A team of final year project students are designing a range of heating systems f

Question

A team of final year project students are designing a range of heating systems for an eco-house - including a heat pump. Heat pumps can be used to transfer heat against a temperature gradient: e.g. from outside to inside a house. This is achieved by upgrading the heat using the vapour compression cycle, a refrigerant, and heat exchangers. Heat loss calculations show that the house requires 2kW of heating power to maintain the inside temperature at a comfortable temperature. This is provided through a hot air flow, recirculating the air inside the building. The eco-house heat pump uses two "circular tube circular fin compact air blast heat exchangers" as highlighted above; one for the condenser and one for the evaporator. Your assignment is to carry-out design calculations for the condenser heat exchanger, and to recommend design changes to maximise performance, and minimise size. The air exis temperature, T_4 (degree C) The mass flow rate of the refrigerant, m_R (kg/s) The overall heat transfer coefficient, U (W/m^2 K) The depth of the heat exchanger (m) Quantify the main resistances to heat transfer Propose design changes to maximise performance and minimize size Illustrate your proposals with calculations, estimating the performance gains he best design proposals that are backed-up by accurate calculations will be incorporated into the eco-house design and build project heat-pump design!

Explanation / Answer

solution:

1)here for cross flow unmixed type counter flow heat exchanger ahs refrigerant flowing through tube and air on outside of tube

2)refrigerant enthalphy at inlet is

h1=hg1=241kj/kg

h2=hf1+x2(hg-hf)=67.6 kj/kg

2)total number of tube are 5,hence heat transfer through single tube is

Q=2/5=.4 kw

3)mass of refrigerant to remove heat of .4 kw is

Q=mr'*(h1-h2)

mr'=2.306*106-3 kg/s

4)where on heat balance

mr'*(h1-h2)=density*va*cp(T3-T4)

on putting value we get

T4=23.067 celsius

5)by LMTD method mean temperature difference for single tube is

Q=UAF(dTm)

here

for counter flow

dT1=23.067+10.9=33.967

dT2=28.9

hence dTm=(dT1-dT2)/ln(dT1/dT2)

dTm=31.3653 degree

hence from above relation

.4=UA*.43*31.3653

UA=.02965

6) for calculating overall heat transfer coefficient,we need to calculate heat transfer coefficient for refrigerant as follows

mr'=density*Q'

Q'=1.7286*10^-6 m3/s

Q'=A*V

Vi=.01155 m/s

reynold number is

Re=VDi/nu=21262.62

prandtl nimber is=mu*Cp/K r=.1338

where nusselt number is

Nu=.023*Re^.8*Pr^.3

we get

Nu=36.45=hoDi/k

hi=257.8337 w/m2k

7)outer surface overall heat transfer coefficient is

Uo=1/((r2/r1)(1/hi)+((r2/k)(ln(r2/r1))+(1/ho))

on putting all value we get

Uo=104.0158 w/m2c

8)where surface area of outer surface is

UoAo=UA=.02965

Ao=2.8505*10^-4 m2=pi*Do*L

L=5.5326*10^-3 m

9)where Ao/V=269

V=1.0596*10^-6 m3

10)where depthis

d=V/Af=5.2983*10^-6 m

11)here resistance at inner surface=R1=16.16

resistace due to tube=R2=.02095

resistance at outer surface=R3=17.54

12)hence resistance for convection at outer surface is maximum and to minimize size and maximize performance is achieved by reducing resistance at outer surface,by inducing fan for force convection

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