166. A large hydraulic turbine is to generate 300 kW at 1000 rpm under a head of 40 m. For initial testing, a 1: 4 scale model of the turbine operates under a head of 10 m. The power generated by the model (in kW)will be

1. 1. 2.34

   
   
   

   2. 4.68

   
   
   

   3. 9.38

   
   
   

   4. 18.75

   
   
   

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1. View Hint View Answer Discuss in Forum

   PM = Pp	=		Nm		3		Dm		5
   			Np				Dp

   
   

   = 300		2000		3		1		5	= 2.34 kW
   		1000				5

   
   

   Correct Option: A

   PM = Pp	=		Nm		3		Dm		5
   			Np				Dp

   
   

   = 300		2000		3		1		5	= 2.34 kW
   		1000				5

   
   

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167. The head loss for a laminar incompressible flow through a horizontal circular pipe is h₁. Pipe length and fluid remaining the same, if the average flow velocity doubles and the pipe diameter reduces to half its previous value, the head loss is h₂. The ratio h₂ /h₁. is

1. 1. 1

   
   
   

   2. 4

   
   
   

   3. 8

   
   
   

   4. 16

   
   
   

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1. View Hint View Answer Discuss in Forum

   Head loss, h ∝	uavg
   	D2

   
   

   ∴	h2	=		D1		2	uavg,2	= 22 = 2
   	h1			D2			uavg,1

   
   

   h2	= 8
   h1

   Correct Option: C

   Head loss, h ∝	uavg
   	D2

   
   

   ∴	h2	=		D1		2	uavg,2	= 22 = 2
   	h1			D2			uavg,1

   
   

   h2	= 8
   h1

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168. Three parallel pipes connected at the two ends have flow-rates Q₁, Q₂ and Q₃ respectively, and the corresponding frictional head losses are h_L1, h_L2 and h_L3 respectively. The correct expressions for total flow rate (Q) and frictional head loss across the two ends (h_L) are

1. 1. Q = Q₁ + Q₂ + Q₃; h_L = h_L1 + h_L2 + h_L3

   
   
   

   2. Q = Q₁ + Q₂ + Q₃; h_L = h_L1 = h_L2 = h_L3

   
   
   

   3. Q = Q₁ = Q₂ = Q₃; h_L = h_L1 + h_L2 + h_L3

   
   
   

   4. Q = Q₁ = Q₂ =Q₃; h_L = h_L1 = h_L2 = h_L3

   
   
   

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1. View Hint View Answer Discuss in Forum

   Total flow rate Q = Q₁ + Q₂ + Q₃
   Head loss h = h_L1 = h_L2 = h_L3

   Correct Option: B

   Total flow rate Q = Q₁ + Q₂ + Q₃
   Head loss h = h_L1 = h_L2 = h_L3

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169. 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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170. 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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