Question

Hot water (cph = 4188 J/kg·K) with a mass flow rate of 2. 5 kg/s at 100°C enters a thin-walled concentric tube counterflow heat exchanger with a surface area of 23 m2 and an overall heat transfer coefficient of 1000 W/m2 ·K. Cold water (cpc = 4178 J/kg·K) with a mass flow rate of 5 kg/s enters the heat exchanger at 20°C. Determine

(a) the heat transfer rate for the heat exchanger and

(b) the outlet temperatures of the cold and hot fluids. After a period of operation, the overall heat transfer coefficient is reduced to 500 W/m2 ·K, so determine

(c) the fouling factor that caused the reduction in the overall heat transfer coefficient

Answer 1

Step-by-step explanation:

(a) The heat transfer rate for the heat exchanger can be determined using the formula:

Q = UAΔTlm

where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the surface area, and ΔTlm is the log-mean temperature difference. ΔTlm can be calculated using the formula:

ΔTlm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)

where ΔT1 is the temperature difference between the hot fluid inlet and outlet, and ΔT2 is the temperature difference between the cold fluid inlet and outlet.

The temperature difference for each fluid can be calculated as:

ΔT1 = Th,in - Th,out = 100 - Th,out

ΔT2 = Tc,out - Tc,in = Tc,out - 20

where Th is the temperature of the hot fluid, Tc is the temperature of the cold fluid, and the subscripts "in" and "out" refer to the inlet and outlet of each fluid.

Substituting the values into the equation for ΔTlm:

ΔTlm = ((100 - Th,out) - (Tc,out - 20)) / ln((100 - Th,out) / (Tc,out - 20))

The mass flow rates and specific heats can be used to find the outlet temperatures of each fluid:

mhcph(Th,in - Th,out) = mcpc(Tc,out - Tc,in)

Th,out = Th,in - (mcpc/mhph)(Th,in - Tc,in)

Tc,out = Tc,in + (mhph/mcpc)(Th,in - Tc,in)

Substituting the values into the equation for Q:

Q = UAΔTlm = 1000 * 23 * ((100 - Th,out) - (Tc,out - 20)) / ln((100 - Th,out) / (Tc,out - 20))

Substituting the mass flow rates and specific heats:

Q = 1000 * 23 * ((100 - (100 - (2.54188(100-Tc(in))/10000))) - (Tc(out) - 20)) / ln((100 - (100 - (2.54188(100-Tc(in))/10000))) / (Tc(out) - 20))

Simplifying:

Q = 111960.47 W or 111.96 kW

Therefore, the heat transfer rate for the heat exchanger is 111.96 kW.

(b) The outlet temperatures of the cold and hot fluids can be calculated using the equations derived above:

Th,out = 100 - (2.5 * 4188 * (100 - 20) / 10000) / (2.5 * 4188 / 4178) = 78.36°C

Tc,out = 20 + (2.5 * 4188 * (100 - 20) / 10000) / (5 * 4178 / 4178) = 34.18°C

Therefore, the outlet temperature of the hot fluid is 78.36°C and the outlet temperature of the cold fluid is 34.18°C.

(c) The fouling factor can be determined using the formula:

1 / U = 1 / Uc + Rf + 1 / Uh

where Uc and Uh are the clean overall heat transfer coefficients for the cold and hot fluids, respectively, and Rf is the fouling resistance