Question

Vent 10 ft Fully open globe valve (KL = 10) TT 6 ft 1500 ft with 15 (including the one shown) 90° elbows (K1 = 1.5) For the water tower shown above, the pressure at section (2) should not fall be- low 50 psi. The top of the water tower is vented to the open air, and all pipes are smooth 6-in. diameter plastic (€/D –0). 4.1 (4 pts) For a pipe of this size, what is the maximum discharge that can be achieved while still maintaining laminar flow. (ANS) 4.2 (7 pts) For this maximum laminar flowrate, determine whether or not the pres- sure at section (2) stays above the critical pressure of 50 psi if the height of the water tower is h = 50 ft. IncluDe only major losses. (ANS) 4.3 (9 pts) For a flowrate of 0.95 cfs, determine the minimum height, h, of the water tower that is required to maintain the presSure of 50 psi at section (2). Include both major and minor losses. (ANS)

Answer 1

Ans 4.1) We know, Reynold number (Re) = V D / $\nu$

where, V = Flow velocity

D = Pipe diameter = 6 in or 0.5 ft

  $\nu$ = Kinematic viscosity of water = 1.13 x $10^{-5}$ ft2/s

Also, for Laminar flow, Re $\leq$ 2000, so using limiting case of Re = 2000

=> 2000 = V (0.5) / 1.13 x $10^{-5}$

=> V = 0.0452 ft/s

Also, discharge Q = Area x velocity

=> Q = = ($\pi$/4)$(0.5)^2$ x 0.0452

=> Q = 0.0088 cfs

Hence, maximum discharge while maintaining laminar flow is 0.0088 cfs

Ans 4.2) Apply Bernoulli equation between point 1 located at water surface elevation of tank and point 2 respectively,

P1/$\gamma$ + $V1^2$ / 2 g + Z1 = P2/$\gamma$ + $V2^2$ / 2 g + Z2 + Hf

Since, point 1 is open to atmosphere, Pressure is only atmospheric hence gauge pressure P1 = 0

Velocity at surface is negligible so V1 = 0

Hf is fricton head loss (major loss)

Elevation assuming point 2 as datum, Z1 = h + 10 + 6 = 50 + 16 = 66 ft and Z2 = 0

Hence, above equation reduces to,

66 = P2/$\gamma$ + $V2^2$ / 2 g + Hf

According to above part , Velocity = 0.0452 ft/s

Since, velocity is so small, neglect velocity head,

=> 66 = P2/$\gamma$ + Hf

  Also, Hf = f L $V^2$ / (2 g D )

where, f = friction factor

L = pipe length = 1500 ft

D = Pipe diameter = 6 in or 0.50 ft

According to Moody disgram,for Re = 2000 and e/D = 0 , friction factor = 0.049

=> Hf = (0.049)(1500)$(0.0452)^2$ / (2 x 32.2 x 0.5 )

=> Hf = 0.0046 ft

Putting values in equation 1,

66 = P2/$\gamma$ + 0.0046

=> P2/$\gamma$ = 65.99 ft

=> P2 = 62.4 x 65.99 = 4118 psf or 28.6 psi < 50 psi

Since, pressure at point 2 does not stays more than critical pressure of 50 psi

Ans 4.3) Again, apply Bernoulli equation between point 1 located at water surface elevation of tank and point 2 respectively,

P1/$\gamma$ + $V1^2$ / 2 g + Z1 = P2/$\gamma$ + $V2^2$ / 2 g + Z2 + Hf + Hm

Since, point 1 is open to atmosphere, Pressure is only atmospheric hence gauge pressure P1 =0

Required Pressure at point 2, P2 = 50 psi or 7200 psf

Velocity at surface is negligible so V1 = 0

Hf is fricton head loss (major loss)

Hm is minor loss

Elevation assuming point 2 as datum, Z1 = h + 10 + 6 = (h + 16) ft and Z2 = 0

Hence, above equation reduces to,

h + 16 = P2/$\gamma$ + $V2^2$ / 2 g + Hf + Hm..............................(2)

Given, Q = 0.95 cfs

=> Velocity at point 2, $V_2$ = Q / A = 0.95 / [ ($\pi$/4)$(0.5)^2$] = 4.84 ft/s

=> Re = V D /$\nu$ = 4.84 x 0.5 / (1.13 x $10^{-5}$ ) = 214160

According to Moody disgram,for Re = 214160 and e/D = 0 , friction factor = 0.015

  Also, Hf = f L $V^2$ / (2 g D )

=> Hf = (0.015)(1500)$(4.84)^2$ / (2 x 32.2 x 0.5 )

=> Hf = 16.37 ft

Now,

Hm =  $\sum$ K $V^2$ / 2 g

where, $\sum$ K = Sum of loss coefficients due to bends and valves

=> $\sum$ K = 15(1.5) + 10 = 32.5

=> Hm = 32.5 $(4.84)^2$ / (2 x 32.2)

=> Hm = 11.82 ft

  Putting values in equation 2,

=> h + 16 = [(7200)/62.4] +  [$(4.84)^2$ / (2 x 32.2)] + 16.37 + 11.82

=> h + 16 = 143.94

=> h = 127.94 ft $\approx$ 128 ft

Hence, minimum required height of water tower (h) = 128 ft