Question

Steam is to be condensed on the shell side of a oneshell-pass and eight-tube- passes condenser, with 60 tubes in each pass at $90^{\circ} \mathrm{F}\left(h_{f g}=1043 \mathrm{Btu} / \mathrm{lbm}\right)$. Cooling water $\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\( enters the tubes at \)55^{\circ} \mathrm{F}$ and leaves at \(70^{\circ} \mathrm{F}\). The tubes are thin walled and have a diameter of $3 / 4\( in and a length of \)5 \mathrm{ft}$ per pass. If the overall heat transfer coefficient is $600 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2},{ }^{\circ} \mathrm{F}\(, determine \)(a)$ the rate of heat transfer, (b) the rate of condensation of steam, and (c) the mass flow rate of the cold water.

Short answer

Answer: The rate of heat transfer is 1,318,000 Btu/h, the rate of condensation of steam is 1,264.04 lbm/h, and the mass flow rate of cold water is 87,866.67 lbm/h.

Step-by-step solution

(Step 1: Calculate the rate of heat transfer)

First, we need to find the rate of heat transfer (\(Q\)) in the condenser using the overall heat transfer coefficient (\(U\)), the heat transfer area (\(A\)), and the temperature difference between the steam and cooling water (\(\Delta T\)). The formula for this is: \(Q=U \times A \times \Delta T\) We are given the overall heat transfer coefficient as U = \(600 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2},{}^{\circ} \mathrm{F}\), and we can calculate the heat transfer area from the given dimensions: \(A = 60 \text{ tubes} \times 8 \text{ passes} \times \pi (3 / 4 \text{ in}) \times 5 \text{ft} \times \dfrac{1 \text{ ft}}{12 \text{ in}} = 62.83 \mathrm{ft}^2\) The temperature difference between the steam and the cooling water inlet is given as \(\Delta T = 90 - 55 = 35^{\circ} \mathrm{F}\). Now we can calculate the rate of heat transfer: \(Q = 600 \times 62.83 \times 35 = 1,318,000 \mathrm{Btu} / \mathrm{h}\) So, (a) the rate of heat transfer is \(1,318,000 \mathrm{Btu} / \mathrm{h}\).

(Step 2: Calculate the rate of condensation of steam)

Next, we need to calculate the rate of condensation of steam (mass flow rate of steam, \(\dot{m}_{s}\)) using the heat transfer: \(\dot{m}_{s} = \dfrac{Q}{h_{fg}}\) where \(h_{fg}\) is the enthalpy difference between the liquid water and water vapor at \(90^{\circ}\mathrm{F}\), which is given as \(1043 \mathrm{Btu} / \mathrm{lbm}\). \(\dot{m}_{s} = \dfrac{1,318,000}{1043} = 1,264.04 \mathrm{lbm} / \mathrm{h}\) So, (b) the rate of condensation of steam is \(1,264.04 \mathrm{lbm} / \mathrm{h}\).

(Step 3: Calculate the mass flow rate of cold water)

Finally, we will determine the mass flow rate of the cooling water (\(\dot{m}_{w}\)) using the heat balance equation between the steam and cooling water: \(Q = \dot{m}_{w} \times c_p \times \Delta T_w\) where \(c_p\) is the specific heat of water, given as \(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\), and \(\Delta T_w\) is the temperature difference between the cooling water inlet and outlet, given as \(70 - 55 = 15^{\circ} \mathrm{F}\). \(\dot{m}_{w} = \dfrac{Q}{c_p \times \Delta T_w} = \dfrac{1,318,000}{1.0 \times 15} = 87,866.67 \mathrm{lbm} / \mathrm{h}\) So, (c) the mass flow rate of the cold water is \(87,866.67 \mathrm{lbm} / \mathrm{h}\).