Topic: Heat Transfer
A shell-and-tube heat exchanger (1 shell pass, 2 tube passes) is to be used to condense $2.73~kg/s$ of saturated steam at 340 K. Condensation occurs on the outer tube surfaces and the corresponding convection coefficient is $10,000~W/m^{2}K$. The temperature of the cooling water entering the tubes is $15^{\circ}C$ and the exit temperature is not to exceed $30^{\circ}C$. Thin-walled tubes of 19 mm diameter are to be used and the mean velocity of water flow through the tubes is to be maintained at $0.5~m/s$. Determine the minimum number of tubes that should be used? Water properties: $\rho=998~kg/m^{3}$, $\mu=0.000959~Pa.s$, $h_{fg}=2.34\times10^{6}J/kg$, $k=0.606~W/mK$, $C_{p}=4181~J/_{k}gK$
- A. 680
- B. 720
- C. 460
- D. 920
Answer: B. 720
Explanation: 1. Calculate heat transferred from steam: $Q = \dot{m}_s h_{fg} = 2.73 \times 2.34 \times 10^6 = 6.3882 \times 10^6~W$.
2. Determine mass flow rate of cooling water required: $Q = \dot{m}_w C_p \Delta T_w$.
$6388200 = \dot{m}_w \times 4181 \times (30 - 15) \Rightarrow \dot{m}_w = \frac{6388200}{4181 \times 15} = 101.86~kg/s$.
3. Calculate the mass flow rate through a single tube: $\dot{m}_{tube} = \rho \times A_{tube} \times V$.
$\dot{m}_{tube} = 998 \times \frac{\pi}{4}(0.019)^2 \times 0.5 = 998 \times 0.0002835 \times 0.5 = 0.1414~kg/s$.
4. The number of tubes per pass ($N_p$) is the total mass flow divided by the mass flow per tube:
$N_p = \frac{101.86}{0.1414} \approx 720.3$.
The minimum number of tubes per pass (effectively representing the tube bank size requirement to handle the flow rate at the specified velocity) is roughly 720.