1. $1m^3/s$ of air is conveyed through a straight, horizontal duct of uniform cross-section and a length of 40 m. If the velocity of air through the duct is 5 m/s, find the required fan power input when a) A circular duct is used, and b) A rectangular duct of aspect ratio 1:4 is used. Take the efficiency of the fan to be 0.7. If a GI sheet of 0.5 mm thick with a density of 8000 kg/m3 is used to construct the duct, how many kilograms of sheet metal is required for circular and rectangular cross sections? Assume standard conditions and the static pressure at the inlet and exit of the duct to be same.

Answer:

From the continuity equation; the required cross-sectional area of the duct is given by:

$A_{cs}$=(Flow rate, Q/Velocity, V)=(1.0/5.0)=0.2m$^2$

(a) Circular duct:

The required diameter of the duct, $D = (4A_{cs}/\pi) = 0.50463 m$

Then using the equation for frictional pressure drop;

$\Delta p_f=\dfrac{0.022243Q_{air}^{1.852}L}{D^{4.973}}=\dfrac{0.022243(1.0)^{1.852}40}{(0.50463)^{4.973}}=26.7N/m^2$

Since the duct is straight, the dynamic pressure drop is zero in the absence of any fittings, hence:

Fan Total Pressure, FTP = ΔPf = 26.7 N/m$^2$

Hence the required fan power input, $W_{fan}$ is:

$W_{fan}=\dfrac{Q_{air}.FTP}{\eta_{fan}}=\dfrac{1.0\times26.7}{0.7}=38.14W$ (ans)

The required mass of the duct is given by:

$m_{circular} = (πD)\times(\text{thickness})\times(\text{density}) = 6.34$ kg (Ans.)

b) Rectangular duct: of aspect ratio (a:b) = (1:4)

As before, cross sectional area, $A_{cs} = 0.2 m^2 = a \times b =a\times4a$

⇒ a = 0.22361 m and b = 4a = 0.89440 m

The equivalent diameter, $D_{eq}$ is given by:

$D_{eq}=1.3\dfrac{(ab)^{0.625}}{(a+b)^{0.25}}=1.3\dfrac{(0.2)^{0.625}}{(0.22361+0.8944)^{0.25}}=0.46236m$

Using the friction equation, the frictional pressure drop is:

$\Delta p_f=\dfrac{0.022243Q_{air}^{1.852}L}{D^{4.973}}=\dfrac{0.022243(1.0)^{1.852}40}{(0.46236)^{4.973}}=41.24N/m^2$

Hence, the required fan power is:

$W_{fan}=\dfrac{Q_{air}.FTP}{\eta_{fan}}=\dfrac{1.0\times41.24}{0.7}=58.91W$ (ans)

The required mass of the rectangular duct is given by:

$m_{rectangular}= 2(a+b)\times(\text{thickness})\times(\text{density}) = 8.94$ kg (Ans.)

Thus for the same flow rate and velocity, a rectangular duct consumes 54.5% higher fan power and weighs 41% compared to a circular duct.

2. Air at a flow rate of 1.2 kg/s flows through a fitting with sudden enlargement. The area before and after the enlargements are 0.1$m^2$ and 1$m^2$, respectively. Find the pressure drop due to sudden enlargement using Borda-Carnot Equation. What is the pressure drop if the same amount of air flows through a sudden contraction with area changing from 0.1 $m^2$ and 1 $m^2$.

Answer:

Assuming standard air conditions, the density of air is approximately equal to 1.2 kg/m$^3$. Hence the volumetric flow rate of air, Q is given by:

Q=(mass flow rate/density)=(1.2/1.2)=1.0m$^3$/s

Sudden Enlargement:

From Borda-Carnot Equation; pressure drop due to sudden enlargement is given by:

$\Delta p_{d, enl}=\left(\dfrac{\rho V_1^2}{2}\right)\left(1-\dfrac{A_1}{A_2}\right)^2$

Velocity $V_1 = Q/A_1 = 1.0/0.1 = 10 m/s$ and $A_1/A_2 = 0.1/1.0 = 0.1$

Substituting the above values in the equation, we get:

$\Delta p_{d, enl}=\left(\dfrac{\rho V_1^2}{2}\right)\left(1-\dfrac{A_1}{A_2}\right)^2=\left(\dfrac{1.2\times10^2}{2}\right)\left(1-0.1\right)^2=48.6N/m^2$

Sudden contraction:

Pressure drop due to sudden contraction is given by Borda-Carnot equation:

$\Delta p_{d, con}=\left(\dfrac{\rho V_2^2}{2}\right)\left(\dfrac{A_2}{A_{1'}}-1\right)^2=\left(\dfrac{\rho V_2^2}{2}\right)\left(\dfrac 1{C_c}-1\right)^2$

From Table 37.3, for an area ratio ($A_2/A_1$) of 0.1, the contraction coefficient $C_c$ is 0.624. The velocity after contraction ($V_2$) is 10 m/s. Hence substituting these values in the above equation:

$\Delta p_{d, con}=\left(\dfrac{\rho V_2^2}{2}\right)\left(\dfrac 1{C_c}-1\right)^2=\left(\dfrac{1.2\times10^2}{2}\right)\left(\dfrac 1{0.624}-1\right)^2=36.15N/m^2$ (ans)

It can be seen from the example that for the same area ratio and flow rate, the pressure drop due to sudden enlargement is larger than that due to sudden contraction by about 34.4%.

3. Air at a flow rate of 1 $^3/s$ flows through a fitting whose cross-sectional area increases gradually from 0.08 $m^2$ to 0.12 $m^2$. If the static regain factor (R) of the fitting is 0.8, what is the rise in static pressure (static regain) and total pressure loss as air flows through the fitting?

Answer:

The velocity of air at the inlet and exit of the fitting are:

$V_{in}=1/A_{in}=1/0.08=12.5m/s$ and $V_{out}=1/A_{out}-1/0.12=8.33m/s$

Taking a value of 1.2kg/m$^3$ for the density of air, the velocity pressure at the inlet and exit are given by:

$P_{v, in} = (\rho V_{in}^2)/2 = (1.2 \times 12.5^2)/2 = 93.75 N/m^2$

$P_{v,out} = (\rho V_{out}^2)/2 = (1.2 \times 8.332)/2 = 41.63 N/m^2$

Static pressure rise through the fitting (static regain) is given by:

$(P_{s,out} − P_{s,in}) = R(P_{v,in} − P_{v,out}) = 0.7 \times (93.75 − 41.63) = 36.484 N/m^2$ (Ans.)

The loss in total pressure is given by:

$ΔP_{t,loss} = (1−R) (P_{v,in} − P_{v,out}) = 0.3 \times (93.75 − 41.63) = 15.636 N/m^2$ (Ans.)

4. Find the dimensions of a rectangular duct of aspect ratio (1:2) when 0.2 $m^3/s$ of air flows through it. The allowable frictional pressure drop is 3 Pa/m.

Answer:

For a flow rate of 0.2 $m^3/s$ and an allowable frictional pressure drop of 3 Pa/m, the equivalent diameter is found to be 0.2 m from friction chart or friction equation.

Then taking an aspect ratio of 1:2, the dimensions of the rectangular duct are found to be :

a ≈ 0.13 m and b ≈ 0.26 m. (Ans.)

5. The following figure shows a typical duct layout. Design the duct system using a) Velocity method, and b) Equal friction method. Take the velocity of air in the main duct (A) as 8 m/s for both the methods. Assume a dynamic loss coefficient of 0.3 for upstream to downstream and 0.8 for upstream to branch and for the elbow. The dynamic loss coefficients for the outlets may be taken as 1.0.Find the FTP required for each case and the amount of dampering required.

Answer:

a) Velocity method: Select a velocity of 5 m/s for the downstream and branches. Then the dimensions of various duct runs are obtained as shown below:

Segment A: Flow rate, $Q_A = 4 m^3/s$ and velocity, $V_A = 8 m/s$

⇒ cross-sectional area $A_A = Q_A/V_A = 4/8 = 0.5 m^2 ⇒ D_{eq,A} = 0.798 m$ (Ans.)

Segment B: Flow rate, $Q_B = 1 m^3/s$ and velocity, $V_B = 5 m/s$

⇒ cross-sectional area $A_B = Q_B/V_B = 1/5 = 0.2 m^2 ⇒ D_{eq,B} = 0.505 m$ (Ans.)

Segment C: Flow rate, $Q_C = 3 m^3/s$ and velocity, $V_C = 5 m/s$

⇒ cross-sectional area $A_C = Q_C/V_C = 3/5 = 0.6 m^2 ⇒ D_{eq,A} = 0.874 m$ (Ans.)

Segment D: Flow rate, $Q_D = 2 m^3/s$ and velocity, $V_D = 5 m/s$

⇒ cross-sectional area $A_D = Q_D/V_D = 2/5 = 0.4 m^2 ⇒ D_{eq,D} = 0.714 m$ (Ans.)

Segments E&F: Flow rate, $Q_{E,F} = 1 m^3/s$ and velocity, $V_{E,F} = 5 m/s$

⇒ cross-sectional area $A_{E,F} = Q_{E,F}/V_{E,F} = 1/5 = 0.2 m^2 ⇒ D_{eq,A} = 0.505 m$ (Ans.)

Calculation of pressure drop:

Section A-B:

$ΔP_{A-B} = ΔP_{A,f} + ΔP_{b,f} + ΔP_{u-b} + ΔP_{exit}$

where $ΔP_{A,f}$ and $ΔP_{B,f}$ stand for frictional pressure drops in sections A and B, respectively, $ΔP_{u-b}$ is the dynamic pressure drop from upstream to branch and $ΔP_{exit}$ is the dynamic pressure loss at the exit 1.

The frictional pressure drop is calculated using the equation:

$\Delta p_{A,f}=\dfrac{0.022243Q_{air}^{1.852}L}{D^{4.973}}=\dfrac{0.022243(4)^{1.852}\times15}{(0.798)^{4.973}}=13.35Pa$

$\Delta p_{B,f}=\dfrac{0.022243Q_{air}^{1.852}L}{D^{4.973}}=\dfrac{0.022243(1)^{1.852}\times6}{(0.505)^{4.973}}=3.99Pa$

The dynamic pressure drop from upstream to branch is given by:

$\Delta p_{u-b}=C_{u-b}\left(\dfrac{\rho V_d^2}{2}\right)=0.8\left(\dfrac{1.2\times5^2}{2}\right)=12pa$

The dynamic pressure drop at the exit is given by:

$\Delta p_{exit,1}=C_{exit}\left(\dfrac{\rho V_1^2}{2}\right)=1.0\left(\dfrac{1.2\times5^2}{2}\right)=15pa$

Hence total pressure drop from the fan to the exit of 1 is given by:

$ΔP_{A-B} = ΔP_{A,f} + ΔP_{b,f} + ΔP_{u-b} + ΔP_{exit} = 13.35+3.99+12+15 = 44.34 Pa$

In a similar manner, the pressure drop from fan to 2 is obtained as:

$ΔP_{A-C-D} = ΔP_{A,f} + ΔP_{C,f} + ΔP_{D,f} + ΔP_{u-b} + ΔP_{u-d} + ΔP_{exit}$

$ΔP_{A-C-D} = 13.35+3.99+2.57+12+4.5+15 = 51.41 Pa$

Pressure drop from fan to exit 3 is obtained as:

$ΔP_{A-C-E-F} = ΔP_{A,f}+ΔP_{C,f}+ΔP_{E,f}+ΔP_{F,f}+ΔP_{u-d,C}+ΔP_{u-d,E}+ΔP_{elbow}+ΔP_{exit}$

$ΔP_{A-C-E-F} = 13.35+3.99+11.97+3.99+4.5+4.5+12+15 = 69.3 Pa$

Thus the run with maximum pressure drop is A-C-E-F is the index run. Hence the FTP required is:

$FTP = ΔP_{A-C-E-F} = 69.3 Pa$ (Ans.)

Amount of dampering required at 1 = $FTP - ΔP_{A-B} = 24.96 Pa$ (Ans.)

Amount of dampering required at 2 = $FTP - ΔP_{A-C-D} = 17.89 Pa$ (Ans.)

b) Equal Friction Method:

The frictional pressure drop in segment A is given by:

$\Delta p_{f, A}=\dfrac{0.022243Q_{air}^{1.852}}{D^{4.973}}=\dfrac{0.022243(4)^{1.852}}{(0.798)^{4.973}}=0.89Pa/m$

The frictional pressure drops of B, C, D, E and F should be same as 0.89 Pa/m for Equal Friction Method. Hence, as discussed before:

$\left(\dfrac{Q^{1.852}}{D_{eq}^{4.973}}\right)_A=\left(\dfrac{Q^{1.852}}{D_{eq}^{4.973}}\right)_B=\left(\dfrac{Q^{1.852}}{D_{eq}^{4.973}}\right)_C=...$

From the above equation, we obtain:

$D_{eq, B}=D_{eq, A}\left(\dfrac{Q_B}{Q_A}\right)^\left({\dfrac{1.852}{4.973}}\right)=0.4762m$

$D_{eq, C}=D_{eq, A}\left(\dfrac{Q_C}{Q_A}\right)^\left({\dfrac{1.852}{4.973}}\right)=0.717m$

$D_{eq, D}=D_{eq, A}\left(\dfrac{Q_D}{Q_A}\right)^\left({\dfrac{1.852}{4.973}}\right)=0.6164m$

$D_{eq, E}=D_{eq, F}=D_{eq, A}\left(\dfrac{Q_E}{Q_A}\right)^\left({\dfrac{1.852}{4.973}}\right)=0.4762m$

Calculation of total pressure drop:

From fan to 1:

$ΔP_{A-B} = ΔP_{A,f} + ΔP_{B,f} + ΔP_{u-b} + ΔP_{exit}$

$ΔP_{A-B} = 13.35 + 5.34 + 15.1 + 18.9 = 52.69 Pa$

From fan to 2:

$ΔP_{A-C-D} = ΔP_{A,f} + ΔP_{C,f} + ΔP_{D,f} + ΔP_{u-d,C} + ΔP_{u-b} + ΔP_{exit}$

$ΔP_{A-C-D} = 13.35 + 10.68 + 5.34 + 9.94 + 21.55 + 26.9 = 87.76 Pa$

From fan to exit 3:

$ΔP_{A-C-E-F} = ΔP_{A,f}+ΔP_{C,f}+ΔP_{E,f}+ΔP_{F,f}+ΔP_{u-d,C}+ΔP_{u-d,E}+ΔP_{elbow}+ΔP_{exit}$

$ΔP_{A-C-E-F} = 13.35 + 10.68 + 16.02 + 5.34 + 9.94 + 5.67 + 15.1 + 18.9 = 95 Pa$

As before, the Index run is from fan to exit 3. The required FTP is:

$FTP = ΔP_{A-C-E-F} = 95 Pa$ (Ans.)

Amount of dampering required at 1 = $FTP - ΔP_{A-B} = 42.31 Pa$ (Ans.)

Amount of dampering required at 2 =$ FTP - ΔP_{A-C-D} = 7.24 Pa$ (Ans.)

From the example, it is seen that the Velocity method results in larger duct diameters due to the velocities selected in branch and downstream. However, the required FTP is lower in case of velocity method due to larger ducts.

Equal Friction method results in smaller duct diameters, but larger FTP.

Compared to velocity method, the required dampering is more at outlet 1 and less at outlet 2 in case of equal friction method.

6. A fan is designed to operate at a rotative speed of 20 rps. At the design conditions the airflow rate is 20 m3/s, the static pressure rise is 30 Pa and the air temperature is 20$^o$C. At these conditions the fan requires a power input of 1.5 kW. Keeping the speed constant at 20 rps, if the air temperature changes to 10$^o$C, what will be the airflow rate, static pressure and power input?

Answer:

At design condition 1,Rotative speed, $\Omega_1 = 20 rps$

Air temperature, $T_1 = 20^oC = 293$ K

Airflow rate,$Q_1 = 20 m^3/s$

Static pressure rise, $Δp_{s,1} = 30 Pa$

Power input, $W_1 = 1.5$ kW

At off-design condition 2, the rotative speed is same as 1, but temperature changes to 10$^o$C (283 K), which changes the density of air. To find the other variables, the fan law 2 has to be applied as density varies; i.e.,

i) Airflow rate $Q_1 = Q_2 = 20 m^3/s$

ii) Static pressure rise at 2,

$Δp_{s,2} = Δp_{s,1} (\rho_2/\rho_1) = Δp_{s,1}(T_1/T_2) = 30(293/283) = 31.06 Pa$ (Ans.)

iii) Power input at 2,

$W_2 = W_1(\rho_2/\rho_1) = W_1(T_1/T_2) = 1.5(293/283) = 1.553$ kW (Ans.)