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Chapter 13: Problem 144

13-144 In a natural-gas fired boiler, combustion gases pass through 6-m-long 15 -cm-diameter tubes immersed in water at 1 atm pressure. The tube temperature is measured to be \(105^{\circ} \mathrm{C}\), and the emissivity of the inner surfaces of the tubes is estimated to be \(0.9\). Combustion gases enter the tube at 1 atm and \(1200 \mathrm{~K}\) at a mean velocity of \(3 \mathrm{~m} / \mathrm{s}\). The mole fractions of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) in combustion gases are 8 percent and 16 percent, respectively. Assuming fully developed flow and using properties of air for combustion gases, determine \((a)\) the rates of heat transfer by convection and by radiation from the combustion gases to the tube wall and \((b)\) the rate of evaporation of water.

Short Answer

Expert verified
1. Find the Nusselt number, Reynolds number, and Prandtl number using correlations for fully developed flow in a tube and properties of air at the average temperature. 2. Calculate the outlet temperature of gases using mass flow rate of the combustion gases and conservation of energy principle. 3. Calculate the rate of heat transfer by convection using the log mean temperature difference (LMTD) method and the heat transfer coefficient. 4. Calculate the rate of heat transfer by radiation using the Stefan-Boltzmann law and the gas film temperature (average of inlet and outlet temperatures). 5. Calculate the rate of evaporation of water using the total heat transfer rate and the specific enthalpy of vaporization.

Step by step solution

01

Find the Nusselt number, Reynolds number, and Prandtl number

We will use the below correlation for a fully developed flow in a tube to find the Nusselt number: Nu = 0.023*Re^0.8*Pr^0.4 To find the Reynolds number, we need the viscosity and properties of air at the inlet temperature: Re = (ρ*u*D)/μ Pr = (cp*μ)/k We will use the properties of air at the average temperature, (1200+373)/2 = 786.5 K. Since the temperature is given in Celsius, we must convert to Kelvin: T_w = 105+273.15=378.15 K. We can find the necessary properties in tables: For air at 786.5 K: ρ = 0.37 kg/m^3, cp = 1087 J/(kg*K), k = 0.040 W/(m*K), μ = 36.8 * 10^{-6} Pa*sec.
02

Calculate the outlet temperature of gases

To find the outlet temperature of the gases, we will first need to find the mass flow rate of the combustion gases, which can be found using the formula: m_dot = ρ*u*A where A = π*(D^2)/4 is the cross-sectional area of the tube. At a flow rate of 3 m/s, we have: m_dot = 0.37*3*(π*(0.15^2)/4)=0.0524 kg/s We can calculate the heat transfer coefficient (h) using the Nusselt number (Nu) calculated in step 1: h = (Nu*k)/D
03

Calculate the rate of heat transfer by convection (Q_conv) from the gases to the tube wall using the LMTD method.

We can use the formula for the log mean temperature difference (LMTD) as follows: LMTD = ((T_inlet - T_wall) - (T_outlet - T_wall))/ln((T_inlet - T_wall)/(T_outlet - T_wall)) We will need T_outlet to calculate the LMTD, which can be found by using the conservation of energy principle (ignoring radiation for now): Q_conv=m_dot*cp*(T_inlet-T_outlet) Now we can solve for T_outlet. Once we have the LMTD, we can use the formula for Q_conv: Q_conv = h*A*LMTD
04

Calculate the rate of heat transfer by radiation (Q_rad) from the gases to the tube wall.

To find the rate of heat transfer by radiation, we can use the Stefan-Boltzmann law: Q_rad = (σ*emissivity*A*(Tg^4 - Tw^4))/((1/emissivity)-1+(A*g/((1/albedo)*A₀))) where σ = 5.67*10^{-8} W/(m²*K^4) is the Stefan-Boltzmann constant, g is the gap between the tubes (0.15 m for these tubes), and albedo is the reflectance of the tube walls, which we can assume to be 1-emissivity. Note that we need to find the gas film temperature (Tg) to calculate Q_rad. For that, we can use the average of the inlet and outlet temperatures of the gas: Tg = (T_inlet+T_outlet)/2
05

Calculate the rate of evaporation of water (m_evp) using the total heat transfer rate (Q_total).

The total heat transfer rate (Q_total) is the sum of the heat transfer rates by convection and radiation: Q_total = Q_conv + Q_rad Since we know the specific enthalpy of vaporization (h_fg) at 1 atm pressure, we can now determine the rate of evaporation of water (m_evp) using the total heat transfer rate: m_evp = Q_total / h_fg where h_fg = 2257 kJ/kg (using steam tables at 1 atm pressure). And now we have determined both the rates of heat transfer by convection and radiation from the combustion gases to the tube wall, as well as the rate of evaporation of water.

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Most popular questions from this chapter

Two parallel disks of diameter \(D=3 \mathrm{ft}\) separated by \(L=2 \mathrm{ft}\) are located directly on top of each other. The disks are separated by a radiation shield whose emissivity is \(0.15\). Both disks are black and are maintained at temperatures of \(1200 \mathrm{R}\) and \(700 \mathrm{R}\), respectively. The environment that the disks are in can be considered to be a blackbody at \(540 \mathrm{R}\). Determine the net rate of radiation heat transfer through the shield under steady conditions.

A flow-through combustion chamber consists of 15 -cm diameter long tubes immersed in water. Compressed air is routed to the tube, and fuel is sprayed into the compressed air. The combustion gases consist of 70 percent \(\mathrm{N}_{2}\), 9 percent \(\mathrm{H}_{2} \mathrm{O}, 15\) percent \(\mathrm{O}_{2}\), and 6 percent \(\mathrm{CO}_{2}\), and are maintained at \(1 \mathrm{~atm}\) and \(1500 \mathrm{~K}\). The tube surfaces are near black, with an emissivity of \(0.9\). If the tubes are to be maintained at a temperature of \(600 \mathrm{~K}\), determine the rate of heat transfer from combustion gases to tube wall by radiation per \(m\) length of tube.

13-59 This question deals with steady-state radiation heat transfer between a sphere \(\left(r_{1}=30 \mathrm{~cm}\right)\) and a circular disk \(\left(r_{2}=120 \mathrm{~cm}\right)\), which are separated by a center-to- center distance \(h=60 \mathrm{~cm}\). When the normal to the center of disk passes through the center of the sphere, the radiation view factor is given by $$ F_{12}=0.5\left\\{1-\left[1+\left(\frac{r_{2}}{h}\right)^{2}\right]^{-0.5}\right\\} $$ Surface temperatures of the sphere and the disk are \(600^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\), respectively; and their emissivities are \(0.9\) and \(0.5\), respectively. (a) Calculate the view factors \(F_{12}\) and \(F_{21}\). (b) Calculate the net rate of radiation heat exchange between the sphere and the disk. (c) For the given radii and temperatures of the sphere and the disk, the following four possible modifications could increase the net rate of radiation heat exchange: paint each of the two surfaces to alter their emissivities, adjust the distance between them, and provide an (refractory) enclosure. Calculate the net rate of radiation heat exchange between the two bodies if the best values are selected for each of the above modifications.

Consider a person who is resting or doing light work. Is it fair to say that roughly one-third of the metabolic heat generated in the body is dissipated to the environment by convection, one-third by evaporation, and the remaining onethird by radiation?

A furnace is of cylindrical shape with a diameter of \(1.2 \mathrm{~m}\) and a length of \(1.2 \mathrm{~m}\). The top surface has an emissivity of \(0.70\) and is maintained at \(500 \mathrm{~K}\). The bottom surface has an emissivity of \(0.50\) and is maintained at \(650 \mathrm{~K}\). The side surface has an emissivity of \(0.40\). Heat is supplied from the base surface at a net rate of \(1400 \mathrm{~W}\). Determine the temperature of the side surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and side surfaces.
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