Given: Hot fluid oil, Cold fluid - water (T 1, T 2 ) (t 1, t 2 ) Water

Transcription

1 . In a counter flow double pipe eat excanger, oil is cooled fro 85 to 55 by water entering at 5. Te ass flow rate of oil is 9,800 kg/ and specific eat f oil is 000 J/kg K. Te ass flow rate of water is 8,000 kg/ and specific eat of water is 480 J/kg K. Deterine te eat excanger area and eat transfer rate for an overall eat transfer co-efficient of 80 W/ K. Given: Hot fluid oil, old fluid - water (T, T ) (t, t ) Water Entry teperature of oil, T = 85 Oil Exit teperature of oil T = 55 Water Mass flow rate of oil (Hot fluid), = 9,800 kg/ 9,800 kg/ s 600 =.7 kg/s Specific eat of oil, p = 000 J/kg K Mass flow rate of water (cold fluid), c = 8,000 kg/ 8,000 kg/ s 600 c =. kg/s Specific eat of water, pc = 480 J/kg K Overall eat transfer co-efficient, U = 80 W/ K

2 To find: Solution:. Heat excanger area, (A). Heat transfer rate, (Q) Heat lost by oil (Hot fluid) = Heat gained by water (cold fluid) t Q = Q c p T T c pct t t t t.9 0 Exit teperature of water, t 4.5 Heat transfer, Q c pc Q. Q 6 0 t t or pt T W Q UA T.. () Heat transfer, Were T - Logaritic Mean Teperature Difference. (LMTD) For ounter flow,

3 T T T t T t T In T t t In 55 5 T = 5. 8 Substitute T U and Q values in Equn () () Q UA T A 5.8 ==> A = 6.6 Result:. Heat excanger area, A = 6.6. Heat transfer, Q = 6 0 W. Water flows at te rate of 65 kg/in troug a double pipe, counter flow eat excanger. Water is eated fro 50 to 75 by oil flowing troug te tube. Te specific eat of te oil is.780 kj/kg K. Te oil enters at 5 and leaves at 70. Te overall eat transfer coefficient is 40 W/ K. alculate te following Given:. Heat excanger area. Rate of eat transfer Hot fluid oil, old fluid - water (T, T ) (t, t ) Mass flow rate of oil (cold fluid), c = 65 kg/in 65 kg/ s 60 c =.08 kg/s

4 To find: Solution: Entry teperature of water, t = 50 Exit teperature of water, t = 75 Specific eat of oil (Hot fluid) p =.780 kj/kg K Entry teperature of oil, T = 5 Exit teperature of oil, T = 70 =.7800 J/ K Overall eat transfer co-efficient, U = 40 W/ K. Heat excanger area, (A). Heat transfer rate, (Q) Heat transfer, Q t t or T Q c pc pc Q T p T T Specific eat of water, pc = 486 J/kgK Q U A T.() Heat transfer, Were T - Logaritic Mean Teperature Difference. (LMTD) For ounter flow,

5 T T T t T t T In T t t In Substitute T U and Q and U values in Equn () () Q UA T 0 40 A 8.8 T = 8. 8 ==> A =.54 Result:. Heat excanger area, A =.54. Heat transfer, Q = 0 W. In a counter flow single pass eat excanger is used to cool te engine oil fro 50 to 55 wit water, available at as te cooling ediu. Te specific eat of oil is 5 J/kg K. Te flow rate of cooling water troug te inner tube of 0.4 diaeter is. kg/s. Te flow rate of oil troug te outer tube of 0.75 diaeter is.4 kg/s. If te value of te overall eat transfer co-efficient is 40 W/ K, ow long ust te eat excanger be to eet its cooling requireent? Given: Hot fluid oil, old fluid - water (T, T ) (t, t ) Entry teperature of oil, T = 50 Exit teperature of oil, T = 55 Entry teperature of water, T = Specific eat of oil (Hot fluid) p = 5 kj/kg K

6 Inner diaeter, D = 0.4 Flow rate of water (cooling fluid), c =. kg/s Outer diaeter, D = 0.75 Flow rate of oil (Hot fluid), =.4 kg/s Overall eat transfer co-efficient, U = 40 W/ K To find: Lengt of te eat excanger, L Solution: Heat lost by oil (Hot fluid) = Heat gained by water (cold fluid) Q Q p c T T c pct t t t Heat transfer, Q t t or T c Q pc [ Specific eat of water, pc = 486 J/kgK] p T Q W 0 Heat transfer Q UAT () Were T - Logaritic Mean Teperature Difference. (LMTD)

7 For ounter flow, T T T t T t T t In T t In 55 Substitute T, U and Q values in Equn () () Q UA T T = A 50. ==> A = 40.0 Area, A = D L 40.0 = 0.4 L ==> L =.9 Result: Lengt of te eat excanger, L = In an oil cooler for a lubrication syste, oil is cooled fro 70 to 40 by using a cooling water flow at 5. Te ass flow rate of oil is 900 kg/ and te ass flow rate of water is 700 kg/. Give your coice for a parallel flow or counter flow eat excanger, wit reasons. If te overall eat transfers co-efficient is 0 W/ K, find te area of te eat excanger. Take specific eat of oil is kj/ kg. Given: Hot fluid oil, (T, T ) old fluid - water (t, t ) Entry teperature of oil, T = 70 Exit teperature of oil, T = 40

8 Entry teperature of water, T = 5 Te ass flow rate of oil, 900kg/ 900 kg/ s kg/ s Te ass flow rate of water, c 700kg/ 700 kg/ s kg/ s Overall eat transfer co-efficient, U = 0 W/ K Specific eat of oil, p = kj/kg = 0 J/kg To find:. oice of eat excanger (Weter parallel flow or counter flow). Area of eat excanger. Solution: Heat lost by oil (Hot fluid) = Heat gained by water (old fluid) Q Q p c T T c pct t t [ Specific eat of water, pc = 486 J/kg K] 5, t 0,0.0 ==> t = 4.47

9 Exit teperature of water, t = 4.47 > T Since t > T, counter flow arrangeent sould be used. Heat transfer Q UAT For ounter flow, T T () T t T t T In T t t In 40 5 T = () Heat transfer, Q p Q 0.5 T T or c pct t pt T Q = 5,000 J/s.. () Substitute, Q, U, and T values in equation () 5,000 0 A Q UA T 0.6 ==> A = 7.0

10 Result:. oice of eat excanger counter flow arrangeent. Surface area, A = In a refrigerating plant water is cooled fro 0 to 7 by brine solution entering at - and leaving at. Te design eat load is 5500 W and te overall eat transfer co-efficient is 800 W/ K. Wat area required wen using a sell and tube eat excanger wit te water aking one sell pass and te brine aking two tube passes. Given: Hot fluid water, old fluid water (T, T ) (t, t ) To find: Solution: Were Entry teperature of water, T = 0 Exit teperature of water, T = 7 Entry teperature of water, T = - Exit teperature of brine solution, t = Heat load, Q = 5500 W Overall eat transfer co-efficient, U = 800 W/ K Area required (A) Sell and tube eat excanger One sell pass and two tube passes For sell and tube eat excanger (or) cross flow eat excanger. F orrection factor Q FUA T [ounter flow].. () T - Logaritic ean teperature difference for counter flow.

11 For ounter flow, T T t T t T In T In 7 t t Fro grap, T =.57 X axis t t value, P T t 0 5 P = 0. urve value, T P t T t X axis value is 0., curve value is.6, and corresponding Y axis value is Substitute T Q, U and F value in Equn () () Q FUA T A.57 R =.6 ==> A = 0.58 Result: Area of eat excanger, A = 0.58

12 6. A parallel flow eat excanger is used to cool 4. kg/in of ot liquid of specific eat.5 kj/kg K at 0. A cooling water of specific eat 4.8 kj/kg K is used for cooling purpose at a teperature of 5. Te ass flow rate of cooling water is 7 kg/in. calculate te flowing. Take Given:. Outlet teperature of liquid. Outlet teperature of water. Effectiveness of eat excanger Overall eat transfer co-efficient is 00 W/ K. Heat excanger area is 0.0 Mass flow rate of ot liquid, = 4. kg/in = 0.07 kg/s Specific eat of ot liquid, p =.5 kj/kg K p.5 0 J / kgk Inlet teperature of ot liquid, T = 0 Specific eat of water, pc = 4.8 kj/kg K Inlet teperature of cooling water, t = 5 Mass flow rate of cooling water, c = 7 kg/in Overall eat transfer co-efficient, U = 00 W/ K Area, A = 0.0 To find:. Outlet teperature of liquid, (T ). Outlet teperature of water, (t ) pc 4.80 J / kgk c = 0.8 kg/s

13 . Effectiveness of eat excanger, Solution: apacity rate of ot liquid, p = 45 W/K.. () apacity rate of water, c p = 70.4 W/K () Fro () and (), in = 45 W/K ax = 70.4 W/K in ax in ax = () Nuber of transfer units, NTU = UA in NTU 45 NTU =.4.. (4) Fro grap, Xaxis NYU =.4

14 urve in ax orresponding Y axis value is 64% i.e., 0.64 Maxiu possible eat transfer Q ax = in (T t ) = 45 (0 5) Q ax = 8,75 W Actual eat transfer rate Q Q ax ,75 Q = 8,0 W Heat transfer, Q c pc t t

15 8, ,0 70.4t t 0.40 t 7556 t 5 Outlet teperature of cold water, t 0.40 Heat transfer, Q t pc T T 8, , T T Outlet teperature of ot liquid, T 56.4 Result:. T = t = = In a counter flow eat excanger, water at 0 flowing at te rate of 00 kg/. it is eated by oil of specific eat 00 J/kg K flowing at te rate of 50 kg/ at inlet teperature of 95. Deterine te following Take Given:. Total eat transfer. Outlet teperature of water. Outlet teperature of oil Overall eat transfer co-efficient is 00 W/ K. Heat excangers are is. apacity rate of ot oil, = p

16 = 0.4 W/K () apacity rate of water, = p W / K () Inlet teperature of ot liquid, T = 0 Specific eat of water, pc = 4.8 kj/kg K pc 4.80 J / kgk Fro Equn () and (), in = 0.4 W/K ax = 8. W/K in ax in ax = 0.8. () Nuber of transfer units, NTU = UA in 00 NTU 0.4 NTU =... (4)

17 Fro grap, X axis NYU =. urve in 0. 8 ax orresponding Y axis value is 0.95 i.e., 0.95 in ax 0.8 Maxiu possible eat transfer Q ax = in (T t ) = 0.4 (95 0) Q ax =,680 W Actual eat transfer rate Q Q ax 0.95,680 Q =,546 W

18 Heat transfer, Q c pc t t, t 0 pc 486J / kgk, t 7,67.6 ==> t = 5.5 Outlet teperature of water, t 5.5 Heat transfer, Q pc T T, T,546 8,78 0.4T ==> T =.75 Outlet teperature of oil, T.75 Result:. T =,546W. T =.75. t = In a counter flow eat excanger, water is eated fro 0 to 80 by an oil wit a specific eat of.5 kj/kg K and ass flow rate of 0.5 kg/s. Te oil is cooled fro 0 to 40. If te overall eats transfer co-efficient is 400 W/K, find te following by using NTU etod. Mass flow rate of water. Effectiveness of eat excanger. Surface area

19 Given: Hot fluid oil, old fluid - water (T, T ) (t, t ) Inlet teperature of water, t = 0 Outlet teperature of water, t = 80 Specific eat of oil, p =.5 kj/kg - K =.5 0 J/kg - K Te ass flow rate of oil, 0.5kg/ s To find: Solution: Inlet teperature of oil, T = 0 Outlet teperature of oil, T = 40 Overall eat transfer co-efficient, U = 400 W/ K. Mass flow rate of water, c. Effectiveness of eat excanger,. Surface area, A Heat lost by oil = Heat gained by water Q Q p c T T c pct t c [ Specific eat of water, pc = 486 J/kg K] ==> c = 0.48 kg/s Mass flow rate of oil (Hot fluid), = p

20 = 50 W/K () apacity rate of water (old fluid), = c pc = W/K () Fro Equn () and (), in = 50 W/K ax = W/K in ax in ax = () T Effectiveness, T T T Fro grap, X axis urve in ax

21 orresponding X axis value is.4, i.e., NTU =.4 NTU = UA in NTU UA in ==> A =.0 Result:. c = 0.48 kg/s. = A =.0