Question

The condenser of a steam power plant operates at a pressure of 0.95 psia. The condenser consists of 144 horizontal tubes arranged in a \(12 \times 12\) square array. Steam condenses on the outer surfaces of the tubes whose inner and outer diameters are 1 in and 1.2 in, respectively. If steam is to be condensed at a rate of \(6800 \mathrm{lbm} / \mathrm{h}\) and the temperature rise of the cooling water is limited to \(8^{\circ} \mathrm{F}\), determine \((a)\) the rate of heat transfer from the steam to the cooling water and ( \(b\) ) the average velocity of the cooling water through the tubes.

Short answer

Answer: The rate of heat transfer from the steam to the cooling water is 57.03 kW, and the average velocity of the cooling water through the tubes is 1.22 m/s.

Step-by-step solution

Convert the mass flow rate of steam from lbm/h to kg/s

First, let's convert the mass flow rate of steam from lbm/h to kg/s. We have the following conversion factors: 1 lbm = 0.453592 kg 1 h = 3600 s So, $$ m_s = 6800 \frac{\mathrm{lbm}}{\mathrm{h}}\cdot\frac{0.453592\mathrm{kg}}{1\mathrm{lbm}}\cdot\frac{1\mathrm{h}}{3600\mathrm{s}} = 3.088\frac{\mathrm{kg}}{\mathrm{s}} $$.

Calculate the heat transfer rate from the steam to the cooling water using Q = mcΔT formula.

In this step, we'll find the heat transfer rate (Q) using the mass flow rate of steam (converted to kg/s) and the given temperature difference (\(\Delta T\)): The specific heat capacity for water (C_p) is approximately 4.186 kJ/kg·K. Since the problem is in °F and not °C, we need to convert ΔT in °F to K; we can use the relation 1°F = 5/9 K. Hence, \(\Delta T = 8^{\circ}\mathrm{F}\cdot\frac{5}{9} = 4.44\mathrm{K}\) Now, we can find the heat transfer rate: $$ Q = m_s \cdot C_p \cdot \Delta T = 3.088\frac{\mathrm{kg}}{\mathrm{s}} \cdot 4.186\frac{\mathrm{kJ}}{\mathrm{kg}\cdot\mathrm{K}} \cdot 4.44\mathrm{K} = 57.03\mathrm{kW} $$ So, the rate of heat transfer from the steam to the cooling water is 57.03 kW.

Calculate the average velocity of the cooling water through the tubes.

In this step, we'll calculate the average velocity (v) of the cooling water. We'll use the average velocity formula and calculate the flow rate Q first. First, let's convert the inner diameter to m (it's given in in). We have the following conversion factor: 1 inch = 0.0254 m So, \(D = 1\mathrm{in}\cdot\frac{0.0254\mathrm{m}}{1\mathrm{in}} = 0.0254\mathrm{m}\) Now, we can calculate the flow rate Q using the mass flow rate (m_s) and density of water (ρ): $$ Q = m_s\div \rho_{\mathrm{water}} = 3.088\frac{\mathrm{kg}}{\mathrm{s}}\div 1000\frac{\mathrm{kg}}{\mathrm{m}^3} = 0.003088\mathrm{m}^3\mathrm{/s} $$ Finally, we can find the average velocity in the tubes: $$ v = \frac{4\cdot Q}{\pi D^2} = \frac{4\cdot 0.003088\mathrm{m}^3\mathrm{/s}}{\pi(0.0254\mathrm{m})^2} = 1.22\frac{\mathrm{m}}{\mathrm{s}} $$ So, the average velocity of the cooling water through the tubes is 1.22 m/s.