Memorial University of NewfoundlandFaculty of Engineering and Applied Science

ENGI-7903, Mechanical Equipment, Spring 2011Assignment 2Vandad Talimi

Attempt all questions. The assignment may be done individually or in groups of two (2). No groupshall have more than two members. You may use Maple, Mathematica, MatLab, or Excel as needed toassist you.

Problem 1 - For a given pump, show the effect of a fluid change i.e. ρ1 6= ρ2 for the same rotationalspeed (ω1 = ω2). Another case (same powers) has been solved in class.

Solution - We know that ω1 = ω2, D1 = D2, and ρ1 6= ρ2, so:

H2

H1=

(ω2

ω1

)2 (D2

D1

)2

= (1)2 (1)2 = 1 (1)

and

Q2

Q1=

(ω2

ω1

)(D2

D1

)3

= (1) (1)3 = 1 (2)

and

W2

W1

=

(ρ2

ρ1

)(ω2

ω1

)3 (D2

D1

)5

=

(ρ2

ρ1

)(1)3 (1)5 =

ρ2

ρ1(3)

Problem 2 - The 11.25 (in) impeller option of the Taco Model 4013 Fl Series centrifugal pump of Fig. 1is used to pump water at 25 (oC) from a reservoir whose surface is 4.0 (ft) above the centerline of the pumpinlet as shown in Fig. 2. The properties of water at 25 (oC) are: ρ = 997.0(kg/m3), µ = 8.91×10−4(Pa.s),and Pv = 3.169(kPa). Standard atmospheric pressure is Patm = 101.3(kPa).

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Fig. 1- Pump curve, Taco 4013 Fl series

The piping system from the reservoir to the pump consists of 10.5 (ft) of cast iron pipe with an ID of4.0 (in) and an average inner roughness height of 0.02 (in). There are several minor losses: a sharp-edgedinlet (Kin = 0.5), three flanged 90o standard elbows (Kel = 0.3 each), and a fully open flanged globe valve(Kv = 6.0). Estimate the maximum volume flow rate (in units of (GPM)) that can be pumped withoutcavitation. Discuss how you might increase the maximum flow rate while still avoiding cavitation.

Pump

Inlet piping system

Globe Valve

Reservoir Z1

Flow Z2

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Fig. 2- Inlet piping system from the reservoir (point 1) to the pump inlet (point 2)

Solution - For the maximum flow rate without cavitation see the excel file on the course webpage. Theanswer is around 640 (GPM). This maximum flow rate can be increased in different ways. For example wecan use a simpler piping route in which only one elbow is used. (go to the excel file and set the quantity ofelbows to 1 and see the change in the graph!). We can also use a larger pipe diameter (again go to the excelfile and change the ID). Another way is using a gate valve instead of the globe valve if the application letus to do so. We can also place the pump closer to the reservoir or increase the vertical distance between Z1

2

and Z2. (You can see the effect of each of the proposed changes in the excel file by adjusting the necessarycells)

Problem 3 - Re-solve the example 5-3 using LMTD method. Compare your answer to what we gotin the class.

Solution - The energy balance gives us:

Q = mccp,c (Tc,o − Tc,i) = mhcp,h (Th,i − Th,o) (4)

So

Q = 1.2× 4180 (80− 20) = 2× 4310 (160− Th,o) (5)

This gives:

Q = 300960(W ) (6)

and

Th,i = 125.1(oC) (7)

The log mean temperature difference is: (counter flow)

∆TLMTD =∆T2 −∆T1

ln(

∆T2∆T1

) =(125.1− 20)− (160− 80)

ln(

125.1−20160−80

) = 91.98(oC) (8)

The heat transfer surface area is:

A =Q

U∆TLMTD=

300960

640× 91.98= 5.11(m2) (9)

And the heat exchanger length is:

L =A

πd=

5.11

3.1416× 0.015= 108.43(m) (10)

Which is close to our answer using ε−NTU method.

Problem 4 - A cross flow heat exchanger with both fluids unmixed has an overall heat transfercoefficient of 200 (W/m2.K), and a heat transfer surface area of 400 (m2). The hot fluid has a heat capacityof 40,000 (W/K), while the cold fluid has a heat capacity of 80,000 (W/K). If the inlet temperatures ofboth hot and cold fluids are 80 (oC) and 20 (oC), respectively, determine the exit temperature of the hotand cold fluids.

Solution - The minimum and maximum heat capacities are:

Cmin = Ch = 40000(W/K) (11)

Cmax = Cc = 80000(W/K) (12)

The capacity ratio is:

Cr =Cmin

Cmax= 0.5 (13)

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The maximum possible heat transfer rate is:

Qmax = Cmin (Th,i − Tc,i) = 40000 (80− 20) = 2.4(MW ) (14)

And the number of transfer units is:

NTU =UA

Cmin=

200× 400

40000= 2 (15)

Using the appropriate graph in the handouts we can read ε = 0.73. So the actual heat transfer rate is:

Qact = εQmax = 0.73× 2400000 = 1752000(W ) (16)

So the outlet cold temperature is:

Tc,o = Tc,i +Qact

Cc= 41.9(oC) (17)

And the outlet hot temperature is:

Th,o = Th,i −Qact

Ch= 36.2(oC) (18)

Problem 5 - In a 1-shell and 2-tube heat exchanger, cold water with inlet temperature of 20 (oC) isheated by hot water supplied at the inlet at 80 (oC). The cold and hot water flow rates are 5000 (kg/h) and10,000 (kg/h), respectively. If the shell and tube heat exchanger has a UA value of 11,600(W/K), determinethe cold water and hot water outlet temperatures. Assume cp,c = 4178(J/kg.K) and cp,h = 4188(J/kg.K).

Solution - The cold and hot streams heat capacities are:

Cc = mccp,c =5000

3600× 4178 = 5803.2(W/K) (19)

and

Ch = mhcp,h =10000

3600× 4188 = 11634.3(W/K) (20)

So the minimum and maximum heat capacities are:

Cmin = Cc = 5803.2(W/K) (21)

Cmax = Ch = 11634.3(W/K) (22)

The capacity ratio is:

Cr =Cmin

Cmax= 0.499 = 0.5 (23)

The maximum possible heat transfer rate is:

Qmax = Cmin (Th,i − Tc,i) = 5803.2 (80− 20) = 348192(W ) (24)

And the number of transfer units is:

NTU =UA

Cmin=

11600

5803.2= 1.99 (25)

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Using the appropriate graph in the handouts we can read ε = 0.7. So the actual heat transfer rate is:

Qact = εQmax = 0.7× 348192 = 243734.4(W ) (26)

So the outlet cold temperature is:

Tc,o = Tc,i +Qact

Cc= 62(oC) (27)

And the outlet hot temperature is:

Th,o = Th,i −Qact

Ch= 59(oC) (28)

Problem 6 - A counter flow double pipe heat exchanger with A = 9.0(m2) is used for cooling a liquidstream (cp = 3.15(kJ/kg.K)) at a rate of 10.0 (kg/s) with an inlet temperature of 90 (oC). The coolant(cp = 4.2(kJ/kg.K)) enters the heat exchanger at a rate of 8.0 (kg/s) with an inlet temperature of 10(oC). The plant data gave the following equation for the overall heat transfer coefficient in (W/m2.K):

U =600

1(mc)0.8

+ 2(mh)0.8

(29)

where mc and mh are the cold stream and hot stream flow rates in (kg/s), respectively.

a) Calculate the rate of heat transfer and 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 identicalheat exchangers that are presently stocked in its warehouse, each with A = 5.0(m2). Because the tubediameters in the existing and new units are the same, the above heat transfer coefficient is expected tobe valid for the new units as well. The vendor is proposing that the two new units could be operated inparallel, such that each unit would process exactly one half the flow rate of each of the hot and cold streamsin a counter flow manner; hence, they together would meet (or exceed) the present plant heat duty. Giveyour recommendation, with supporting calculations, on this replacement proposal.

Solution - Part a - The overall heat transfer coefficient is:

U =600

1(mc)0.8

+ 2(mh)0.8

=600

180.8

+ 2100.8

= 1184.73(W/m2.K) (30)

The cold and hot streams heat capacities are:

Cc = mccp,c = 8× 4200 = 33600(W/K) (31)

and

Ch = mhcp,h = 10× 3150 = 31500(W/K) (32)

So the minimum and maximum heat capacities are:

Cmin = Ch = 31500(W/K) (33)

Cmax = Cc = 33600(W/K) (34)

The capacity ratio is:

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Cr =Cmin

Cmax= 0.94 (35)

The maximum possible heat transfer rate is:

Qmax = Cmin (Th,i − Tc,i) = 31500 (90− 10) = 2.52(MW ) (36)

And the number of transfer units is:

NTU =UA

Cmin=

1184.73× 9

31500= 0.338 (37)

The ε can be calculated as follows:

ε =1− exp(−NTU(1− Cr))

1− Crexp(−NTU(1− Cr))=

1− exp(−0.338(1− 0.94))

1− 0.94× exp(−0.338(1− 0.94))= 0.253 (38)

So the actual heat transfer rate is:

Qact = εQmax = 0.253× 2520000 = 637560(W ) (39)

So the outlet cold temperature is:

Tc,o = Tc,i +Qact

Cc= 28.9(oC) (40)

And the outlet hot temperature is:

Th,o = Th,i −Qact

Ch= 69.8(oC) (41)

Part b - The overall heat transfer coefficient is:

U =600

1(mc)0.8

+ 2(mh)0.8

=600

140.8

+ 250.8

= 680.45(W/m2.K) (42)

The cold and hot streams heat capacities are:

Cc = mccp,c = 4× 4200 = 16800(W/K) (43)

and

Ch = mhcp,h = 5× 3150 = 15750(W/K) (44)

So the minimum and maximum heat capacities are:

Cmin = Ch = 15750(W/K) (45)

Cmax = Cc = 16800(W/K) (46)

The capacity ratio is:

Cr =Cmin

Cmax= 0.94 (47)

The maximum possible heat transfer rate is:

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Qmax = Cmin (Th,i − Tc,i) = 15750 (90− 10) = 1.26(MW ) (48)

And the number of transfer units is:

NTU =UA

Cmin=

680.45× 5

15750= 0.216 (49)

The ε can be calculated as follows:

ε =1− exp(−NTU(1− Cr))

1− Crexp(−NTU(1− Cr))=

1− exp(−0.216(1− 0.94))

1− 0.94× exp(−0.216(1− 0.94))= 0.177 (50)

So the actual heat transfer rate is:

Qact = εQmax = 0.177× 1260000 = 223020(W ) (51)

So the outlet cold temperature is:

Tc,o = Tc,i +Qact

Cc= 23.2(oC) (52)

And the outlet hot temperature is:

Th,o = Th,i −Qact

Ch= 75.8(oC) (53)

So the hot stream outlet temperature is 75.8 (oC) if we use the smaller heat exchangers in parallel. Thiswas 69.8 (oC) using the existing large heat exchanger so the new system of two parallel heat exchanger cannot do the desired cooling for us. Note that the difference is only about 6 (oC) and could be acceptablebased on the project target and applications.

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