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. Air flow data: T_3 = 18 degree C V = 240m^3/hr h_air = 200W/m^2K C_p = 1006J/kgK rho = 1.177kg/m^3 Part A. determining the following data: The air exit temperature, T_4 (degree C) The mass flow rate of the refrigerant, m^dot _R(kg/s) The overall heat transfer coefficient, U (W/m^2K) The depth of the heat exchanger (m) Refrigerant flow data: T_1 = T_2 = - 10.09 degree C x_1 = x_2 = 0.15 h_f = 37kJ/kg h_g = 241 kJ/kg C_P = 1308J/kgK mu = 0.01 times 10^-3Pa.s rho = 1334 kg/m^3 k = 0.0976W/mK Part B. Design Challenge: Quantify the main resistances to heat transfer Propose design changes to maximise performance and minimise size Illustrate your proposals with calculations, estimating the performance gains Heat exchanger data: Q = 2kW F = 0.43 A_f = 0.2m^2 alpha = 269m^2/m^3 D_i = 13.8mm D_0 = 16.4mm R_tubes = 5 k_tube = 237W/mK

Explanation / Answer

solution:

1)here we are designingcircular tube type of heat exchanger with cross flow of fluid where both fluidare unmixed with each other

2)here refrigerant is vapour state initially and during condensation it just lose heat to air by means of latent heat,hence heat loss by heat exchanger is

h1=hg1=241 kj/kg

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

heat loss by refrigerant is

Q=mr'(h1-h2)

as here are five tube means heat exchange by single is

Q=2/5=.4 kw

hence we get

mr'=2.306*10^-3 kg/s

3)where total heat loss by refrigerant is taken by air hence

density of air*V'*Cp*(T3-T4)=mr'(h1-h2)

T4=23.067 C

4)by Log mean temperature difference method for counter flow heat exchanger iswe have

dT1=Th1-Tc2=33.967 c

dT2=Th2-Tc1=28.9 c

hence mean temperature is

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

dTm=31.3653 c

5)for counterflow cross heat exchanger heat flow rate is givenas follows

Q=UAFdTm

hence we get UA=.02965

6)heat overall transfer coefficient at outer surface of tube is given by

Uo=(1/(Ro/Ri*hi)+(Roln(Ro/Ri)/k)+(1/ho))

here we dont hi,hence we have to find it by flow properties as follows

mass flow rate=density*V'

V'=1.7280*10^-6 m3/s

by continuity we get V'=A*v

v=.01155 m/s

reynold number=Re=v*Di/kinematic viscosity=21262.62

prandtl number=mu*Cp/k=.1338

hence nusselt number

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

Nu=36.45

Nu=hi*Di/k

hi=257.8337 w/m2c

hence on putting value in outer surface overall heat transfer coefficient we get

Uo=104.01

where UA=UoAo=.02965

Ao=2.85*10^-4 m2

Ao=pi*Do*L

L=5.5326*10^-3 m

6)Ao/v=269

Volme=1.0596*10^-6 m3

depth is=volume/Af=5.29*10^-6 m

7)resistance to convection at inner and outer surface are

R1=1/hiAi=16.69

R3=1/hoAo=17.54

by tube thickness,R2=(Ro-Ri)/kA=.02095

hence here more dominating resistane is resistance to convection at outer surface

7)one way to improve performance is to increase mass flow rate of refrigerant or change suitable refrigerant to enhance hi,change air to suitable fluid to enhance ho,as welll tube of conductive nature should be used

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