Question

Problem (25 Points - Chapter 11) A concentre tube heat exchanger for cooling lubricating oil is comprised of a thin-walled me tube of 25-mm diameter carrying water and an our tube of 45mm diameter coming the The heat exchanger operates in counterflow with an overall heat transfer coefficient of W K and the average property as given in the table below. If the outlet temperature of the oil is 60°C, determine the following (a) total heat transfer rate, (b) outlet temperature of the water, (c) length required for the heat exchanger, (d) its effectiveness, and (e) number of transfer units, NTU. T = 100°C Oil = 0.1 kg's -0.1 kg's 7.-30°C Properties Water Oil p (kg/m) c, (J/kg-K) v (m”/s) k (W/m.K) Pr 1000 4200 7 x 10-7 0.64 4.7 800 1900 1 x 10- 0.134 140

Answer 1

To water U= Go w/m2 11 Thi inlet temp.ch oil. - sx The = exit temp. of oil. Tei= inlet temp. of water h for of) Tee = Exit temp. of water cy for water, Dia 25mm Do - 45 mm. na make flow or oil mc = mass flow of walter This jou oC mc = aills/s The = Go o n Tci = 30°c. mm = 0.1kg/s water.

a) Total heat transfer rate Da mn Cph Thi The ... = 0.1X 1900 x(100-60) Q. = 7600 W 6 Heat lust by oil - Heat gained by water, mn con Thi The) : me coc (Tee-Tal 0.18 4900 X (lou-60) Tea 70°C 0 1x1900 X (Tce-30) lengan fess heat exchanger. logrithnic mean temp. difference. Coma PT; - pre DTi = Thi-Toe = 100-10-30 ore - The Tci = 60-30=30 (DT) m2 30-30 .0 indeles minate form oo info) Ô DT) m = either DTi or ore I DTM 30°

so rength of heat exchangey, 1 0 - 7600 E UITDi ATM GTX 25 X 10 3 430 L.53175 m) NTU AS Q = UA 6im. 76002 GOX AX 30 A = 7600 = 4.22 m2 60X 30 So NTU = UA - Go X 4.22 (mCp) smaller 0.1 X 1900 1:33 Effective ness Capacity ratio c= (MC) smaller @ goto c (MC) smaller 2 (Map) on) (map) larger MCP) wales C = 0.181900 0.1X 4200 20.45

80 m Effectiveness for parallel Ez e-NTU (1-6). - I-ce-NTU (1-0) I-e-133C 1-0.45 ) 1-01 45 e 1.33 (1-0.45) TE - 0.662 Dry