156. Consider steady, incompressible and irrotational flow through a reducer in a horizontal pipe where the diameter is reduced from 20 cm to 10 cm. The pressure in the 20 cm pipe just upstream of the reducer is 150 kPa. The fluid has a vapour pressure of 50 kPa and a specific weight of 5 kN / m³ . Neglecting frictional effects, the maximum discharge (in m³ / s) that can pass through the reducer without causing cavitation is
     

1. 1. 0.05
      

   
   
   

   2. 0.16
      

   
   
   

   3. 0.27
      

   
   
   

   4. 0.38

   
   
   

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1. View Hint View Answer Discuss in Forum

   Considering potential head difference = 0,
   i.e z₁ = z₂
   Apply Bernoulli’s theorem
   

   	p	+	v2	= C
   	ρg		2g

   
   
   

   	p1	+	v1²	=	p2	+	v2²
   	w1		2g		w2		2g

   
   But w₁ = w₂ = w = 5 (incompressible flow)
   

   ∴	150	+	v1²	=	50	+	v2²
   	5		2g		5		2g

   
   

   or	v2² - v1²	=	150 - 50
   	2g		5

   
   

   or	v2² - v1²	= 20 m ....(1)
   	2g

   
   

   Also, discharge, Q =		π	d1²		v1 =		π	d2²		v2
   		4					4

   
   

   or	v1	=		d2		²	=		10		²
   	v2			d1					20

   
   

   or v1 =	v2	......(2)
   	4

   
   From equations (1) and (2)
   

   		v2² -	v2²
   			16		= 20
   		2g

   
   

   or	15v2²	= 20
   	32g

   
   or v₂ = √(20 × 32g) / 15 = 20.45 m /s
   

   ∴ Discharge , Q =		π	d22		v2
   		4

   
   

   =	π	(0.1)2 × 20.45 = 0.16 m3 /sec
   	4

   
   

   Correct Option: B

   Considering potential head difference = 0,
   i.e z₁ = z₂
   Apply Bernoulli’s theorem
   

   	p	+	v2	= C
   	ρg		2g

   
   
   

   	p1	+	v1²	=	p2	+	v2²
   	w1		2g		w2		2g

   
   But w₁ = w₂ = w = 5 (incompressible flow)
   

   ∴	150	+	v1²	=	50	+	v2²
   	5		2g		5		2g

   
   

   or	v2² - v1²	=	150 - 50
   	2g		5

   
   

   or	v2² - v1²	= 20 m ....(1)
   	2g

   
   

   Also, discharge, Q =		π	d1²		v1 =		π	d2²		v2
   		4					4

   
   

   or	v1	=		d2		²	=		10		²
   	v2			d1					20

   
   

   or v1 =	v2	......(2)
   	4

   
   From equations (1) and (2)
   

   		v2² -	v2²
   			16		= 20
   		2g

   
   

   or	15v2²	= 20
   	32g

   
   or v₂ = √(20 × 32g) / 15 = 20.45 m /s
   

   ∴ Discharge , Q =		π	d22		v2
   		4

   
   

   =	π	(0.1)2 × 20.45 = 0.16 m3 /sec
   	4

   
   

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157. The following data about the flow of liquid was observed in a continuous Chemical process plant:
     
     Mean flow rate of the liquid is
     

1. 1. 8.00 liters/s
      

   
   
   

   2. 8.06 liters/s
      

   
   
   

   3. 8.16 liters/s
      

   
   
   

   4. 8.26 liters/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   ∴ Mean flow rate =	∑fx	=	652.8	= 8.16
   	∑f		80

   Correct Option: C

   
   

   ∴ Mean flow rate =	∑fx	=	652.8	= 8.16
   	∑f		80

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158. Navier Stoke's equation represents the conservation of
     

1. 1. Energy
      

   
   
   

   2. Mass
      

   
   
   

   3. Pressure
      

   
   
   

   4. Momentum

   
   
   

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1. View Hint View Answer Discuss in Forum

   Momentum

   Correct Option: D

   Momentum

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159. Bernoulli's equation can be applied between any two points on a stream line for a rotational flow field. State:

1. 1. TRUE

   
   
   

   2. FALSE

   
   
   

   3. A and B

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   TRUE

   Correct Option: A

   TRUE

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160. The radial component of the fluid acceleration at r = R is
     

1. 1. 	3V2R
      	4h2

      
      

   
   
   

   2. 	V2R
      	4h2

      
      

   
   
   

   3. 	V2R
      	2h2

      
      

   
   
   

   4. 	V2h
      	2R2

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Radial acceleration
   

   ar = Vr ×	∂Vr	+	∂Vr
   	∂r		∂t

   
   

   ar =	V.r	×	∂		V.r		+	∂		V.r
   	2h		∂r		2h			∂t		2h

   
   

   -∂h	= V
   ∂t

   
   

   ∴ ar =	V.r	×	V.r	+	V.r	×		-1	∂h
   	2h		2h		2			h2	∂t

   
   

   ∴ ar =	V2r	+	2V2r
   	4h2		4h2

   
   

   ar =	3V2r
   	4h2

   
   

   Correct Option: A

   Radial acceleration
   

   ar = Vr ×	∂Vr	+	∂Vr
   	∂r		∂t

   
   

   ar =	V.r	×	∂		V.r		+	∂		V.r
   	2h		∂r		2h			∂t		2h

   
   

   -∂h	= V
   ∂t

   
   

   ∴ ar =	V.r	×	V.r	+	V.r	×		-1	∂h
   	2h		2h		2			h2	∂t

   
   

   ∴ ar =	V2r	+	2V2r
   	4h2		4h2

   
   

   ar =	3V2r
   	4h2

   
   

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