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Hot engine oil with a heat capacity rate of 4440 W/K (product of mass flow rate and specific heat) and an inlet temperature of 150ºC flows through a double pipe heat exchanger. The heat exchanger is constructed using a 1.5-m-long copper pipe (k = 250 W/m·K) with an inner tube of inside diameter 2 cm and outside tube diameter of 2.25 cm. The inner diameter of the outer tube is 6 cm. Oil flowing at a rate of 2 kg/s through the inner tube exits the heat exchanger at a temperature of 50ºC. The cold fluid, i.e., water, enters the heat exchanger at 20ºC and exits at 108ºC. Assume the fouling factor on the oil side and water side to be 0.00015 m²·K/W and 0.0001 m²·K/W, respectively.

The properties of water evaluated at the average inlet and exit temperatures of 20ºC and 70ºC (average 45ºC) are:

- [tex]\rho = 990.1 \, \text{kg/m}^3[/tex]
- [tex]C_p = 4180 \, \text{J/kg}·K[/tex]
- [tex]k = 0.637 \, \text{W/m}·K[/tex]
- [tex]\mu = 0.596 \times 10^{-3} \, \text{kg/m}·s[/tex]
- [tex]\text{Pr} = 3.91[/tex]

Calculate the mass flow rate of water. (Round the answer to three decimal places.)

Answer :

Answer:

The mass flow rate of water is 1.21 kg/s

Explanation:

according to the conditions given in the exercise:

mhcp=4440 W/K

Th1=150°C

Th2=50°C

Tc1=20°C

Tc2=108°C

di=2 cm

do=2.25 cm

Di=6 cm

mh=2 kg/s

Rfo=0.00015 m^2K/W

Rfi=0.0001 m^2K/W

The properties of the water at 45°C are:

p=990.1 kg/m^3

Cp=4180 J/kgK

From the energy balance, we have:

mc*Cpc(Tc2-Tc1)=mh*Cph(Th1-Th2)

Clearing mc:

[tex]m_{c}=\frac{m_{h}C_{ph}(T_{h1}-T_{h2}) }{C_{pc}(Tc_{2}-Tc_{1} }=\frac{4440(150-50)}{4180(108-20)} =1.21 kg/s[/tex]

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