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To determine the effect of flowrate on rate of heat transfer.

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Introduction

OBJECTIVE: To determine the effect of flowrate on rate of heat transfer INTRODUCTION: A concentric tube or double pipe heat exchanger is one that is composed of two circular tubes. One fluid flows in the inner tube, while the other fluid flows in the annular space between the two tubes. In counter flow, the two fluids flow in parallel, but opposite directions. In parallel-flow the two fluids flow in parallel and in the same direction. Fig. (1) PROCEDURE: The work was carried out the lower exchanger of a pair of concentric tube heat exchange. The lower exchanger had water passing through the inner tube, being heated by steam which flows into the outer tube and condenses. The condensate usually returns to the steam boiler, but there was a sampling facility so it may be collected and timed to determine condensate flowrate. The water to the lower exchanger was turned on and the flow was set to the desired rate. The condensate return line was checked if it was open to drain (floor),then the stem was turned on to the lower exchanger and the pressure was set at 10 psig. ...read more.

Middle

(Tc out - Tc in) .................................... (1) For the steam side: Qh =mh.?v+mh.Cph. (Ts-Th out) ............................... (2) Where mh can be calculated from mh = Fh* ?c And mc = Fc * ?c Symbols:- Cpc= specific heat capacity of cold water, J/kg.K Cph= specific heat capacity of hot condensate, J/chg. Fc = volumetric flowrate of cold water, m3/s Fh = volumetric flowrate of hot condensate, m3/s mc = mass flowrate of cold water, kg/s mh = mass flowrate of hot condensate, kg/s Qc = energy gained by cold water, J/s Qh = energy lost by steam, J/s Tc in = temperature of cold water entering Ts = temperature of steam entering exchanger, K Th out = temperature of hot condensate leaving exchanger, K ?v = latent heat of vaporisation of steam, J/kg ?c = density of cold water, kg/m3 ?h = density of hot condensate, kg/m3 RESULTS: The results which took in the laboratory: Runs Cold water flowrate (Litters/min) Steam in temp. Steam out temp. Could water in temp.(C) Could water out temp.(C) Condensate flowrate (millilitres/min) 1 17 113 112 0.7 41 1800 1840 2 2 113 112 0.7 84 1280 3600 3 15 113 112 0.7 41 2040 1680 4 5 112 111 0.7 63 1360 1280 5 10 113 112 0.7 48 1440 1566 CALCULATION: Note: Conversion: (Litters/min) ...read more.

Conclusion

Example: (I used the first run in this example): Fc = 17 * (0.001/60) = 0.000283 (m3/s) mc=0.000283 * 996= 0.2822 kg/s Qc =0.2822 * 4180 (314 - 280) = 40106.26 J/s * To worked out the value of Qh we used equation (2), but we had first to got the value of mh from this equation: mh = Fh* ?c To get value of Fh we took the average of the condensate floweate values and then we converted it from (millilitres/min) to (m3/s). Example: (I used the first run in this example): Fh = (1800+1840)/2 * (0.000001/60) = 3.03333E-05(m3/s) mh=3.03333E-05 * 958 = 0.029059333 kg/s Qh = 0.029059333*2.22 * 106 + 0.029059333 * 996 (386 - 385) =64634.35 J/s DISSCUSSING THE RESULT& CONCOLUTION: * When the flowrate is increased the % difference between the energy lost by steam and the energy gained by cold water will decrease. * The relation between Qc and Fc is direct proportion, therefore to transfer heat faster, we must increase the flowrate. * The value of the % difference in the first run is 125.7, which I think it is wrong and that happened because the flowarete value was too small, and that make the value of Qc small as well, therefore the difference is high. ...read more.

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