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Chapter 4.4 - Heat Exchangers

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Selected questions Chapter 4.4: Heat Exchangers Short type Questions 1. Define effectiveness of heat exchanger? Ans. Effectiveness = to ti (T i ti ) = Cold fluid temp. range Inlet temp. diff . of hot & cold fluid Where, to Cold fluid outlet temperature ti = Cold fluid inlet temperature Ti 2. = = Hot fluid inlet temperature What do you mean by heat duty of a heat exchanger? Ans. 3. Heat duty of a heat exchanger is defined as the heat that is transferred from hot fluid to cold fluid per unit time. Mention 2 basic parameters that have to be measured at site to evaluate the performance of a heat exchanger? Ans. Hot and cold fluid in let and outlet temperatures 2. 4. 1. Mass flow rate of cold and hot fluids What do you mean by capacity ratio in a heat exchanger? Ans. R= Ti To Temp. range of hot fluid = t o t i Temp. range of Cold fluid Where, to Cold fluid outlet temperature ti = Cold fluid inlet temperature Ti = Hot fluid inlet temperature To 5. = = Hot fluid outlet temperature Find the Overall heat transfer coefficient for a shell and tube counter flow heat exchanger where the heat exchanged is 11.27 GJ/hour with the heat transfer area of 96 m2. Assume the LMTD as 76 oC. Ans. Heat Transfer Co-efficient Overall HTC, U = Q/ (A x LMTD) = 11.27 x 106 / (3600 x 96 x 76) = 0.43 kW/m2. K 6. Give the equation to calculate the Overall heat transfer coefficient. Ans. Heat exchanger performance is normally evaluated by the overall heat transfer coefficient U that is defined by the equation U=Q/ (A x LMTD) Where Q = Heat transferred in kCal/hr A = Heat transfer surface area in m2 LMTD = Log Mean Temperature Difference in 0C U = Overall heat transfer Coefficient kCal/hr/m2/0C 4.4 - Heat exchangers (table format) 22 Selected questions 7. Give the equation to Calculate LMTD for a counter flow heat exchanger Ans. LMTD Counter current Flow = ((Ti-to) (To-ti)) / ln ((Ti-to)/(To-ti)) Where, Ti- Hot fluid inlet temperature, ti- Cold fluid inlet temperature To- Hot fluid outlet temperature, to- Cold fluid outlet temperature 8. Give the equation to Calculate LMTD for a parallel flow heat exchanger Ans. LMTD Parallel Flow = ((Ti-ti) (To-to)) / ln ((Ti-ti)/(To-to)) Where, Ti- Hot fluid inlet temperature, ti- Cold fluid inlet temperature To- Hot fluid outlet temperature, to- Cold fluid outlet temperature 9. Write the equation to calculate heat duty for a heat exchanger. Ans. ..Eqn-1 Heat Duty, Qc = Wx Cpc x ( ti-to) 10. Heat Duty, Qh = Wx Cph x (Ti-To) ...Eqn-2 What do you mean by overall heat transfer coefficient? Ans. The ratio of heat flux per unit difference in approach across heat exchange equipment considering the individual coefficient and heat exchanger metal surface conductivity. The magnitude indicates the ability of heat transfer for a given surface. Higher the coefficient lesser will be the heat transfer surface requirement Long type questions 1. Explain the following a) Over all heat transfer coefficient b) Log mean temp difference c) Co-current and counter current flow d) Heat duty of heat exchanger Ans. a) Over all heat transfer coefficient: Heat exchanger performance is normally evaluated by the overall heat transfer coefficient U that is defined by the equation Q=U x A x LMTD Where Q = Heat transferred in kCal/hr A = Heat transfer surface area in m2 LMTD = Log Mean Temperature Difference in 0C U = Overall heat transfer Coefficient kCal/hr/m2/0C When the hot and cold stream flows and inlet temperatures are constant, the heat transfer coefficient may be evaluated using the above formula. It may be observed that the heat pick up by the cold fluid starts reducing with time. b) Log mean temp difference 4.4 - Heat exchangers (table format) 23 Selected questions a) Counter current Flow b) Co-current flow Ti To Ti To to ti ti to LMTD Counter current Flow = ((Ti-to) (To-ti)) / ln ((Ti-to)/(To-ti)) LMTD Co current Flow = ((Ti-ti) (To-to)) / ln ((Ti-ti)/(To-to)) c) co current and counter current flow If the flow of hot and cold fluids is in the same direction then the flow is described as co current flow. If the flow is in opposite direction then it is called as counter current flow. d) Heat duty of heat exchanger The capacity of the heat exchanger equipment expressed in terms of heat transfer rate, viz. magnitude of energy or heat transferred per time. It means the exchanger is capable of performing at this capacity in the given system. 2. Explain with equations the procedure for calculating the overall heat transfer co-efficient for heat exchangers Ans. 1. Heat Duty, Q = qs + ql qs is the sensible heat and ql is the latent heat qs = Wx Cph x(Ti- To)/1000/3600 (or) qs = w x Cpc x (to-ti)/1000/3600 in kW ql= W x h , h Latent heat of Condensation of a hot condensing vapour (or) = w x c , c - Latent heat of Vaporization 2. Hot Fluid Pressure Drop, Ph = Pi Po 3. Cold fluid pressure drop, Pc = pi- po 4. Temperature range hot fluid, T = Ti- To 5. Temperature range cold fluid, t = to ti 6. Capacity ratio, R = W x CPh / w Cpc (or) (Ti- To) / (to- ti) 7. Effectiveness, S = (to- ti) / (Ti ti) 8. LMTD a) Counter current Flow b) Co-current flow Ti To Ti To to ti ti to 4.4 - Heat exchangers (table format) 24 Selected questions LMTD Counter current Flow = ((Ti-to) (To-ti)) / ln ((Ti-to)/(To-ti)) LMTD Co current Flow = ((Ti-ti) (To-to)) / ln ((Ti-ti)/(To-to)) b) Correction factor for LMTD to account for Cross flow 1 SR (R + 1) 1 / 2 x ln 1 S F = 1/2 2 S ( R + ! (R + 1 ) (1 R ) x ln 1/2 2 S (R + 1) + (R + 1 ) { } 9. Corrected LMTD = F x LMTD 10. Heat Transfer Co-efficient U = Q / (A x Corrected LMTD) Numerical type questions 1. In a double pipe heat exchanger hot fluid is entering at 220 C and leaving at 115 C. Cold fluid enters at 10 deg c and leaves at 75 C. Mass flow rate of hot fluid 100 kg/hr, cp of hot fluid 1.1 kcal/kg C. Cp of cold fluid 0.95kcal/kg C. Calculate LMTD i) If the flow is parallel ii) If the flow is counter current. iii) Find the mass flow rate of cold fluid if the heat loss during the exchange is 5%. i. LMTD Parallel flow t1 t2 t1 ln t2 LMTD = t1 = 210 C t2 = 40 C = 102.5 C LMTD = ii. 210 40 210 ln 40 LMTD Counter current flow t1 = 145 C t2 = 105 C 4.4 - Heat exchangers (table format) 25 Selected questions LMTD = = t1 t2 t1 ln t2 145 105 145 ln 105 123.9 C m. Mass flow rate of hot fluid = 100 kg/hr cph. = 1.1 kcal/kg C CPC = 0.95 kcal/kg C Cold fluid inlet temperature = 10 C Cold fluid outlet temperature = 75 C Hot fluid inlet temperature = 220 C Hot fluid outlet temperature = 115 C Q = m.hxCPh X th x 0.95 = mc x CPC x tc Mass flow rate of cold fluid mc = = 100 x 1.1 x (220 115) x 0.95 0.95 x (75 10) = iii. = 177.7 kg/hr Mass flow rate of cold fluid Data: 2. m h cPh Th x 0.95 cPh x t c When a heat exchanger was designed its overall heat transfer co-efficient was 600 kcal/hr mt C. The heat transfer area provided = 10mt . Over a period of time, its overall heat transfer coefficient has fallen to 450 kcal/hr mt C due to fouling. Data: Specific heat of hot fluid = 1 kcal/kg C Hot fluid entering temperature = 80 C Hot fluid leaving temperature = 60 C Cold fluid entering temperature = 25 C Cold fluid leaving temperature = 40 C Calculate: How much additional area is to be added to maintain the same rate of heat transfer? Assumption: Flow is assumed in counter current direction Q = A U TLM TLM = = t1 t2 t1 ln t2 37.4 C 4.4 - Heat exchangers (table format) 26 Selected questions T1 = 40 C T2 = 35 C Q = 10 X 600 X 37.4 = After fouling area required = A = 224400 kcal/hr 224400 = 13.3 mt 2 450 X 37.4 Additional heat transfer area to be provided is 3.3 m2 3. An after cooler of shell and tube type with single pass is used for cooling compressed air from 85 to 35 oC. The compressed air generated is 1350 m3/h at mean air temperature. Calculate: oC 1) The amount of cooling water to be circulated at a temperature of 30 oC. Assume the cooling water outlet temperature as 35 oC. 2) The LMTD and no of tubes if the dia of the tube is 25.4 mm and 2500 mm length. Assume overall heat transfer coefficient for air to water is 150 W/m2.K. 3) The hp of the pump required if the pressure required is 3.5 kg/cm2.g. Indicate all assumptions clearly. Ans: 1) Inlet compressed air temperature = 85 oC Exit compressed air temperature = 35 oC Cooling water temperature = 30 oC Cooling water outlet temperature = 35 oC Density of air at 60 oC = 1.06 kg/m3 Mass of compressed air = 1350 x 1.06 = 1431 kg/hr = 4.178 kJ/kg oK = m Cp deta.T = 1431 x 1.005 x ( 85-35)/ 3600 = 19.974 kW = 19.974/(4.178 x (35-30) = 3442 kg/hr (Ans) 2) LMTD = (85 35) (35 30) = 19.54oC ( Ans) In (50 / 5) Heat transfer area = 19.974/( 0.15 x 19.54) = 6.81 m2 = 6.81/ ( 3.14 x 0.0254 x 2.5 ) = 35 tubes ( Ans) Specific heat of water at 30 oC Heat duty: Hot fluid: Mass of water to be circulated No of tubes 3) Theoretical power required for the pump= = 3442 x 9.81 x 35 / ( 3600 x 1000 ) 0.33 kW Assuming pump efficiency of 70% and motor efficiency of 85%, the required pump Motor power 0.33/ (0.7 x 0.85) = 4.4 - Heat exchangers (table format) = 0.555 kW 27 Selected questions Motor required 4. = 1 hp (Ans) Hot air is to be generated at the rate of 30,000 m3/h at 120 0C from 35 oC for pocket ventilation purpose in a high speed paper machine. Steam is used in indirect heat exchanger steam coils to produce hot air at 120 oC. Steam is available at a pressure of 5.0 bar at the user side header. Find out the quantity of steam required. Calculate LMTD. Data: Total enthalpy of steam at 5 bar = Latent heat of steam at 5 bar = 2757 kJ of kg 2086 kJ/kg Note: Assume that condensate is leaving at a temperature of 100 C. Ans. Inlet air temperature = 35 oC Outlet air temperature = 120 oC Steam pressure = 5.0 bar Total heat of steam at 5 bar = 2757 kj/kg Latent heat of steam at 5 bar = 2086 kJ/kg Saturation temperature of steam = 159 oC Mass of hot air required = 30,000 x 0.898 = 26940 kg/hr = 1.009 kJ/kg oK = m Cp delta.T = 26940 x 1.009 x (120-35)/ 3600 = 641.8 kW = 641.8 x 3600/ 2086 = 1107 kg/hr (Ans) = (159 35) (159 120) = 26.73 oC In (124 / 39) Specific heat of air at 120 oC Heat duty: Cold fluid: Mass of steam to be circulated LMTD 4.4 - Heat exchangers (table format) 28

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