Saturated vapor at 200 °C condenses to saturated liquid at the rate of 150 kg/s on the shell side of a heat exchanger (enthalpy of condensation h_fg = 2400 kJ/kg). A fluid with C_p = 4 kJ-kg^-1K^-1. enters at 100 °C on the tube side. If the effectiveness of the heat exchanger is 0.9, then the mass flow rate of the fluid in the tube side is __________ kg/s (in integer).

The correct answer is 1000.

Step-by-Step Explanation:

1. Identify the parameters given in the problem statement and the diagram:

From the heat exchanger schematic shown in the image, we can identify the following parameters:

- Condensation rate of saturated vapor (hot fluid) on the shell side: ${ṁ}_{\mathrm{cond}}=150\text{kg/s}$

- Enthalpy of condensation (latent heat of vaporization): $h_{\mathrm{fg}}=2400\text{kJ/kg}$

- Shell-side condensing temperature: $T_h=200\text{°C}$

- Specific heat capacity of the tube-side fluid: $C_p=4\text{kJ·kg}{\mathrm{–1}}^{}\text{K}{\mathrm{–1}}^{}$

- Inlet temperature of the tube-side fluid (cold fluid): $T_{c,i}=100\text{°C}$

- Effectiveness of the heat exchanger: $\epsilon =0.9$

2. Calculate the outlet temperature of the cold fluid using the effectiveness definition:

For a condenser where phase change occurs at a constant temperature, the heat capacity rate of the condensing fluid behaves as if it is infinitely large:

$C_h\to \infty$

Thus, the heat capacity ratio is $C_r=\frac{C_{\min }}{C_{\max }}\to 0$, making the cold fluid capacity rate the minimum capacity rate ($C_{\min }=C_c$). The effectiveness ($\epsilon$) simplifies to:

$\epsilon =\frac{T_{c,o}-T_{c,i}}{T_h-T_{c,i}}$

Substituting the values from the problem and image:

$0.9=\frac{T_{c,o}-100}{200-100}$

$0.9=\frac{T_{c,o}-100}{100}$

Solving for the cold fluid outlet temperature ($T_{c,o}$):

$T_{c,o}-100=90$

$T_{c,o}=190\text{°C}$

3. Use the Energy Balance principle:

The total heat rate released by the condensing steam must equal the heat rate absorbed by the cold fluid flowing through the tube side:

$Q_{\mathrm{released}}=Q_{\mathrm{absorbed}}$

${ṁ}_{\mathrm{cond}}\times h_{\mathrm{fg}}={ṁ}_c\times C_p\times (T_{c,o}-T_{c,i})$

Substituting the values into this energy balance equation:

$150\times 2400={ṁ}_c\times 4\times (190-100)$

$360000={ṁ}_c\times 4\times 90$

$360000={ṁ}_c\times 360$

Solving for the mass flow rate of the cold fluid (${ṁ}_c$):

${ṁ}_c=\frac{360000}{360}=1000\text{kg/s}$

Thus, the mass flow rate of the fluid on the tube side is 1000 kg/s.