A pressure drop Δp provides a measure of the frictional losses of a fluid as it flows through a pipe, Fig. 8–5. Determine how Δp is related to the variables that influence it, namely, the pipe diameter D, its length L, the fluid density ρ, viscosity μ, velocity V, and the relative roughness factor

Question

A pressure drop Δp provides a measure of the frictional losses of a fluid as it flows through a pipe, Fig. 8–5. Determine how Δp is related to the variables that influence it, namely, the pipe diameter D, its length L, the fluid density ρ, viscosity μ, velocity V, and the relative roughness factor ε/D, which is a ratio of the average size of the surface irregularities to the pipe’s diameter D.

Accepted Answer

Define the Physical Variables. In this case, [latex] \Delta p=f_1(D, L, ho, \mu, V, \varepsilon / D) [/latex] or [latex] f_2(\Delta p, D, L, ho, \mu, V, \varepsilon / D)=0 [/latex], and so n = 7. Using the M-L-T system in Table 8–1, the primary dimensions of the variables are

[latex] \begin{array}{ll} ext { Pressure drop, } \Delta p & M L^{-1} T^{-2} \ ext { Diameter, } D & L \ ext { Length, } L & L \ ext { Density, } ho & M L^{-3} \ ext { Viscosity, } \mu & M L^{-1} T^{-1} \ ext { Velocity, } V & L T^{-1} \ ext { Relative roughness, } \varepsilon / D & L L^{-1} \end{array} [/latex]

Since all three primary dimensions are involved, m = 3, and there will be (n - m) = (7 - 3) = 4 Π terms.

Select the Repeating Variables. Here D, V, ρ will be selected as the m = 3 repeating variables. Notice that ε/D cannot be selected because it is already dimensionless. Also, we cannot select the length in place of velocity since both the diameter and the length have the same dimensions and time T would not be represented.

Π Terms and Dimensional Analysis. The Π terms are constructed using our repeating variables D, V, ρ, along with q = Δp for [latex] \Pi_1 [/latex], [latex] q=L ext { for } \Pi_2, q=\mu ext { for } \Pi_3 ext {, and } q=\varepsilon / D ext { for } \Pi_4 ext {. Thus, for } \Pi_1 [/latex],

[latex] \begin{aligned} \Pi_1 & =D^a V^b ho^c \Delta p \ & =\left(L^a ight)\left(L^b T^{-b} ight)\left(M^c L^{-3 c} ight)\left(M L^{-1} T^{-2} ight)=M^{c+1} L^{a+b-3 c-1} T^{-b-2} \end{aligned} [/latex]

For M: 0 = c + 1
For L: 0 = a + b - 3c - 1
For T: 0 = -b - 2

Solving yields a = 0, b = -2, c = -1, and so

[latex] \Pi_1=D^0 V^{-2} ho^{-1} \Delta p=\frac{\Delta p}{ ho V^2} [/latex]

Next, for [latex] \Pi_2 [/latex],

[latex] \begin{aligned} \Pi_2 & =D^d V^e ho^h L \ & =\left(L^d ight)\left(L^e T^{-e} ight)\left(M^h L^{-3 h} ight)(L)=M^h L^{d+e-3 h+1} T^{-e} \end{aligned} [/latex]

For M: 0 = h
For L: 0 = d + e - 3h + 1
For T: 0 = -e

Solving yields d = -1, e = 0, h = 0, and so

[latex] \Pi_2=D^{-1} V^0 ho^0 L=\frac{L}{D} [/latex]

Now, for [latex] \Pi_3, [/latex],

[latex]\begin{aligned} \Pi_3 & =D^i V^j ho^k \mu \ & =\left(L^i ight)\left(L^j T^{-j} ight)\left(M^k L^{-3 k} ight)\left(M L^{-1} T^{-1} ight)=M^{k+1} L^{i+j-3 k-1} T^{-j-1} \end{aligned} [/latex]

For M: 0 = k + 1
For L: 0 = i + j - 3k - 1
For T: 0 = -j - 1

Solving yields i = -1, j = -1, k = -1, so

[latex] \Pi_3=D^{-1} V^{-1} ho^{-1} \mu=\frac{\mu}{D V ho} [/latex]

This is the inverse of the Reynolds number for the pipe, and so we will consider

[latex] \Pi_3^{-1}=\frac{ ho V D}{\mu}= Re [/latex]

Finally, for [latex] \Pi_4 [/latex],

[latex] \begin{aligned} \Pi_4 & =D^l V^m ho^n(\varepsilon / D) \ & =\left(L^l ight)\left(L^m T^{-m} ight)\left(M^n L^{-3 n} ight)\left(L L^{-1} ight)=M^n L^{l+m-3 n+1-1} T^{-m} \end{aligned} [/latex]

For M: 0 = n
For L: 0 = l + m - 3n + 1 - 1
For T: 0 = -m

Solving yields l = 0, m = 0, n = 0, and so

[latex] \Pi_4=D^0 V^0 ho^0\left(\frac{\varepsilon}{D} ight)=\frac{\varepsilon}{D} [/latex]

Realize that at the start, we could have saved some time in solving this problem and determined [latex] \Pi_2=L / D ext { and } \Pi_4=\varepsilon / D [/latex] simply by inspection, because each is a dimensionless ratio representing length over length. Had this inspection been done at the start, then only two Π terms would have to be determined.
In either case, our results indicate

[latex] f_2\left(\frac{\Delta p}{ ho V^2}, \operatorname{Re}, \frac{L}{D}, \frac{\varepsilon}{D} ight)=0 [/latex]

This equation can be solved for the ratio Δp/ρV² and then written in the form

[latex] \Delta p= ho V^2 f_3\left(\operatorname{Re}, \frac{L}{D}, \frac{\varepsilon}{D} ight) [/latex]

In Chapter 10, we will show how this relationship has important applications in the design of pipe systems.

Question 8.4: A pressure drop Δp provides a measure of the frictional loss...

Related Answered Questions

The pump in Fig. 8–14 is used to supply water through the 100-mm-diameter pipe at 3 m/s. If the measured power loss through the pipe is 3 kW, determine the power loss under similar conditions when a 75-mm-diameter pipe is used. ...

Verified Answer:

The spillway in Fig. 8–13 is to be built such that the estimated average flow over its crest will be Q = 3000 m³/s. Determine the required flow over the crest of a model built to a scale of 1/25. ...

Verified Answer:

A model of the car in Fig. 8–12 is to be constructed to a scale of 1/4 and is to be tested at 20°C in a water tunnel. Determine the required velocity of the water if the actual car is traveling at 30 m/s in air at this same temperature. ...

Verified Answer:

Verified Answer:

Establish the Reynolds number for a fluid flowing through the pipe in Fig. 8–2 using dimensional analysis, realizing that the flow is a function of the density ρ and viscosity μ of the fluid, along with its velocity V and the pipe’s diameter D. ...

Verified Answer:

Verified Answer:

Verified Answer: