Q3. A stream of cold fluid is heated from 70F to 150F by contacting with a hot c
ID: 1718786 • Letter: Q
Question
Q3. A stream of cold fluid is heated from 70F to 150F by contacting with a hot counter flow double pipe heat exchanger. Cold fluid is in the inner pipe and hot fluid is in the annular region. No phase change occurs in any fluid stream. Estimate the length of the heat exchanger needed using the following data:
(30pts)
Cold Fluid (Properties at mean bulk temperature)
Mass flow rate = 1200 Ib/hr Viscosity = 2.4191 Ib/ft-hr
Density = 60 Ib/ft^3 Cp = 0.8 Btu/lb-F
Kf=0.4 Btu/hr-ft-F
Hot Fluid (Properties at mean bulk temperature)
Mass flow rate = 1200 Ib/hr Cp = 1.0 Btu/lb-F
Q3,(cont.) Hot Fluid
Inlet temperature =200F Outlet temperature = unknown
HTC on the outside surface of the inner pipe = 400 Btu/hr-ft^2-F
Inner Pipe: OD =1.05” ; ID = 0.824” and k pipe material = 26 Btu/hr-ft-°F
HTC equations available are:
Nux = 0.664Rex ^0.5.Pr^0.3 for laminar flow
Nui = 0.026 ReD ^0.8. Pr^0.3 for turbulent flow
HINT: Write energy balance equations for both streams
Explanation / Answer
Solution:
Hot fluid gives up heat Qh = mhch(th1-th2)
Cold fluid picks up heat Qc = mccc(tc2-tc1)
The structure of heat exchanger transfers the heat from the hot fluid to coolant Qex = UA (qm)
From energy balance equation:
Qh = Qc = Qex
Mean temperature difference qm =q1-q2/ln(q1/q2)
We know,
U=(ho*hi)/(ho+hi)
To find hi :
Because the flow is obviously turbulent use the equation:
Nu=0.026(Re)0.8(Pr)0.3
Assume Pr =5.83
Claculte Reynolds number Re = (4m/pi*d*mu)
d=0.824 in= 0.068ft
m=1200lb/hr
mu=2.4191
Re = 9288.14
So, Nu= 65.9
Since, hi =Nu * k/d = 25204Btu/hr-ft2-F
And U = 393 Btu/hr ft2 F
From Qc=mccc(tc1-tc2) = 76800 Btu
Qc =Qh = mhch(200-th2)
Th2= 136F
Qm = 57.6
Q=UA(qm)
A = 2*pi*(do-di)*l
Substituting values in equation:
Length of heat exchanger (l) = 53.9 ft.
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