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Water flows through a 2.5 cm pipe at115 L/minand o.15 bar pressure. It vents vertically 1 mabove the pipe. How high will the water shootfrom the vent?

Q4/ In Fig. all pipes are 8-cm-diameter cast iron. Determine the flow rate from reservoir (1) if valve C is (a) closed; and (b) open, with Kvalve = 0.5. Oz= 25 m 2 z = 0m Water 10 m at 20°C L= 50 m B. L= 100 m 30 m L= 70 m Valve

(b) Use Bernoulli's energy balance equation (E = PV + 0.5*mU²), assume volume flow rate remains constant along the pipe such that (Q=UA), and assume that pressure is balanced manometer to perform the following derivations: (i) (ii) (iii) 1- P2 in the pipe (substitute U2 in this step). an expression for P₁ an expression for P₁ P2 in the U-tube manometer. Set the expressions in parts (i) and (ii) equal to one another and derive the flow equation: U₁ = 2 (Pm -1) gh BA - 1

A pump receives water from a tank. If the pump needs 24 m of NPSHA, how high must the water level be above the pump centerline if the water is saturated at 1724 kPag? The friction loss in the pipe averages 0.3 m for each 6 m of height and flow is 23 lps in a 15.24 cm ID pipe.

A horizontal 12-in pipe contracts to a 6-in diameter. The flow is 4.5 ft3 /s ofoil, s.g. = 0.83. If the pressure in the larger pipe is 42 psi and the pressure in thesmaller pipe is 38 psi, what resultant force is exerted on the contraction? Reportthe answer in lbs.

Water at 25°C is pumped at 1800 gal/min from the lower to upper reservoir, as shown in Figure below (Zı = 30 ft and z2 = 180 ft). Pipe friction losses are approximated by hf = 35V/2g, where V is the average velocity in the pipe. The pump is 80% efficient. (a) Find the head loss due to friction, hẹ in ft. in horse power (b) Compute the pump head, h, in ft and the pump power needed, Ppump (hp). Given: Specific weight of water, ywater = 62.4 lb/ft'; 1 ft/s = 448.8 gal/min. Show all calculations. Z2 = 180 ft Z¡ = 30 ft D= 6 in Pump

A necked-down section in a pipe flow, called a venturi,develops a low throat pressure that can aspirate fluidupward from a reservoir, as in Fig. UsingBernoulli’s equation with no losses, derive an expressionfor the velocity V 1 that is just sufficient to bring the reservoirfl uid into the throat.