A fluid with a specific weight of 78.5 lb/ft³ and dynamic viscosity of 0.638 lbs/s-ft is flowing at a rate of 20 cfs. The section lengths of the pipe and minor loss coefficients for the 90° bends are given below; the pipe is made of galvanized iron. Assuming there are no losses due to the pump connections, what is the required head of the pump? State all minor losses. L₁ = 7 ft L₂ = 6 ft L3 = 3 ft L4 = 4 ft 1 √f -2.0 log10 24 in. P = == L₁ 50 psi L₂ PUMP Solve for an initial friction factor (f) either using the Moody diagram (attach image so I can provide feedback) to achieve an initial guess of f to 3 significant digits, then use that initial guess into the Colebrook equation to determine f to 4 significant digits. You can do calculations by hand or using software (i.e., MATLAB or Excel). Colebrook equation (implicit): €/D 2.51 + 3.7 R√f (turbulent flow) L4 3 ft 24 in. P = 45 psi

Elements Of Electromagnetics
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A fluid with a specific weight of 78.5 lb/ft³ and dynamic viscosity of 0.638 lbs/s-ft is flowing at a
rate of 20 cfs. The section lengths of the pipe and minor loss coefficients for the 90° bends are
given below; the pipe is made of galvanized iron. Assuming there are no losses due to the pump
connections, what is the required head of the pump? State all minor losses.
L₁ = 7 ft
L₂= 6 ft
L3 = 3 ft
L4 = 4 ft
LI
1
√f
24 in.
P = 50 psi
L₂
L3
PUMP
+
Re
=-2.0 log₁0 3.7RNf (turbulent flow)
L4
3 ft
Solve for an initial friction factor (f) either using the Moody diagram (attach image so I can
provide feedback) to achieve an initial guess of f to 3 significant digits, then use that initial guess
into the Colebrook equation to determine f to 4 significant digits. You can do calculations by
hand or using software (i.e., MATLAB or Excel).
Colebrook equation (implicit):
€/D 2.51
24 in.
P = 45 psi
Transcribed Image Text:A fluid with a specific weight of 78.5 lb/ft³ and dynamic viscosity of 0.638 lbs/s-ft is flowing at a rate of 20 cfs. The section lengths of the pipe and minor loss coefficients for the 90° bends are given below; the pipe is made of galvanized iron. Assuming there are no losses due to the pump connections, what is the required head of the pump? State all minor losses. L₁ = 7 ft L₂= 6 ft L3 = 3 ft L4 = 4 ft LI 1 √f 24 in. P = 50 psi L₂ L3 PUMP + Re =-2.0 log₁0 3.7RNf (turbulent flow) L4 3 ft Solve for an initial friction factor (f) either using the Moody diagram (attach image so I can provide feedback) to achieve an initial guess of f to 3 significant digits, then use that initial guess into the Colebrook equation to determine f to 4 significant digits. You can do calculations by hand or using software (i.e., MATLAB or Excel). Colebrook equation (implicit): €/D 2.51 24 in. P = 45 psi
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