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Contents Question 1..................................................................................................................................1 Part A.....................................................................................................................................1 Part B......................................................................................................................................2 Part C......................................................................................................................................2 Part D.....................................................................................................................................3 Question 2..................................................................................................................................4 Part A.....................................................................................................................................4 Part B......................................................................................................................................5 Question 3..................................................................................................................................5 Part A.....................................................................................................................................5 Part B......................................................................................................................................6 Part C......................................................................................................................................6 Question 4..................................................................................................................................7 Part A.....................................................................................................................................7 References..................................................................................................................................8
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Question 1 Part A Heat transfer occurs from higher temperature to lower temperature. Thus heat flow direction is from the surface at which temperature is T2 to the surface at which temperature is T1. ‘k’ is the thermal conductivity of the material, ‘A’ is the area perpendicular to the direction of heat transfer and ‘dx’ is the thickness of the solid body. According to Fourier law of heat conduction heat transfer by conduction is expressed as Q = -kA∆T ∆xand∆T= T1-T2 Part B Thermal resistance is the resistance to heat flow. The thermal resistance can be compared with electrical resistance. The modification of the Fourier law can be written as Q = - ∆T (∆x kA). Thus the heat transfer is written as the ratio of driving force and thermal resistance. Comparing with electrical resistance current is the ratio of voltage and electric resistance. Thue thermal resistance is written as R_thermal =(∆x kA) Part C
The total thermal resistance of the following system is ∑Rthermal=∑Rconduction+∑Rconvection ∑Rthermal=1 h1A1 +L1 k1A1 +L2 k2A2 +1 h2A2 Since the area is the same ∑Rthermal=1 h1 +L1 k1 +L2 k2 +1 h2 Part D The thermal resistance of the whole system is expressed as R1 =∆x kA= 0.01/2.5/0.12 = 0.033 C/W R2 =∆x kA= 0.05/18/0.04 = 0.156 C/W R3 =∆x kA= 0.05/7/0.04 = 0.1785 C/W
R4 =∆x kA= 0.05/18/0.04 = 0.069 C/W R5 =∆x kA= 0.1/16.5/0.06 = 0.101 C/W R6 =∆x kA= 0.1/34/0.06 = 0.049 C/W R7 =∆x kA= 0.06/2.5/0.12 = 0.2 C/W Combined effect of R2. R3 and R4 is 1 R234 =1 R2 +1 R3 +1 R4 =1 0.156+1 0.1785+1 0.069leads to R123 = 0.0377 C/W Combined effect of R5 and R6 is 1 R56 =1 R5 +1 R6 =1 0.101+1 0.049leads to R56 = 0.0329C/W The total thermal resistance is ∑Rthermal=R1+R234+R56+R7= 0.033+0.0377+0.0329+0.2 = 0.3036C/W Rate of heat transfer =∆T (R)= (375-88)/0.3036 = 945W Same heat transfer is passing through the first four material 945 =∆T (R1+R234)therefore∆T=945∗(0.0330−.0377)= 66.81C The temperature at the point = 375-66.81 = 308.188C The temperature drop at material ‘F’ is ∆T=Q∗R7=945∗0.2=189C Question 2 Part A Assumption in the analysis are no fouling factor, kinetic energy an dpotential enrgy changesa re neglected, insulated heat exchanger so that no heat is moving outside, and steady state condition The heat transfer taken place in the heat exchanger is calculated as absed on water Q =˙mcp(Tout−T¿)= 0.2 * 4.18 * (60-25) = 29.3KW
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Heat removed from one fluid is same as heat accepted by another fluid. Thus the same amount of heat is accepted by geothermal water adnteh exit temperature og getherma water can be calculated as T_exit = Q/mcp = 140 – 29.3/.3/4.3 = 117C The temperature relationship in an heat exchanger is logarithmic, therefore the logarithmic temperature difference in calculated as ΔT1 = left side temperature difference = 140-25 = 115C ΔT2 = rightside temperatue difference = 117.4-60 = 57.4C LMTD = (140-25)/ln(115/57.4) = 83C. To find the surface area over all heat transfer coefficient isused Q = A * U * LMTD A = Q/ U / LMTD = 29.3/0.55/83 = 0.65m2 Length of the tube = A/3.14/D = 0.65/pi/0.008 = 25m Part B Assumption in the analysis are no fouling factor, kinetic energy an dpotential enrgy changesa re neglected, insulated heat exchanger so that no heat is moving outside, and steady state condition Heat transfer rate is calculated as (oil) Q =˙mcp(Tout−T¿)= 2.27 * 2198 * (150-40) = 548KW The exit temperature of the water can be calculated as T_exit =T_in+Q/mcp = 22+ 548/1.36/4187 = 118C The temperature relationship in an heat exchanger is logarithmic, therefore the logarithmic temperature difference in calculated as ΔT1 = left side temperature difference = 150-118 = 115C ΔT2 = rightside temperatue difference = 40-21= 19C LMTD = (115-19)/ln(115/19) = 53C. Over all heat transfer coefficient U = Q /A/LMTD = 548 / (3.14*0.127*60)/53 = 432W/m2-K
Question 3 Part A In this problem ther is no external heat transfer coefficient is given. Thus the heat transfer to the surrounding air is by radiation Heat transfer by radiation is calculated as Q = ε A σ(Ts4−Ta4) = 0.72 * (3.14*0.0603*70) *5.67 × 10−8*¿¿) = 17.915KW Part B Efficiency of furnace =heatusedforsteamgeneraton heatsupplied=Qused Qin= 0.8 Qin−Qlost Qin= 0.8 Qin−17.915 Qin= 0.8, thre fore Qin = 85.97KW Total energy used from the natural gas is 85.97 * 8760 * 3600 = 2711.3GJ Total therm = 2711.3/0.1055=25699.6 Cost = 25699.6 * 1.128269.5 = 28269.55£ Part C Energy balance Heat transfer by conduction is same as heat transfer by radiation Q_cond = Q_rad 2∗314∗k∗L∗(Ts−T) ln(r2 r1)= ε A σ(T4−Ta4) Solve for T
2∗314∗0.038∗70∗((175+23)−T) ln(80.15 30.15)= 0.72 * (3.14*0.0603*70) *5.67 × 10−8*(T4−2914) T = 324.4K Heat loss is = Q = ε A σ(Ts4−Ta4) = 0.72 * (3.14*0.0603*70) *5.67 × 10−8*¿¿) = 2.083KW Saved energy = Q_lost _old-Q_lost_new = 17.915KW-2.083KW = 15.832KW Cost saved for an year = 15.832 * 8760 * 3600 = 499.27GJ Total therm = 499.27/0.1055=4732.4 Cost saved in year= 4732.4* 1.128269.5 = 5339.1£ Coast saved per day = 5339.1/365 = 14.62£ Number of days needed to cover 750£ = 750/14.62=51.3days Question 4 Part A In parallel flow heat exchagr both fluid will flows in the same direction in thecounter flow the fluids flow in oppsite direction. Part B
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Part C Since the temperature at inlet is high there will be chances of thermal stress developemtn and may spol the material. Temperature at the exit of the cold fluid will never be higher than hot exit temperature The design is desireable if the temperature at exit needs to be same Part D Counter flow heat exchanger shows uniform temperature profile The exit cold fluid temperature can be more thant eh exit hot fluid temperauere Heat transfer is maximum for same fluid flow temperauters compare to parallel flow References Holman, J., 2010.HeatTransfer. Boston [Mass.]: McGraw Hill Higher Education. Nellis, G. and Klein, S., 2012.HeatTransfer. Cambridge: Cambridge University Press.