A crossflow heat exchanger consists of 80 thinwalled tubes o

A cross-flow heat exchanger consists of 80 thin-walled tubes of 3-cm diameter located in a duct of 1 m times 1 m cross section. There are no fins attached to the tubes. Cold water (c_p = 4180 J/kg middot K) enters the tubes at 18 degree C with an average velocity of 3 m/s, while hot air (c_p = 1010 J/kg middot K) enters the channel at 130 degree C and 105 kPa at an average velocity of 12 m/s. If the overall heat transfer coefficient is 130 W/m^2 middot K, determine the outlet temperatures of both fluids and the rate of heat transfer.

Solution

Assumptions: The thickness of the tube is negligible.

Analysis: The mass flow rates of the hot and the cold fluids are

m_c = VAc = (1000 kg/m )(3 m/s)[80(0.03 m) /4] =169.6 kg/s

_air = P/RT = 105 kPa/ (0.287 kPa.m / kg.K) (130 + 273 K) = 0.908 kg / m³

m_h = VAc = (0.908 kg / m3)(12 m/ s)(1 m)2 = 10.90 kg / s

A_s = n*pi*DL = 80*pi*(0.03 m)(1 m) = 7.54 m²

C_c = m_cC_pc = (169.6 kg/s)(4.18 kJ/kg. C) = 708.9 kW/ deg.C

C_h = mhC_ph = (10.9 kg/s)(1.010 kJ/kg. C) = 11.01 kW/ deg.C

C_min = C_h =11.01 kW/deg.C

c = C_min/C_{max =} 11.01/ 708.9 = 0.01553
Q_max = C_min (T_h,in T_c,in ) = (11.01 kW/°C)(130°C18°C) =1233 kW

= NTU = UA_s/C_{min =} (130 W/m . C) (7.54 m )/ 11,010 W/ C

Q = *Q_max = (0.08513)(1233 kW) = 105 kW (Answer)

Q = C_c (T_c,out - T _c,in)

T_{c,out =} T _c,in + Q/C_{c =} 18 deg.C+ 105 kW/708.9 kW / C = 18.15 deg.C (Answer)

Q = C_h (T_h,out - T _h,in)

T_{h,out =} T _h,in - Q/C_{h =} 130 C - 105 kW/11.01 kW/ C = 120.5 deg.C (Answer)