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Solution of Problem 1. Assumption:

1) The flow is uniform and pressure distribution is due to hydrostatic before and after the jump. 2) Losses due to the friction on the surface of the bed of the channel are small and neglected. 3) Slope of the bed of the channel is small, so that the component of the weight of the fluid in the direction of the fluid is negligible. 1

2

1

2

The CV cuts through sections 1 and 2 and surrounds the jump, as shown. Wall shear is neglected. Since there are no obstacles. The only forces are due to hydrostatic pressure, considering unit width channel, forces acting on the mass of water between sections 1-1 and 2-2 From conservation of mass:

=> => =>

q = (h1*V1 ) = (h2*V2 ) ρ*(h1*1)*V1 = ρ*(h2*1)*V2 V2 = V1 (h1/ h2)

1) Pressure forces Pf1 on channel section 1-1 and Pf2 on channel section 2-2.

Pf1 = ρ*g*h1*(1* h1/2) and Pf2 = ρ*g*h2*(1* h2/2) so the net forces acting on the mass of the water is = Pf2-Pf1

= ρ*g (h2^2- h1^2)/2 But from momentum principle,

net force acting on the mass of fluid must be equal to rate of change of momentum in the same section, assuming 'q' as discharge per unit width then,

=>ρ*g (h2^2- h1^2)/2 = ρ*g * (q*1) *(V1- V2) = ρ*g * (q*1)*((q/ h1)-(q/ h2)) =>(h2+ h1) = 2*q2 /( h1 h2) , => h2^2+ h1 h2-(2q^2/(g* h1)) = 0 => h2 = (-h1±sqrt(h1^2-(4*1*(-2*q^2/(g* h1)))))/2

h2/ h1 = -0.5±0.5*sqrt (1+(8* V1^2/(g* h1)))

Solution of problem 2: Given: D1 = 27cm & D2 = 13 cm. and 1 gal = 3600 cm^3 Q = AV =4000*3600/10^6 (m^3/min) = (14.4/60) (m^3/sec) Q = 4000 gal/min = 0.24 m^3/sec.

Required to Find: Torque at point 'B'' Solution:

Q = A1*V1 = A2*V2

=> 0.25*pi* (0.27)^2*V1 = 0.24 => V1 = 4.2 m/sec and similarly V2 = 18.1 m/sec Rate of change of momentum = Force = m_dot*Velocity + v_dot*m => m_dot = rho * A2* V2 = 1000*0.25*pi*0.13^2*18.1 => m_dot = 240.23 kg/sec.

Ans. = TB = m_dot*r2*V2 =240.23*0.5*18.1 = 2.2 kN-m

Solution of problem 3:

CV

y

Co-ordinates considered as

x

V0 = (Q/2A), is the exit velocity relative to the pipe walls V1 =V0cos(ϴ) V2 = V0 sin(ϴ) & V3 = ΩR0 Vupper = V1.i + V2.j + V3 .k Vlower = V1.i - V2.j - V3 .k Then the moments about the x-axis are related to angular momentum is given by, Mx = T.i = (ρQ/2)Vupper × R0 .j + (ρQ/2). Vlower × (-R0 .j) - (ρQ . Vin . Rin ) T.i = (ρQ/2)( V1 R0.k + V3. R0.(-i) ) + (ρQ/2).( -V1 R0.k + V3. R0.(-i) ) = (ρQ) (V3. R0.(-i)) = (ρQ) ( ΩR0. R0.(-i)) = -ρQΩ(R0^2)(i)

T = -ρQΩ(R0^2)

Solution of problem 4:

From Continuity equation; ρ Vin hin = ρ Vout hout Vout = (hin/ hout) Vin Momentum force in x-direction FX = Body Force + Surface force - Restoring force (F) ............................ (1) Surface force = (pressure + viscous) force Assumption: neglecting viscous force and body force Pressure force = ρ g h1 (b*h1/2) - ρ g h2 (b*h2/2)

................ (2)

Momentum equation in x-direction FX = Vin ρ(-Vin) h1 b + Vout ρ(Vout) h2 b .... (3) F = ρ g h1 (b*h1/2) - ρ g h2 (b*h2/2) –( Vin ρ(-Vin) h1 b + Vout ρ(Vout) h2 b) F/b = 0.5 ρ g (h1^2- h2^2) + ρ(h1Vin^2 – h2Vout^2) Where,

(F/b) is the reaction force on the gate in the x-direction per unit width of the gate. To eliminate the velocities from this relation we use Bernoulli’s equation PA + 0.5 ρ Vin^2 + ρ g h1 = PA + 0.5 ρ Vout^2 + ρ g h2 And

Vout = (hin / hout) Vin

 Vin^2 = (2 g h2^2/(h2+ h1))

 F/b = 0.5 ρ g (h1- h2)^3 / (h2+ h1) ......Ans.