Chapter 7, Problem 40EP

A lightweight parachute is being designed for military use (Fig. P7-42E). Its diameter D is 20 ft and the total weight W of the falling payload, parachute, and equipment is 145 lbf. The design terminal settling speed V_t, of the parachute at this weight is 18 ft/s. A one-twelfth scale model of the parachute is tested m a wind tunnel. The wind tunnel temperature and pressure are the same as those of the prototype, namely 60°F and standard atmospheric pressure, (a) Calculate the drag coefficient of the prototype. (Hint: At terminal settling speed, weight is balanced by aerodynamic drag.) (b) At what wind tunnel speed should the wind tunnel be run in order to achieve dynamic similarity? (c) Estimate the aerodynamic drag of the model parachute in the wind tunnel (in lbf)

To determine

(a)

The drag coefficient of the prototype.

Answer to Problem 40EP

The drag coefficient of the prototype is1.202.

Explanation of Solution

Given information:

A lightweight parachute is being designed for military use.

Diameter = 20 ft

Total weight = 145 lbf

Design terminal speed (V_t) = 18 ft/s

  $\begin{array}{l}\text{For}\text{air}\text{at}\text{60}\text{°F:}\\ \text{ρ}\text{=}\text{0}\text{.07633}\text{lbm/}{\text{ft}}^3\\ \text{μ}\text{=}\text{1}\text{.213}\text{×}{\text{10}}^{\text{-5}}\text{lbm/ft}\text{.s}\end{array}$

Hint:

At terminal settling speed, weight is balanced by aerodynamic drag.

Drag coefficient is given as,

  ${\text{C}}_{\text{D}}\text{= }\frac{{\text{F}}_{\text{D}\text{.P}}}{\frac{\text{1}}{\text{2}}{\text{ρV}}_{\text{P}}{}^{\text{2}}{\text{A}}_{\text{P}}}$

Where,

  $\begin{array}{l}\text{ρ}\text{=}\text{pressure}\\ \text{V}\text{= velocity}\\ \text{A}\text{=}\text{area}\\ {\text{F}}_{\text{D}}\text{=}\text{aerodynamic}\text{drag}\end{array}$

  $\text{μ}=\text{Viscosity}$

On substituting the values,

  ${\text{C}}_{\text{D}}\text{= }\frac{\text{145}\text{lbf}}{\frac{\text{1}}{\text{2}}\left(\text{0}\text{.07633}\text{lbm/}{\text{ft}}^3\right){\left(\text{18}\text{ft/s}\right)}^{\text{2}}\left(\text{π}\frac{{\left(\text{20}\text{ft}\right)}^{\text{2}}}{\text{4}}\right)}\left(\frac{\text{32}\text{.2lbm}\text{ft}}{\text{lbf}{\text{s}}^{\text{2}}}\right)$

  $\begin{array}{l}{\text{C}}_{\text{D}}\text{= }\frac{\text{145}\text{lbf}}{\left(12.365\right)\left(\text{314}\text{.159}\right)}\left(\frac{\text{32}\text{.2lbm}\text{ft}}{\text{lbf}{\text{s}}^{\text{2}}}\right)\\ {\text{C}}_{\text{D}}\text{= }\frac{4669}{3884.57604}\\ {\text{C}}_{\text{D}}\text{=}1.2019\\ {\text{C}}_{\text{D}}\text{=}1.202\end{array}$

So, the value of drag coefficient is 1.202

To determine

(b)

The speed of wind tunnel to achieve dynamic similarities.

Answer to Problem 40EP

The air speed to run wind tunnel to achieve dynamic similarity is 216 ft/s.

Explanation of Solution

Given information:

A lightweight parachute is being designed for military use.

Diameter = 20 ft

Total weight = 145 lbf

Design terminal speed (V_t) = 18 ft/s

  $\begin{array}{l}\text{For}\text{air}\text{at}\text{60}\text{°F:}\\ \text{ρ}\text{=}\text{0}\text{.07633}\text{lbm/}{\text{ft}}^3\\ \text{μ}\text{=}\text{1}\text{.213}\text{×}{\text{10}}^{\text{-5}}\text{lbm/ft}\text{.s}\end{array}$

For similarity of model and prototype can be understood by calculating Reynold's number of both.

  $\begin{array}{l}\boxed{{\text{Re}}_{\text{model}}=\frac{{\text{ρ}}_{\text{m}}{\text{V}}_{\text{m}}{\text{L}}_{\text{m}}}{{\text{μ}}_{\text{m}}}}\\ \boxed{{\text{Re}}_{\text{prototype}}=\frac{{\text{ρ}}_{\text{P}}{\text{V}}_{\text{P}}{\text{L}}_{\text{P}}}{{\text{μ}}_{\text{P}}}}\\ \text{where,}\\ \text{ρ}\text{=}\text{pressure}\\ \text{V}\text{= velocity}\\ \text{L}\text{=}\text{length}\end{array}$

  $\text{μ}=\text{Viscosity}$

For dynamic similarity of model and prototype,

  $\boxed{{\text{Re}}_{\text{model}}=\frac{{\text{ρ}}_{\text{m}}{\text{V}}_{\text{m}}{\text{L}}_{\text{m}}}{{\text{μ}}_{\text{m}}}={\text{Re}}_{\text{prototype}}=\frac{{\text{ρ}}_{\text{P}}{\text{V}}_{\text{P}}{\text{L}}_{\text{P}}}{{\text{μ}}_{\text{P}}}}\mathrm{...}\text{(i)}$

On rearranging the equation (i) for speed of wind tunnel,

  ${\text{Re}}_{\text{model}}={\text{Re}}_{\text{prototype}}=\frac{{\text{ρ}}_{\text{m}}{\text{V}}_{\text{m}}{\text{L}}_{\text{m}}}{{\text{μ}}_{\text{m}}}=\frac{{\text{ρ}}_{\text{P}}{\text{V}}_{\text{P}}{\text{L}}_{\text{P}}}{{\text{μ}}_{\text{P}}}$

  $\begin{array}{l}\Rightarrow \frac{{\text{ρ}}_{\text{m}}{\text{V}}_{\text{m}}{\text{L}}_{\text{m}}}{{\text{μ}}_{\text{m}}}=\frac{{\text{ρ}}_{\text{P}}{\text{V}}_{\text{P}}{\text{L}}_{\text{P}}}{{\text{μ}}_{\text{P}}}\\ \Rightarrow {\text{V}}_{\text{m}}=\frac{{\text{μ}}_{\text{m}}}{{\text{ρ}}_{\text{m}}{\text{L}}_{\text{m}}}\left(\frac{{\text{ρ}}_{\text{P}}{\text{V}}_{\text{P}}{\text{L}}_{\text{P}}}{{\text{μ}}_{\text{P}}}\right)\end{array}$

On rearranging,

  ${\text{V}}_{\text{m}}={\text{V}}_{\text{P}}\left(\frac{{\text{ρ}}_{\text{P}}}{{\text{ρ}}_{\text{m}}}\right)\left(\frac{{\text{μ}}_{\text{m}}}{{\text{μ}}_{\text{P}}}\right)\left(\frac{{\text{L}}_{\text{P}}}{{\text{L}}_{\text{m}}}\right)$

On substituting the values,

  $\begin{array}{l}{\text{V}}_{\text{m}}=\left(\text{18}\text{ft/s}\right)\left(\frac{\text{0}\text{.07633}\text{lbm/}{\text{ft}}^3}{\text{0}\text{.07633}\text{lbm/}{\text{ft}}^3}\right)\left(\frac{\text{1}\text{.213}\text{×}{\text{10}}^{\text{-5}}\text{lbm/ft}\text{.s}}{\text{1}\text{.213}\text{×}{\text{10}}^{\text{-5}}\text{lbm/ft}\text{.s}}\right)\left(\frac{12}{\text{1}}\right)\\ {\text{V}}_{\text{m}}=\left(18\text{ft/s}\right)\left(1\right)\left(1\right)\left(12\right)\\ {\text{V}}_{\text{m}}=216\text{ft/s}\end{array}$

Thus, the air speed to run wind tunnel to achieve dynamic similarity is 216 ft/s.

To determine

(c)

The aerodynamics drag of the model parachute in the wind tunnel (in lbf).

Answer to Problem 40EP

The aerodynamics drag of the model parachute in the wind tunnel is 145lbf.

Explanation of Solution

Given information:

A lightweight parachute is being designed for military use.

Diameter = 20 ft

Total weight = 145 lbf

Design terminal speed (V_t) = 18 ft/s

  $\begin{array}{l}\text{For}\text{air}\text{at}\text{60}\text{°F:}\\ \text{ρ}\text{=}\text{0}\text{.07633}\text{lbm/}{\text{ft}}^3\\ \text{μ}\text{=}\text{1}\text{.213}\text{×}{\text{10}}^{\text{-5}}\text{lbm/ft}\text{.s}\end{array}$

For similarity of model and prototype can be understood by calculating Reynold's number of both.

  $\begin{array}{l}\boxed{{\text{Re}}_{\text{model}}=\frac{{\text{ρ}}_{\text{m}}{\text{V}}_{\text{m}}{\text{L}}_{\text{m}}}{{\text{μ}}_{\text{m}}}}\\ \boxed{{\text{Re}}_{\text{prototype}}=\frac{{\text{ρ}}_{\text{P}}{\text{V}}_{\text{P}}{\text{L}}_{\text{P}}}{{\text{μ}}_{\text{P}}}}\\ \text{where,}\\ \text{ρ}\text{=}\text{pressure}\\ \text{V}\text{= velocity}\\ \text{L}\text{=}\text{length}\end{array}$

  $\text{μ}=\text{Viscosity}$

For similarity of model and prototype,

  $\boxed{{\text{Re}}_{\text{model}}=\frac{{\text{ρ}}_{\text{m}}{\text{V}}_{\text{m}}{\text{L}}_{\text{m}}}{{\text{μ}}_{\text{m}}}={\text{Re}}_{\text{prototype}}=\frac{{\text{ρ}}_{\text{P}}{\text{V}}_{\text{P}}{\text{L}}_{\text{P}}}{{\text{μ}}_{\text{P}}}}\mathrm{...}\text{(i)}$

Reynolds number is similar for both model and prototype. Thus, drag force of prototype and model will also be equal.

  ${\text{Re}}_{\text{model}}={\text{Re}}_{\text{prototype}}=\frac{{\text{F}}_{\text{D}\text{.m}}}{{\text{ρ}}_{\text{m}}{\text{V}}_{\text{m}}{}^2{\text{L}}_{\text{m}}{}^2}=\frac{{\text{F}}_{\text{D}\text{.p}}}{{\text{ρ}}_{\text{P}}{\text{V}}_{\text{P}}{}^2{\text{L}}_{\text{P}}{}^2}$

  $\boxed{\Rightarrow \frac{{\text{F}}_{\text{D}\text{.p}}}{{\text{ρ}}_{\text{P}}{\text{V}}_{\text{P}}{}^2{\text{L}}_{\text{P}}{}^2}=\frac{{\text{F}}_{\text{D}\text{.m}}}{{\text{ρ}}_{\text{m}}{\text{V}}_{\text{m}}{}^2{\text{L}}_{\text{m}}{}^2}}$

  $\boxed{{\text{F}}_{\text{D}\text{.p}}={\text{F}}_{\text{D}\text{.m}}=145\text{lbf}}$

Now,

Drag force of prototype and model will be same. Due to following reasons: -

1. The acting fluid is same
2. Dynamic similarity is present
3. Same viscosity ratio of model and prototype.

Thus, the aerodynamics drag of the model parachute in the wind tunnel is 145 lbf.