MET330 20xx0xE Applied Fluid MechanicsInstructor: Instructors nameWeek 2 Review Assignment - 1I pledge to support the Honor System of ECPI. I will refrain from any form of...

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Answered 1 days AfterJan 30, 2023

Answer To: MET330 20xx0xE Applied Fluid MechanicsInstructor: Instructors nameWeek 2 Review...

Dr Shweta answered on Jan 31 2023
46 Votes
Ans 1.
Known values: sg = 0.90, flow rate = 2.35 × 10−3 m3/s.
Governing equations: ρoil = Sg * ρw
Cal
culations: ρoil = 0.9 * 1000 Kg/m3
Mass flow rate of oil is
m* = ρoil * V = 900 * 2.35 * 10-3 = 2.115 Kg/s
Weight flow rate of oil = W = m*g = 20115 *9.8 = 20.75N/s.
Ans 2.
Known values: flow rate 2000 L/min, inside diameter of 300mm, later reduces to 150mm diameter.
Governing equations: A1. V1=A2.V2=2m3/min
Calculations:
3.14*0.3*0.3/4*V1=2/60
V1=0.47m/s
similarly
V2=1.88m/s
Ans 3.
Known values: 2.80 L/min of oil = 4.66 * 10-5m3/s, maximum velocity of 0.30 m/s.
Governing equations: Q=VA
Calculations: flow area=
A = 4.666 * 10-5m3/s/0.3 = 1.55 * 10-4m2
A = π/4Di2
π/4Di2 = 1.55 * 10-4m2
Di2 = 4 * 1.55 * 10-4m2/ 3.14
Di = 0.014 m or 14 mm
From the standard table:
D0 = 20 mm
t = 1.2 mm
Di =17.6 mm
Flow area (A) = 2.433 * 10-4m2
Ans 4. Known values: T = 10 ͦ C, rate = 0.37 m3/s, pressure at A = 66.2 kPa, density of water at 10 ͦ C =999.77 kg/m3, dB = 0.6 m, dA = 0.3 m, hB = 4.5 m
Governing equations:...
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