An air-conditioning system employs horizontal, round sheet metal ducts of 6 and 12 in. diameter. What is the pressure change in a 400 ft³/min ﬂow of 70°F air through a sudden enlargement from the 6 in. to the 12 in. duct? What is the frictional pressure drop caused by this enlargement? Find the

Question

An air-conditioning system employs horizontal, round sheet metal ducts of 6 and 12 in. diameter. What is the pressure change in a 400 ft³/min ﬂow of 70°F air through a sudden enlargement from the 6 in. to the 12 in. duct? What is the frictional pressure drop caused by this enlargement? Find the corresponding quantities in a 400 ft³/min ﬂow through a sudden contraction between these duct sizes.

Accepted Answer

We are asked to determine the pressure change and frictional pressure drop caused by a speciﬁed sudden enlargement in a piping system and then to compare these values with the corresponding quantities caused by a contraction between the same two duct sizes. Figures 13.23 and 13.24 show the appropriate geometries for the expansion and contraction.

First we calculate the average velocity in the 6 in. and 12 in. ducts, respectively, as

[latex]\bar V_6=\frac{Q}{A}=\frac{400\ \mathsf{ft}^3/min }{(π/4)(0.5\ \mathsf{ft})^2}=2037\ \mathsf{ft}/min[/latex]

and

[latex]\bar V_{12}=\frac{Q}{A}=\frac{400\ \mathsf{ft}^3/min }{(π/4)(1\ \mathsf{ft})^2}=509\ \mathsf{ft}/min[/latex]

To analyze the enlargement, we will apply Eq. 13.29a:

[latex]\left(\frac{p_1}{\rho}+\frac{1}{2}\alpha _1\bar V^2_1+gz_1 \right)-\left(\frac{p_2}{\rho}+\frac{1}{2}\alpha _2\bar V^2_2+gz_2 \right)=\sum{h_L}+\sum{h_M} [/latex]

Assuming turbulent ﬂow, and noting that the duct is horizontal with no major loses, we write [latex](p_1/ρ + \frac{1}{2}\bar V^2 _1 )−(p_2/ρ + \frac{1}{2}\bar V^2 _2) =h_M [/latex], where for the enlargement point 1 is in the smaller duct. After combining Eqs. 13.35 and 13.37, we can write the minor loss as

[latex]h_M=K_E\frac{\bar V^2}{2}=\left[1-\left(\frac{D_1}{D_2} \right)^2 \right]^2 \frac{\bar V^2}{2}[/latex]

Thus the pressure change across the enlargement is given by

[latex]p_1 − p_2 =\frac{1}{2} ρ\left(\bar V^2 _2 −\bar V^2 _1\right)+ρ\left[1-\left(\frac{D_1}{D_2} \right)^2 \right]^2 \frac{\bar V_1^2}{2}[/latex]

The ﬁrst term is the frictionless pressure increase due to the air slowing down; the second term is the pressure drop due to friction. Inserting the data and evaluating each term separately, we have (with [latex]\bar V_1=\bar V_6[/latex] and [latex]\bar V_2=\bar V_{12}[/latex])

[latex]\frac{1}{2} ρ\left(\bar V^2 _2 −\bar V^2 _1\right)=\frac{1}{2}(2.329×10^{−3} slug/\mathsf{ft}^3)[(509\ \mathsf{ft}/min)^2 −(2037\ \mathsf{ft}/min)^2]\left(\frac{1\ min}{60\ s} \right)^2 [/latex]

= − 1.26 lb_f/ft²

[latex]ρ\left[1-\left(\frac{D_1}{D_2} \right)^2 \right]^2 \frac{\bar V^2_1}{2}= (2.329×10^{−3} slug/\mathsf{ft}^3)\left[1-\left(\frac{0.5}{1}\right)^2 \right]^2\left(\frac{1}{2} \right)(2037\ \mathsf{ft}/min)^2 \left(\frac{1\ min}{60\ s} \right)^2 [/latex]

= 0.75 lb_f/ft²

We see that that pressure change across the enlargement is

p₁ − p₂ = − 1.26 lb_f/ft² + 0.75 lb_f/ft² = − 0.51 lb_f/ft² [latex]\left(\frac{1\ in. H_2O}{5.2\ lb_{\mathsf{f}}/\mathsf{ft}^2} \right) [/latex]

= − 0.1 in. H₂O

which is an increase in pressure. To understand this result, note that the effect of the area increase is larger than that of the minor loss. The enlargement loss coefficient is seen to be

[latex]K_E=\left[1-\left(\frac{D_1}{D_2} \right)^2 \right]^2=\left[1-\left(\frac{0.5}{1} \right)^2 \right]^2=0.56[/latex]

To analyze the contraction we again apply Eq. 13.29a, and assume turbulent ﬂow to write [latex](p_1/ρ + \frac{1}{2}\bar V^2 _1 )−(p_2/ρ + \frac{1}{2}\bar V^2 _2) =h_M [/latex], where for the contraction, point 1 is in the larger duct. The minor loss is [latex]h_M = K_C(\bar V^2_2 /2)[/latex], where we read the value of [latex]K_C[/latex] from Figure 13.23. Thus the pressure change across the contraction is given by [latex]p_1 − p_2 = \frac{1}{2} ρ(\bar V^2 _2 − \bar V^2 1 )+ρK_C(\bar V^2 _2 /2)[/latex]. The ﬁrst term is the frictionless pressure decrease due to the air speeding up; the second term is the pressure drop due to friction. Note that the area ratio in this case is [latex]A_2/A_1 = D^2 _2/D^2_1 = (0.5\ \mathsf{ft} )^2/(1\ \mathsf{ft} )^2 =0.25[/latex], hence [latex]K_C[/latex] = 0.41. Inserting the data with [latex]\bar V_1=\bar V_6[/latex] and [latex]\bar V_2=\bar V_{12}[/latex], we have

[latex]\frac{1}{2} ρ\left(\bar V^2 _2 −\bar V^2 _1\right)=\frac{1}{2}(2.329×10^{−3} slug/\mathsf{ft}^3)[(2037\ \mathsf{ft}/min)^2-(509\ \mathsf{ft}/min)^2 ]\left(\frac{1\ min}{60\ s} \right)^2 [/latex]

= 1.26 lb_f/ft²

[latex]ρK_C \frac{\bar V^2_2}{2}= (2.329×10^{−3} slug/\mathsf{ft}^3)(0.41)\left(\frac{1}{2} \right)(2037\ \mathsf{ft}/min)^2 \left(\frac{1\ min}{60\ s} \right)^2 [/latex]

= 0.55 lb_f/ft²

We see that that pressure change across the contraction is

p₁ − p₂ = 1.26 lb_f/ft² + 0.55 lb_f/ft² = 1.81 lb_f/ft² [latex]\left(\frac{1\ in. H_2O}{5.2\ lb_{\mathsf{f}}/\mathsf{ft}^2} \right) [/latex] = 0.35 in. H₂O

which is a decrease due to the ﬂow speeding up, and a decrease due to friction.

Question 13.10: An air-conditioning system employs horizontal, round sheet m...

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