{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

engr2250hwch13

# engr2250hwch13 - 13-1 Hot exhaust gases are used in the...

This preview shows pages 1–3. Sign up to view the full content.

13-1 Hot exhaust gases are used in the reheat section of a Rankine cycle. Consider a commercial steel tube 5-cm outside diameter, 4.5-cm inside diameter used to convey the steam. The air side heat transfer coefficient is 85 W/m 2 ·K, and that of the steam side is 200 W/m 2 ·K. Determine: a. the overall heat transfer coefficient based on the inside tube area (in W/m 2 ·K) b. the overall heat transfer coefficient based on the outside tube area (in W/m 2 ·K) c. the overall heat transfer coefficient based on the inside area if air side fouling is 0.0015 m 2 ·K/W and steam side fouling is 0.0005 m 2 ·K/W (in W/m 2 ·K). Approach: Use Eq. 13-3 and Eq. 13-4, which give the overall heat transfer coefficient in terms of its component parts. Assumptions: 1. There are no fins. 2. There is no fouling. Solution: The overall heat transfer coefficient is calculated with Eq. 13-3 and Eq. 13-4: , , , , 1 1 1 1 i o w i i o o o i i i o i i o o o o o o o R R R U A U A h A A A h A η η η η ′′ ′′ = = + + + + No fins are used, so , , 1 o i o o η η = = . For a circular tube ( ) ( ) ln ln 2 2 o i o i w r r D D R kL kL π π = = From Appendix A-2, the thermal conductivity of commercial steel is k = 60.5 W/mK. The areas are: and i i o o A D L A D L π π = = a) Incorporating the above expressions into the equation for the overall heat transfer coefficient based on the inside area, noting that the length cancels, and ignoring fouling: ( ) ln 1 2 i o i i i i i i o o D L D D A D L U h A kL h D L π π π π = + + ( ) ( ) ( ) 5 2 2 0.045m ln 0.050 0.045 1 1 1 0.045m = + + 0.005 3.92 10 0.0106 2 60.5W mK 0.050m 200W m K 85W m K i U = + × + 2 =64.0W m K i U Answer b) The overall heat transfer coefficient based on the outside area is: 2 2 W 0.045 W 64.0 =57.6 0.050 m K m K i o i o A U U A ⎞⎛ = = ⎟⎜ ⎠⎝ Answer c) With fouling on both sides of the heat exchanger, we obtain ( ) ( ) ( ) 2 2 2 2 0.045m ln 0.050 0.045 1 1 m K m K 0.045m 1 0.045m = +0.0005 + + 0.0005 + W 2 60.5W mK W 0.050m 0.050m 200W m K 85W m K i U ⎞⎛ ⎟⎜ 2 =60.3W m K i U Answer Comment: The overall heat transfer coefficient is low, so the effect of fouling is small. If the overall heat transfer coefficient had been large, then the addition of fouling would have had a much greater effect. 13- 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
13-2 A two shell pass, eight tube pass heat exchanger with a surface area of 8,300 ft 2 is used to heat 1,700 lbm/min of water from 75 ° F to 210 ° F. Hot exhaust gases enter at 570 ° F and exit at 255 ° F. Assuming the exhaust gases have the same properties as air, determine: a. the overall heat transfer coefficient (in Btu/hr·ft 2 · ° F) b. the overall heat transfer coefficient if fouling on both sides equivalent to 0.005 hr·ft 2 · ° F/Btu is present in the heat exchanger (in Btu/hr·ft 2 · ° F). Approach: With all temperatures known, along with the area and one flow rate, sufficient information is given to evaluate the heat transfer rate and the LMTD .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 90

engr2250hwch13 - 13-1 Hot exhaust gases are used in the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online