/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 The condenser of a steam power p... [FREE SOLUTION] | 91Ó°ÊÓ

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The condenser of a steam power plant contains \(N=1000\) brass tubes \(\left(k_{\mathrm{t}}=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), each of inner and outer diameters, \(D_{i}=25 \mathrm{~mm}\) and \(D_{o}=\) \(28 \mathrm{~mm}\), respectively. Steam condensation on the outer surfaces of the tubes is characterized by a convection coefficient of \(h_{o}=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If cooling water from a large lake is pumped through the condenser tubes at \(m_{c}=400 \mathrm{~kg} / \mathrm{s}\), what is the overall heat transfer coefficient \(U_{o}\) based on the outer surface area of a tube? Properties of the water may be approximated as \(\mu=9.60 \times\) \(10^{-4} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\mathrm{Pr}=6.6 .\) (b) If, after extended operation, fouling provides a resistance of \(R_{f, i}^{\prime}=10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), at the inner surface, what is the value of \(U_{o}\) ? (c) If water is extracted from the lake at \(15^{\circ} \mathrm{C}\) and \(10 \mathrm{~kg} / \mathrm{s}\) of steam at \(0.0622\) bars are to be condensed, what is the corresponding temperature of the water leaving the condenser? The specific heat of the water is \(4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Short Answer

Expert verified
\(U_{o}\) without fouling is calculated by finding the inner heat transfer coefficient (\(h_{i}\)) using the Sieder-Tate correlation, and then computing \(U_{o}\) based on the resistance network. With fouling, we include the added resistance to compute \(U_{o}'\). Finally, we apply an energy balance for the condenser, considering the mass flow rate, specific heat, and latent heat of vaporization to find the outlet temperature of the cooling water, \(T_{out}\).

Step by step solution

01

Compute the inner heat transfer coefficient (h_i)

Using the Sieder-Tate correlation, we can estimate the inner heat transfer coefficient (h_i) as: \(h_{i}=0.023\left(\frac{4 m_{c}}{\pi D_{i}}\right)^{0.8} \mathrm{Pr}^{0.3}\frac{k}{D_{i}}\) We can plug in values for mc, Di, and Pr.
02

Compute the overall heat transfer coefficient without fouling (U_o)

Now, we can compute U_o using the resistance network as follows: \(\frac{1}{U_{o}}=R_{i}+R_{m}+R_{o}\) Where, \(R_{i}=\frac{1}{h_{i} A_{i}}\) \(R_{m}=\frac{\ln (D_{o} / D_{i})}{2 \pi k_{t} L}\) \(R_{o}=\frac{1}{h_{o} A_{o}}\) In which Ai and Ao are the inner and outer tube surface areas, respectively. Calculate each resistance term, then find U_o accordingly. #b) Calculating the Overall Heat Transfer Coefficient (U_o) with fouling#
03

Recalculate overall heat transfer coefficient (U_o) with fouling:

Now, we need to account for the added resistance due to fouling on the inner surface (R_f,i). The new overall heat transfer coefficient will be: \(\frac{1}{U_{o}'}=R_{i}+R_{f, i}+R_{m}+R_{o}\) Compute U_o' as described earlier, including the fouling resistance R_f,i. #c) Calculating the temperature of the water leaving the condenser#
04

Apply energy balance to the condenser:

Assuming no heat lost to the surroundings, we can write the energy balance equation as: \(m_{w} C_{p,w}(T_{out} - T_{in}) = m_{s} h_{fg}\) Where: mw = mass flow rate of the cooling water, Cp,w = specific heat of water, Tin = inlet temperature of the cooling water, Tout = outlet temperature of the cooling water, ms = mass flow rate of the steam, hfg = latent heat of vaporization. We are given water mass flow rate, inlet temperature, steam mass flow rate, and can find the latent heat of vaporization using steam tables for the given pressure. Now, solve for Tout.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Condenser Heat Transfer
The transfer of heat in a condenser is an essential aspect of thermal management in systems such as power plants and refrigeration cycles. In the context of a steam power plant, the condenser serves as a device to convert the exhaust steam from the turbine into liquid water by removing heat from the steam.

For an effective heat transfer process, we consider various coefficients that characterize how well heat is transferred from the steam to the cooling water. An important coefficient is the convection coefficient \( h_o \), which describes the heat transfer capability of the steam at the condenser outer surface. This coefficient greatly influences the overall heat transfer coefficient \( U_o \), which is a measure of the total heat transfer resistance including the condenser tube material and the characteristics of the inner fluid flow.

The efficiency of condensers is often limited by fouling, which affects the heat transfer and can be accounted for by adding fouling resistance into the calculation of \( U_o \). This process requires a detailed and multidimensional analysis of each component's effect on the overall efficiency of the heat transfer. The final goal is to ensure steam is effectively condensed while minimizing any losses in energy transfer.
The Impact of Fouling Resistance
Over time, the inner surfaces of the condenser tubes can accumulate deposits, a phenomenon known as fouling. This layer of unwanted material adds thermal resistance to the heat transfer process, described as fouling resistance \( R_{f, i} \). Fouling can drastically reduce the efficiency of a condenser by forming an insulating layer that impedes the flow of heat.

Fouling resistance increases the overall resistance to heat transfer and thus lowers the overall heat transfer coefficient \( U_o \). When calculating the impact of fouling, we add this resistance to the existing network of thermal resistances within the condenser. The greater the fouling resistance, the more significant the effect on the overall performance of the condenser, necessitating regular maintenance or cleaning to maintain heat transfer efficiency.
Energy Balance in Condensers
An energy balance in a condenser accounts for the energy transferred from the steam to the cooling water within the system. In an ideal scenario, the energy lost by the steam, which condenses as it loses heat, equals the energy gained by the cooling water, as it absorbs heat.

The energy balance can be mathematically expressed using the conservation of energy principle. The mass flow rate of the cooling water \( m_w \), its specific heat capacity \( C_{p,w} \), and the difference between its outlet and inlet temperatures \( T_{out} - T_{in} \), should equal the mass flow rate of the steam \( m_s \) multiplied by its latent heat of vaporization \( h_{fg} \).

This balance allows us to predict the exit temperature of the cooling water, \( T_{out} \), based on the amount of steam being condensed. If the amount of condensed steam increases, we expect the exit temperature of the cooling water to rise accordingly, assuming the specific heat and mass flow rate remain constant. Regular monitoring and adjustments based on energy balance are crucial for maintaining the desired performance levels of the condenser.
Applying the Sieder-Tate Correlation
The Sieder-Tate correlation is an empirical relationship used to calculate the internal heat transfer coefficient \( h_i \) for turbulent flow inside tubes. This correlation takes into account fluid properties, tube dimensions, and flow characteristics to provide an estimate of the heat transfer coefficient.

According to the correlation, \( h_i \) is proportional to \(0.023\left(\frac{4 m_c}{\pi D_i}\right)^{0.8} \Pr^{0.3}\frac{k}{D_i}\), where \( m_c \) is the mass flow rate of the cooling water, \( D_i \) is the inner diameter of the tube, \( \Pr \) is the Prandtl number, and \( k \) is the thermal conductivity of the fluid. By plugging in these values, we can calculate \( h_i \) and subsequently determine the overall heat transfer coefficient \( U_o \) for the condenser.

The Sieder-Tate correlation is particularly useful when precise measurements of the heat transfer coefficient are not available, allowing engineers and students alike to estimate the performance of heat exchange equipment with reasonable accuracy. However, it’s important to note that this correlation is valid under certain conditions, such as fully developed turbulent flow and assuming that the physical properties of the fluid are constant across the tube.

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Most popular questions from this chapter

A novel design for a condenser consists of a tube of thermal conductivity \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) with longitudinal fins snugly fitted into a larger tube. Condensing refrigerant at \(45^{\circ} \mathrm{C}\) flows axially through the inner tube, while water at a flow rate of \(0.012 \mathrm{~kg} / \mathrm{s}\) passes through the six channels around the inner tube. The pertinent diameters are \(D_{1}=10 \mathrm{~mm}, D_{2}=14 \mathrm{~mm}\), and \(D_{3}=50 \mathrm{~mm}\), while the fin thickness is \(t=2 \mathrm{~mm}\). Assume that the convection coefficient associated with the condensing refrigerant is extremely large. Determine the heat removal rate per unit tube length in a section of the tube for which the water is at \(15^{\circ} \mathrm{C}\).

The oil in an engine is cooled by air in a cross-flow heat exchanger where both fluids are unmixed. Atmospheric air enters at \(30^{\circ} \mathrm{C}\) and \(0.53 \mathrm{~kg} / \mathrm{s}\). Oil at \(0.026 \mathrm{~kg} / \mathrm{s}\) enters at \(75^{\circ} \mathrm{C}\) and flows through a tube of 10-mm diameter. Assuming fully developed flow and constant wall heat flux, estimate the oil- side heat transfer coefficient. If the overall convection coefficient is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total heat transfer area is \(1 \mathrm{~m}^{2}\), determine the effectiveness. What is the exit temperature of the oil?

Saturated process steam at 1 atm is condensed in a shell-and-tube heat exchanger (one shell, two tube passes). Cooling water enters the tubes at \(15^{\circ} \mathrm{C}\) with an average velocity of \(3.5 \mathrm{~m} / \mathrm{s}\). The tubes are thin walled and made of copper with a diameter of \(14 \mathrm{~mm}\) and length of \(0.5 \mathrm{~m}\). The convective heat transfer coefficient for condensation on the outer surface of the tubes is \(21,800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the number of tubes/pass required to condense \(2.3 \mathrm{~kg} / \mathrm{s}\) of steam. (b) Find the outlet water temperature. (c) Find the maximum possible condensation rate that could be achieved with this heat exchanger using the same water flow rate and inlet temperature. (d) Using the heat transfer surface area found in part (a), plot the water outlet temperature and steam condensation rate for water mean velocities in the range from 1 to \(5 \mathrm{~m} / \mathrm{s}\). Assume that the shell-side convection coefficient remains unchanged.

A cross-flow heat exchanger used in a cardiopulmonary bypass procedure cools blood flowing at \(5 \mathrm{~L} / \mathrm{min}\) from a body temperature of \(37^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\) in order to induce body hypothermia, which reduces metabolic and oxygen requirements. The coolant is ice water at \(0^{\circ} \mathrm{C}\), and its flow rate is adjusted to provide an outlet temperature of \(15^{\circ} \mathrm{C}\). The heat exchanger operates with both fluids unmixed, and the overall heat transfer coefficient is \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The density and specific heat of the blood are \(1050 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3740 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. a) Determine the heat transfer rate for the exchanger. b) Calculate the water flow rate. c) What is the surface area of the heat exchanger? d) Calculate and plot the blood and water outlet temperatures as a function of the water flow rate for the range 2 to \(4 \mathrm{~L} / \mathrm{min}\), assuming all other parameters remain unchanged. Comment on how the changes in the outlet temperatures are affected by changes in the water flow rate. Explain this behavior and why it is an advantage for this application.

A counterflow, concentric tube heat exchanger is designed to heat water from 20 to \(80^{\circ} \mathrm{C}\) using hot oil, which is supplied to the annulus at \(160^{\circ} \mathrm{C}\) and discharged at \(140^{\circ} \mathrm{C}\). The thin-walled inner tube has a diameter of \(D_{i}=20 \mathrm{~mm}\), and the overall heat transfer coefficient is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The design condition calls for a total heat transfer rate of \(3000 \mathrm{~W}\). (a) What is the length of the heat exchanger? (b) After 3 years of operation, performance is degraded by fouling on the water side of the exchanger, and the water outlet temperature is only \(65^{\circ} \mathrm{C}\) for the same fluid flow rates and inlet temperatures. What are the corresponding values of the heat transfer rate, the outlet temperature of the oil, the overall heat transfer coefficient, and the water- side fouling factor, \(R_{f,}^{n}\) ?

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