the answer has given above but need some steps or equations to show how to get i
ID: 2086785 • Letter: T
Question
the answer has given above but need some steps or equations to show how to get it
The overall heat transfer coefficient (U) may be taken as 259 W/m*-°K. Problem 2 Repeat Problem 2, except that you must estimate the overall heat transfer coefficient (U) in units of W/m'- K given the following information: i) Film coefficient from handout table 14.4 for lubricating oil stock. ii Film coefficient from handout table 14.4 for untreated cooling tower water. i) The value for x/k for the heat transfer surface interface is negligible and can be neglected iv) The fouling factor for the lubricating oil is 0.005 ft -hr-°F/BTU. v) The fouling factor for the untreated cooling tower water is 0.005 ft-hr-F/BTU.Explanation / Answer
Solution:
Light lubricating oil
Flow rate = ?1 = 0.5kg/s = 0.5 * 3600 kg/hr = 1800 kg/hr
Heat Capacity (C p) = 2090 J / kg °K
T1 = 375 °K
T2 = 350 °K
Rate of heat transfer = Q = ? * C p * (T1 – T2) = 0.5kg/s * 2090 J / kg °K * (375 – 350 ) °K
= 26.125kW
= 26.125 kJ/s
= 26.125 * 3600 kJ/hr
= 94050 kJ/hr
Cold fluid balance:
Fluid: Water
Flow rate = ?2 = 0.201kg/s = 0.201* 3600 kg/hr = 723.60 kg/hr
Heat Capacity = Cp2 = 4220 J/kg °K
Inlet temperature = t1 = 280 °K
Outlet temperature = t2 = 310.7998 °K
Rate of heat transfer = Q water = ?2 * C p2 * (t1 – t2) = 0.201kg/s * 4220 J / kg °K * (310.7998 – 280) °K
= 26.125kW
= 26.125 * 3600 kJ/hr
= 94050 kJ/hr
For Calculating Overall heat transfer coefficient
Given information:
Heat transfer coefficient due to lubricating oil stock = h oil = 55 BTU / sq. Ft. - °F = 1124.2 kJ / m^2 - °K
(Conversion: 1BTU /ft^2 - °F = 20.44 kJ/m^2- °K)
Thermal resistance due to lubricating oil stock = 1/ hoil = 0.00089 sq.m - °K / kJ
Heat transfer coefficient due to untreated tower water = h tower water = 140 BTU / sq. Ft. - °F = 2861.6kJ / m^2 - °K
Thermal resistance due to untreated tower water =1/ h tower water = 0.000349 sq.m - °K / kJ
Heat transfer coefficient due to fouling of oil = h foul oil = 200 BTU / sq. Ft. - °F ( 1/ 0.005 ft^2 – hr - °F / BTU) = 4088 kJ / m^2 - °K
Thermal resistance due to fouling of oil = 1/ h foul oil = 0.000245 sq.m - °K / kJ
Heat transfer coefficient due to fouling of water = h foul water = 200 BTU / sq. Ft. - °F ( 1/ 0.005 ft^2 – hr - °F / BTU) = 4088 kJ / m^2 - °K
Thermal resistance due to fouling of water = 1/ h water = 0.000245 sq.m - °K / kJ
Total thermal resistance,
= 0.00089 sq.m - °K / kJ + 0.000349 sq.m - °K / kJ + 0.000245 sq.m - °K / kJ + =0.000245 sq.m - °K / kJ
= 0.001728 sq.m - °K / kJ
Overall heat transfer coefficient, U = 1/ total thermal resistance
= 1 / 0.001728 sq.m - °K / kJ
= 578.632kJ/ m^2 - °K
LMTD = ?T1 - ?T2/ ln ( (?T1/ ?T2)) = (64.20021 – 70 ) °K / ln ( 64.20021/70) = 67.05831 °K
We know, Q =U * As * LMTD
Hence, requires heat transfer surface area = As = Q / U * LMTD
= 94050 kJ/hr /[ (578.632kJ/ m^2 - °K) * (67.05831 °K)]
= 2.423838 m^2
2. For a Parallel or cocurrent flow heat exchanager
LMTD = ?T1 - ?T2/ ln ( (?T1/ ?T2)) = (95 – 39.20021 ) °K / ln ( 95/39.20021) = 63.03673 °K
We know, Q =U * As * LMTD
Hence, requires heat transfer surface area = As = Q / U * LMTD
= 94050 kJ/hr /[ (578.632kJ/ m^2 - °K) * (63.03673 °K)]
= 2.578472m^2
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