a horizontal pipe of constant diameter and the flow is laminar. If the diameter of the pipe is increased by 50% keeping the volume flow rate constant, then the pressure drop in the pipe due to friction will decrease by

1. 1. 33%

   
   
   

   2. 50%

   
   
   

   3. 70%

   
   
   

   4. 80%

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∆ P =	32μUL
   	∆²

   
   

   Q = AV =	π	d²V
   	4

   
   

   V =	4Q
   	πd²

   
   

   ∆P =	128μQL
   	πD4

   
   Above equation shows that
   

   ∆P ∝	1
   	D4

   
   

   ∆P1 =	C	=	C	where D1 = D
   	D41		D4

   
   

   & ∆P2 =	C	=	C	where D2 = 1.5D
   	D42		(d42)

   
   % change in pressure drop
   

   % =	∆P1 - ∆P2	× 100
   	∆P1

   
   = 80.24% ≃ 80%

   Correct Option: D

   ∆ P =	32μUL
   	∆²

   
   

   Q = AV =	π	d²V
   	4

   
   

   V =	4Q
   	πd²

   
   

   ∆P =	128μQL
   	πD4

   
   Above equation shows that
   

   ∆P ∝	1
   	D4

   
   

   ∆P1 =	C	=	C	where D1 = D
   	D41		D4

   
   

   & ∆P2 =	C	=	C	where D2 = 1.5D
   	D42		(d42)

   
   % change in pressure drop
   

   % =	∆P1 - ∆P2	× 100
   	∆P1

   
   = 80.24% ≃ 80%

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103. The maximum velocity of a one-dimensional incompressible fully developed viscous flow, between two fixed parallel plates, is 6 ms-1. The mean velocity (in ms-1) of the flow is

1. 1. 2

   
   
   

   2. 3

   
   
   

   3. 4

   
   
   

   4. 5

   
   
   

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1. View Hint View Answer Discuss in Forum

   For fully developed incompressible viscous flow,
   

   Vmean =	2	Vmax
   	3

   
   

   =	2	× 6 = 4 m/s
   	3

   Correct Option: C

   For fully developed incompressible viscous flow,
   

   Vmean =	2	Vmax
   	3

   
   

   =	2	× 6 = 4 m/s
   	3

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104. The pressure drop for laminar flow of a liquid in a smooth pipe at normal temperature and pressure is

1. 1. directly proportional to density

   
   
   

   2. inversely proportional to density

   
   
   

   3. Independent of density

   
   
   

   4. proportional to (density)^0.75

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∆ P =	32μUL
   	D²

   
   Above equation shows that ∆P is independent of velocity.

   Correct Option: C

   ∆ P =	32μUL
   	D²

   
   Above equation shows that ∆P is independent of velocity.

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105. The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, is given by the expression
     

     u(r) = −	R²		∂p			1 −	r²
     	4μ		∂x				R²

     
     where ∂p/∂x is a constant
     The average velocity of fluid in the pipe is
     

1. 1. −	R²		∂p
      	8μ		∂x

   
   
   

   2. −	R²		∂p
      	4μ		∂x

   
   
   

   3. −	R²		∂p
      	2μ		∂x

   
   
   

   4. −	R²		∂p
      	μ		∂x

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   u = −	R²		−	dp			1 −	r²
   	4μ			dx				R

   
   dQ =u (2πrdr)
   
   Solving, we get
   

   Q = −	πR²		−	dp
   	8μ			dx

   
   Now Q = Area ´ average velocity
   

   ∴  uav =	Q	=	Q
   	Area		πR2

   
   

   or   uav =	R²		dp
   	8μ		dx

   Correct Option: A

   
   

   u = −	R²		−	dp			1 −	r²
   	4μ			dx				R

   
   dQ =u (2πrdr)
   
   Solving, we get
   

   Q = −	πR²		−	dp
   	8μ			dx

   
   Now Q = Area ´ average velocity
   

   ∴  uav =	Q	=	Q
   	Area		πR2

   
   

   or   uav =	R²		dp
   	8μ		dx

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