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Chapter 3: Problem 80

Steam exiting the turbine of a steam power plant at \(100^{\circ} \mathrm{F}\) is to be condensed in a large condenser by cooling water flowing through copper pipes \(\left(k=223 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) of inner diameter \(0.4\) in and outer diameter \(0.6\) in at anerage temperature of \(70^{\circ} \mathrm{F}\). The heat of vaporization of water at \(100^{\circ} \mathrm{F}\) is \(1037 \mathrm{Btu} / \mathrm{lbm}\). The heat transfer coefficients are \(1500 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the steam side and \(35 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the water side. Determine the length of the tube required to condense steam at a rate of \(120 \mathrm{lbm} / \mathrm{h}\). Answer: \(1150 \mathrm{ft}\)

Short Answer

Expert verified
Based on the analysis and step-by-step solution above, the length of the copper tube required to condense steam at a rate of 120 lbm/h with given heat transfer coefficients and temperature differences is approximately 1150 feet.

Step by step solution

01

Calculate the heat transfer rate

In order to find the length of the copper tube required to condense steam at a specific rate, we first need to calculate the heat transfer rate. We can find the heat transfer rate (Q) by multiplying the mass flow rate of steam (m) and its heat of vaporization (h). So, Q = m * h where: Q = heat transfer rate (Btu/h) m = mass flow rate of the steam (120 lbm/h) h = heat of vaporization of water at 100°F (1037 Btu/lbm) Q = 120 lbm/h × 1037 Btu/lbm = 124440 Btu/h
02

Calculate individual thermal resistances

Now let's analyze the three thermal resistances which are involved in the process: Resistance on the steam side (R1), resistance of the copper tube (R2), and resistance on the water side (R3). The formulas for calculating these are given by: R1 = 1/(h1 * A1) R2 = ln(r2/r1)/(2 * π * k * L) R3 = 1/(h2 * A2) where: R1, R2, R3 = thermal resistances (h·ft·°F/Btu) h1 = heat transfer coefficient on the steam side (1500 Btu / h·ft²·°F) A1 = area of steam side (π * D1 * L) h2 = heat transfer coefficient on the water side (35 Btu / h·ft²·°F) A2 = area of water side (π * D2 * L) r1 = inner radius of the copper tube (0.2 in) r2 = outer radius of the copper tube (0.3 in) k = thermal conductivity of copper (223 Btu / h·ft·°F) D1 = inner diameter of the copper tube (0.4 in) D2 = outer diameter of the copper tube (0.6 in) L = length of the copper tube (ft)
03

Find the overall heat transfer coefficient

After calculating the individual thermal resistances, we can find the overall heat transfer coefficient (U) by taking the reciprocal of the total thermal resistance (R_total). R_total = R1 + R2 + R3 U = 1/R_total
04

Calculate the required length of the copper tube

Since we've already calculated the heat transfer rate (Q) and the overall heat transfer coefficient (U), we can now find the required length of the copper tube (L) by using the formula: Q = U * A * ΔT where: A = heat transfer area = π * D1 * L (ft²) ΔT = driving temperature difference between steam and water (100°F - 70°F) Rearranging the formula to solve for L: L = Q / (U * π * D1 * ΔT) When plugging in the values, make sure to convert the diameters into feet (1 in = 1/12 ft). Simplify and solve for L: L = 124440 Btu/h / (U * π * 0.4/12 ft * 30°F) Upon substituting the value of U obtained in Step 3, we find that: L ≈ 1150 ft Therefore, the length of the copper tube required to condense steam at a rate of 120 lbm/h is approximately 1150 feet.

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Most popular questions from this chapter

A 20-cm-diameter hot sphere at \(120^{\circ} \mathrm{C}\) is buried in the ground with a thermal conductivity of \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the center of the sphere and the ground surface is \(0.8 \mathrm{~m}\) and the ground surface temperature is \(15^{\circ} \mathrm{C}\). The rate of heat loss from the sphere is (a) \(169 \mathrm{~W}\) (b) \(20 \mathrm{~W}\) (c) \(217 \mathrm{~W}\) (d) \(312 \mathrm{~W}\) (e) \(1.8 \mathrm{~W}\)

Consider two identical people each generating \(60 \mathrm{~W}\) of metabolic heat steadily while doing sedentary work, and dissipating it by convection and perspiration. The first person is wearing clothes made of 1 -mm-thick leather \((k=\) \(0.159 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) that covers half of the body while the second one is wearing clothes made of 1 -mm-thick synthetic fabric \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that covers the body completely. The ambient air is at \(30^{\circ} \mathrm{C}\), the heat transfer coefficient at the outer surface is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the inner surface temperature of the clothes can be taken to be \(32^{\circ} \mathrm{C}\). Treating the body of each person as a 25 -cm-diameter, \(1.7-\mathrm{m}\)-long cylinder, determine the fractions of heat lost from each person by perspiration.

Determine the winter \(R\)-value and the \(U\)-factor of a masonry wall that consists of the following layers: \(100-\mathrm{mm}\) face bricks, 100 -mm common bricks, \(25-\mathrm{mm}\) urethane rigid foam insulation, and \(13-\mathrm{mm}\) gypsum wallboard.

An 8-m-internal-diameter spherical tank made of \(1.5\)-cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located in a room whose temperature is \(25^{\circ} \mathrm{C}\). The walls of the room are also at \(25^{\circ} \mathrm{C}\). The outer surface of the tank is black (emissivity \(\varepsilon=1\) ), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\).

Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend attaching fins both inside and outside the tubes?
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