An indoor fireplace is heated by gas and the hot gases move out through a 20m tall chimney. Hot gases are essentially at a constant temperature of 180 °C. Ambient outdoor air is at a temperature of 20 °C. If the inner diameter of the chimney is 203.2mm estimate the flow rate of gases through the

Question

An indoor fireplace is heated by gas and the hot gases move out through a 20m tall chimney. Hot gases are essentially at a constant temperature of [latex]180^{\circ} C[/latex]. Ambient outdoor air is at a temperature of [latex]20^{\circ} C[/latex]. If the inner diameter of the chimney is 203.2mm estimate the flow rate of gases through the chimney in [latex]m^{3} / h[/latex]. Assume, for simplicity, that the molecular weight of gas is the same as that of air. Pressure may be taken as 1 atm

Accepted Answer

Step 1 The given data is written down, using the notation introduced in the text.

[latex]T_{o}=273+20=293 K[/latex] , [latex]T_{i}=273+180=453 K[/latex] , p = 101300 Pa H = 20m

Under the assumption that the molecular weight of indoor and outdoor air are the same, the gas constant may be taken as R = 287 J/kg K. The densities of indoor and outdoor air are then obtained as

[latex] ho_{i}=\frac{101300}{287 imes 453}=0.779 kg / m ^{3}[/latex] and [latex] ho_{o}=\frac{101300}{287 imes 293}=1.205 kg / m ^{3}[/latex]

Pressure developed due to buoyancy is then given by

[latex]\Delta p=\left( ho_{o}- ho_{i} ight) g H=(1.205-0.779) imes 9.81 imes 20=83.5 Pa[/latex] (16.4)

Step 2 In order to calculate the velocity of air in the chimney, we need to know the value of loss coefficient K. This comprises of 3 parts. First one is due to entry loss where the hot air enters the chimney. Assuming that the entry is flush or square edged contraction (no data specified and we take the worst case) the loss coefficient is [latex]K_{i}=0.5[/latex]. The second one is the exit loss that is characterized by [latex]K_{e}=1[/latex]. The third one is due to friction in the chimney. This depends on the Reynolds number and is based on the friction factor calculated by the use of an appropriate relation for tube flow. Since the Reynolds number is based on as yet unknown velocity we need to use an iterative scheme for the solution. We start the solution by altogether ignoring the pressure drop due to fluid friction and use an initial K value of [latex]K^{(1)}=K_{i}+K_{e}=1+0.5=1.5[/latex] to obtain the first guess for the fluid velocity [latex]V^{(1)}[/latex] as

[latex]V^{(1)}=\sqrt{\frac{2 \Delta p}{K^{(1)} ho_{i}}}=\sqrt{\frac{2 imes 83.5}{1.5 imes 0.779}}=11.95 m / s[/latex]

Step 3 With the first guess for velocity available we estimate the contribution of fluid friction to the pressure drop now. The Reynolds number is calculated, based on inner diameter of chimney of I D = 203.2mm = 0.2032m and the kinematic viscosity of air at temperature of [latex]T_{i}=180^{\circ} C[/latex] given by [latex]v_{i}=[/latex] [latex]3.28 imes 10^{-5} m ^{2} / s[/latex] . Thus we have

[latex]R e_{I D}^{(1)}=\frac{V^{(1)} I D}{v_{i}}=\frac{11.95 imes 0.2032}{3.28 imes 10^{-5}}=74135.3[/latex]

Friction factor is then calculated as (using Equation 14.24(b))

[latex]f=\frac{0.184}{R e_{D}^{0.2}} ext { for } R e_{D}>2 imes 10^{4}[/latex] (14.24)(b)

[latex]f^{(1)}=\frac{0.184}{\left[R e_{I D}^{(1)} ight]^{0.25}}=\frac{0.184}{74135.3^{0.25}}=0.0112[/latex]

Then [latex]K_{f}^{(1)}[/latex] is calculated as

[latex]K_{f}^{(1)}=\frac{4 f^{(1)} H}{I D}=\frac{4 imes 0.0112 imes 20}{0.2032}=4.390[/latex]

Step 4 We may now update the overall K as [latex]K^{(2)}=K_{i}+K_{f}^{(1)}+K_{e}=0.5+4.39 + 1 = 5.89[/latex] . We may revise the velocity calculation and continue the iteration till convergence. The results are tabulated below.

Iteration number	K	V m/s	Change [latex]\Delta V[/latex] , m/s
1	1.5	11.952
2	5.89	6.032	-5.921
3	6.71	5.652	-0.380
4	6.79	5.616	-0.036
5	6.80	5.612	-0.003

Step 5 The converged value for the average velocity of flow through the chimney is taken as V = 5.612 m/s. With this velocity, the draft may be calculated as

[latex]Q=3600 imes \frac{\pi I D^{2} V}{4}=3600 imes \frac{\pi imes 0.2032^{2} imes 5.612}{4}=655.23 m ^{3} / h[/latex]

Question 16.1: An indoor fireplace is heated by gas and the hot gases move ...

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