Saturated steam under pressure of 270.1 kPa is used to heat water at an average
ID: 1718942 • Letter: S
Question
Saturated steam under pressure of 270.1 kPa is used to heat water at an average of 93.3degree C flowing in a horizontal steel pipe line, 1-in. schedule 40 pipe (ID=26.64mm OD=33.mm) a a flow of465 kg/h. Assuming that the outside surface temperature of the metal in contact with steam is at 129degree C, calculate the heat transfer coefficient through the condensing steam, the heat exchanged for 45 cm of pipe, the mass of steam in kg condensed on the pipe per hour due to the heat loss and, the temperature of the wall inside the pipe. The average k for steel is 45 W/mK. Also, check if the outside surface temperature of the metal was assumed correctly.Explanation / Answer
Given :
Saturated steam pressure = 270.1 Kpa
T2 = Temperature = 93.3 oC = 273+ 93.3 = 366.3 K
Internal diameter of steel pipe ID = 26.64 mm =0.02664m =radius r1 = 0.01332 m
External diameter of pipe OD = 33.4 mm = 0.0334m = raidus r2 = 0.0167 m
out side temperature T1= 129 oC = 273 + 129 = 402 K
flow = 465 kg /h = 465 / 3600 = 0.1291 kg /sec
saturated steam at pressure 270.1KPa from steam tables the enthapy = h = 2733.9 kj/Kg
Hence total heat flow =m x h = 2733.9 x 0.1291 = 353.1 KW = 353.1 x 10^3 W
Total heat transfer by conduction Q = 2 x PI x k x L (T1 - T2 ) / ln r2/r1
10^3 x 353.1 = 2 x 3.14 x k x 1 ( 402-366.3) / ln 0.0167 / 0.01332
K = heat transfer oefficient = 370.3 W /mK
Q = 2 x 3.14 x 370.3 x 0.45 x (402- 366.3 ) / ln 0.0167 / 0.01332
= 160338 W
Heat exchanged = 353.1 x 10^3 - 160338
= 192761 W if 45 m Length
mass os steam in kg /hr due to heat loss
192671 = 160338 x m
m = 1.2kg /sec = 1.2 x 3600 = 4327 Kg/hr
Temperature of wall inside the pipe
192761 = 2 x 3.14 x 0.45 x 45 ( T1 - T2 ) /ln r2 / r1
= 2 x 3.14 x 0.45 x 45 ( T1 - T2 ) ln 0.0167 / 0.1332
T 2 = 212 K
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