Respuesta :
Answer:
the rate of heat transfer Q is Q =235.125 kJ/s
the heat transfer surface area A of the heat exchanger is A= 15.30 m²
Explanation:
Assuming negligible loss to the environment, then the heat flow of the hot water goes entirely to the cold water
Denoting a as cold water and b as hot water , then
Q= Fᵃ* cpᵃ * ( T₂ᵃ - T₁ᵃ)
where
F= mass flow
cp = specific heat capacity at constant pressure
T₂= final temperature
T₁ = initial temperature
replacing values
Q = Fᵃ* cᵃ * ( T₂ᵃ - T₁ᵃ) = 1.25 kg/s* 4180 J/kg·K* ( 60°C-15°C) * 1 kJ/1000J= 235.125 kJ/s
if all there is no loss to the surroundings
Qᵃ + Qᵇ = Q surroundings = 0
Fᵃ* cpᵃ * ( T₂ᵃ - T₁ᵃ) + Fᵇ* cpᵇ * ( T₂ᵇ - T₁ᵇ) = 0
T₂ᵇ = T₁ᵇ - [Fᵃ* cpᵃ / (Fᵇ* cpᵇ) ]* ( T₂ᵃ - T₁ᵃ)
replacing values
T₂ᵇ =100°C - [1.25 kg/s* 4180 J/kg·K/ (4 kg/s* 4190 J/kg·K)]* ( 60°C-15°C)
T₂ᵇ = 85.97 °C
the heat transfer surface of the heat exchanger is calculated through
Q = U*A* ΔTlm
where
U= overall heat transfer coefficient
A = heat transfer area of the heat exchanger
ΔTlm = (ΔTend - ΔTbeg)/ ln ( ΔTend - ΔTbeg)
ΔTbeg = temperature difference between the 2 streams at the inlet of the heat exchanger ( hot out - cold in) = 85.97 °C - 15°C = 70.97 °C
ΔTbeg = temperature difference between the 2 streams at the end of the heat exchanger ( hot in - cold out ) = 100°C - 60 °C = 40°C
then
ΔTlm = (ΔTend - ΔTbeg)/ ln ( ΔTend - ΔTbeg) =( 70.97 °C- 40°C)/ ln( 70.97°C/40°C) = 17.455 °C
ΔTlm = 17.455 °C
then
Q = U*A* ΔTlm
A = Q/(U*ΔTlm) = 235.125 kJ/s/(17.455 °C *880 W/m²*K) *1000 J/kJ = 15.30 m²
A= 15.30 m²
