Memorial University of Newfoundland Faculty of Engineering and Applied Science

Transcription

1 Memorial University of Newfoundl Faculty of Engineering Applied Science ENGI-7903, Mechanical Equipment, Spring 20 Assignment 2 Vad Talimi Attempt all questions. The assignment may be done individually or in groups of two 2). No group shall have more than two members. You may use Maple, Mathematica, MatLab, or Excel as needed to assist you. Problem - For a given pump, show the effect of a fluid change i.e. ρ ρ 2 for the same rotational speed ω ω 2 ). Another case same powers) has been solved in class. Solution - We know that ω ω 2, D D 2, ρ ρ 2, so: H 2 H ω2 ω ) 2 ) 2 D2 ) 2 ) 2 ) D Ẇ 2 Ẇ Q 2 Q ρ2 ρ ω2 ω ) ) 3 D2 ) ) 3 2) D ) ) 3 ) 5 ω2 D2 ω D ρ2 ρ ) ) 3 ) 5 ρ 2 ρ 3) Problem 2 - The.25 in) impeller option of the Taco Model 403 Fl Series centrifugal pump of Fig. is used to pump water at 25 o C) from a reservoir whose surface is 4.0 ft) above the centerline of the pump inlet as shown in Fig. 2. The properties of water at 25 o C) are: ρ 997.0kg/m 3 ), µ P a.s), P v 3.69kP a). Stard atmospheric pressure is P atm 0.3kP a).

2 Fig. - Pump curve, Taco 403 Fl series The piping system from the reservoir to the pump consists of 0.5 f t) of cast iron pipe with an ID of 4.0 in) an average inner roughness height of 0.02 in). There are several minor losses: a sharp-edged inlet Kin 0.5), three flanged 90o stard elbows Kel 0.3 each), a fully open flanged globe valve Kv 6.0). Estimate the maximum volume flow rate in units of GP M )) that can be pumped without cavitation. Discuss how you might increase the maximum flow rate while still avoiding cavitation. Z Reservoir ols Inlet piping r rite rw ith d an To system ew Fre Z2 Globe Valve o dit FE Flow Pump D lp Fig. 2- Inlet piping Fil system from the reservoir point ) to the pump inlet point 2) PD Solution - For the maximum flow rate without cavitation see the excel file on the course webpage. The answer is around 640 GP M ). This maximum flow rate can be increased in different ways. For example we can use a simpler piping route in which only one elbow is used. go to the excel file set the quantity of elbows to see the change in the graph!). We can also use a larger pipe diameter again go to the excel file change the ID). Another way is using a gate valve instead of the globe valve if the application let us to do so. We can also place the pump closer to the reservoir or increase the vertical distance between Z 2

3 Z 2. You can see the effect of each of the proposed changes in the excel file by adjusting the necessary cells) Problem 3 - in the class. Re-solve the example 5-3 using LMTD method. Compare your answer to what we got Solution - The energy balance gives us: So This gives: Q ṁ c c p,c T c,o T c,i ) ṁ h c p,h T h,i T h,o ) 4) Q ) T h,o ) 5) The log mean temperature difference is: counter flow) T LMT D T 2 T ) ln T2 T The heat transfer surface area is: A And the heat exchanger length is: Which is close to our answer using ɛ NT U method. Q W ) 6) T h,i 25. o C) 7) ) 60 80) ) 9.98 o C) 8) ln Q U T LMT D m2 ) 9) L A πd m) 0) Problem 4 - A cross flow heat exchanger with both fluids unmixed has an overall heat transfer coefficient of 200 W/m 2.K), a heat transfer surface area of 400 m 2 ). The hot fluid has a heat capacity of 40,000 W/K), while the cold fluid has a heat capacity of 80,000 W/K). If the inlet temperatures of both hot cold fluids are 80 o C) 20 o C), respectively, determine the exit temperature of the hot cold fluids. Solution - The minimum maximum heat capacities are: C min 40000W/K) ) 80000W/K) 2) C r C min 0.5 3) 3

4 The maximum possible heat transfer rate is: And the number of transfer units is: Q max C min T h,i T c,i ) ) 2.4MW ) 4) NT U UA ) C min Using the appropriate graph in the houts we can read ɛ So the actual heat transfer rate is: Q act ɛq max W ) 6) T c,o T c,i + Q act 4.9 o C) 7) T h,o T h,i Q act 36.2 o C) 8) Problem 5 - In a -shell 2-tube heat exchanger, cold water with inlet temperature of 20 o C) is heated by hot water supplied at the inlet at 80 o C). The cold hot water flow rates are 5000 kg/h) 0,000 kg/h), respectively. If the shell tube heat exchanger has a U A value of,w/k), determine the cold water hot water outlet temperatures. Assume c p,c 478J/kg.K) c p,h 488J/kg.K). Solution - The cold hot streams heat capacities are: ṁ c c p,c W/K) 9) 3 ṁ h c p,h W/K) 20) 3 So the minimum maximum heat capacities are: C min W/K) 2) The maximum possible heat transfer rate is: And the number of transfer units is: 634.3W/K) 22) C r C min ) Q max C min T h,i T c,i ) ) 34892W ) 24) NT U UA.99 25) C min

5 Using the appropriate graph in the houts we can read ɛ 0.7. So the actual heat transfer rate is: Q act ɛq max W ) 26) T c,o T c,i + Q act 62 o C) 27) T h,o T h,i Q act 59 o C) 28) Problem 6 - A counter flow double pipe heat exchanger with A 9.0m 2 ) is used for cooling a liquid stream c p 3.5kJ/kg.K)) at a rate of 0.0 kg/s) with an inlet temperature of 90 o C). The coolant c p 4.2kJ/kg.K)) enters the heat exchanger at a rate of 8.0 kg/s) with an inlet temperature of 0 o C). The plant data gave the following equation for the overall heat transfer coefficient in W/m 2.K): U ) ṁ c) 0.8 ṁ h ) 0.8 where ṁ c ṁ h are the cold stream hot stream flow rates in kg/s), respectively. a) Calculate the rate of heat transfer the outlet stream temperatures for this unit. b) The existing unit is to be replaced. A vendor is offering a very attractive discount on two identical heat exchangers that are presently stocked in its warehouse, each with A 5.0m 2 ). Because the tube diameters in the existing new units are the same, the above heat transfer coefficient is expected to be valid for the new units as well. The vendor is proposing that the two new units could be operated in parallel, such that each unit would process exactly one half the flow rate of each of the hot cold streams in a counter flow manner; hence, they together would meet or exceed) the present plant heat duty. Give your recommendation, with supporting calculations, on this replacement proposal. Solution - Part a - The overall heat transfer coefficient is: U The cold hot streams heat capacities are: W/m 2.K) 30) ṁ c) 0.8 ṁ h ) So the minimum maximum heat capacities are: ṁ c c p,c W/K) 3) ṁ h c p,h W/K) 32) C min 3500W/K) 33) 33W/K) 34) 5

6 The maximum possible heat transfer rate is: And the number of transfer units is: The ɛ can be calculated as follows: ɛ C r C min ) Q max C min T h,i T c,i ) ) 2.52MW ) 36) NT U UA ) C min 3500 exp NT U C r)) C r exp NT U C r )) exp )) ) 0.94 exp )) So the actual heat transfer rate is: Part b - The overall heat transfer coefficient is: U The cold hot streams heat capacities are: Q act ɛq max W ) 39) T c,o T c,i + Q act 28.9 o C) 40) T h,o T h,i Q act 69.8 o C) 4) W/m 2.K) 42) ṁ c) 0.8 ṁ h ) So the minimum maximum heat capacities are: ṁ c c p,c W/K) 43) ṁ h c p,h W/K) 44) C min 5750W/K) 45) The maximum possible heat transfer rate is: 6800W/K) 46) C r C min ) 6

7 And the number of transfer units is: The ɛ can be calculated as follows: ɛ Q max C min T h,i T c,i ) ).26MW ) 48) NT U UA ) C min 5750 exp NT U C r)) C r exp NT U C r )) exp )) ) 0.94 exp )) So the actual heat transfer rate is: Q act ɛq max W ) 5) T c,o T c,i + Q act 23.2 o C) 52) T h,o T h,i Q act 75.8 o C) 53) So the hot stream outlet temperature is 75.8 o C) if we use the smaller heat exchangers in parallel. This was 69.8 o C) using the existing large heat exchanger so the new system of two parallel heat exchanger can not do the desired cooling for us. Note that the difference is only about 6 o C) could be acceptable based on the project target applications. 7