Title: Default for 1.5 Formulae and Transposition
Course: HELM Workbooks 2023-2024

Using formulae and substitution

In the study of engineering, physical quantities can be related to each other using a formula. The formula will contain variables and constants which represent the physical quantities. To evaluate a formula we must substitute numbersin place of the variables.

For example, Ohm’s law provides a formula for relating the voltage, \(v\), across a resistor with resistance value, \(R\), to the current through it, \(i\). The formula states \[\begin{aligned} v &= iR\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{aligned}\]We can use this formula to calculate \(v\) if we know values for \(i\) and \(R\). For example, if \(i=13\) \(\textrm{A}\), and \(R=5\,\Omega\), then \[\begin{aligned} v &= iR\\ &= (13)(5)\\ &= 65\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{aligned}\]The voltage is 65 \(\textrm{V}\).

Note that it is important to pay attention to the units of any physical quantities involved. Unless a consistent set of units is used a formula is not valid.

Example

The kinetic energy, \(K\), of an object of mass \(M\) moving with speed \(v\) can be calculated from the formula, \(K=\dfrac{1}{2}Mv^2\). Calculate the kinetic energy of an object of mass 5 \(\textrm{kg}\) moving with a speed of 2 \(\textrm{ms}^{-1}\).

Solution

In this example M = 5 and v = 2. Substituting these values into the formula we find\[\begin{aligned} v &={1\over 2}Mv^2\\ &= {1\over 2}(5)(2^2)\\ &= 10 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{aligned}\]

Example

The volume, \(V\), of a circular cylinder is equal to its cross-sectional area, \(A\), times its length, \(h\). Find the volume of a cylinder having diameter 0.1 \(\textrm{m}\) and length 0.3 \(\textrm{m}\).

Solution

We can use the result of the previous Task to obtain the cross-sectional area \(A=\dfrac{\pi d^2}{4}\). Then\[\begin{aligned} V &= Ah\\ &= \dfrac{\pi d^2}{4}\times0.3\\ &= 0.0024 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{aligned}\]

Rearranging a formula

In the formula for the area of a circle, \(A=\pi r^2\), we say that \(A\) is the subject of the formula. A variable is the subject of the formula if it appears by itself on one side of the formula, usually the left-hand side, and nowhere else in the formula. If we are asked to transpose the formula for \(r\), or solve for \(r\), then we have to make \(r\) the subject of the formula. When transposing a formula whatever is done to one side is done to the other. There are five rules that must be adhered to.

Rearranging a formula:

Key Point

You may carry out the following operations:

- add the same quantity to both sides of the formula
- subtract the same quantity from both sides of the formula
- multiply both sides of the formula by the same quantity
- divide both sides of the formula by the same quantity
- take a ‘function’ of both sides of the formula: for example, find the reciprocal of both sides (i.e. invert).

Example

Transpose the formula \(p=5t-17\) for \(t\).

Solution

We must obtain \(t\) on its own on the left-hand side. We do this in stages by using one or more of the five rules in Key Point 22. For example, by adding 17 to both sides of \(p = 5t − 17\) we find

\(\begin{aligned}\qquad p + 17 &= 5t - 17 + 17\\ &=5t \end{aligned}\)

Example

Transpose the formula \(\sqrt{2q}=p\) for \(q\).

Solution

First we square both sides to remove the square root. Note that \(\sqrt(2q)^2 = 2q\). This gives

\(\begin{aligned}\qquad 2q &= p^2 \end{aligned}\)

Example

Transpose \(x=\dfrac{1}{y}\) for \(y\).

Solution

We must try to obtain an expression for y. Multiplying both sides by y has the effect of removing this fraction: Multiply both sides of \(x=\dfrac{1}{y} \text{ by } y\) to get

\(\begin{aligned}\qquad yx &= y \times \dfrac{1}{y}\\ yx &= 1\end{aligned}\)

Divide both sides by \(x\) to leave \(y\) on its own, \(y = \dfrac{1}{x}\).

Alternatively: simply invert both sides of the equation \(x = \dfrac{1}{y}\) to get \(\dfrac{1}{x} = y\)

Example

Make \(R\) the subject of the formula \(\dfrac{2}{R}=\dfrac{3}{x+y}\)

Solution

In the given form \(R\) appears in a fraction. Inverting both sides gives

\(\begin{aligned}\qquad \dfrac{R}{2} &= \dfrac{x+y}{3} \end{aligned}\)

Thus multiplying both sides by 2 gives

\(\begin{aligned}\qquad R &= \dfrac{2(x+y)}{3} \end{aligned}\)

Example

Make \(R\) the subject of the formula \(\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}\)

Solution

Add the 2 terms on the right: \( \dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{R_2+R_1}{R_1 R_2} \)

Write down the complete formula: \( \dfrac{1}{R} = \dfrac{R_2+R_1}{R_1 R_2} \)

Now invert both sides: \( R = \dfrac{R_1 R_2}{R_2+R_1}\)

Engineering Example: Heat flow in an insulated metal plate

Introduction

Thermal insulation is important in many domestic (e.g. central heating) and industrial (e.g cooling and heating) situations. Although many real situations involve heat flow in more than one dimension, we consider only a one dimensional case here. The flow of heat is determined by temperature and thermal conductivity. It is possible to model the amount of heat \(Q\) (\(\textrm{J}\)) crossing point \(x\) in one dimension (the heat flow in the \(x\) direction) from temperature \(T_2\) \((\textrm{K})\) to temperature \(T_1\) \((\textrm{K})\) (in which \(T_2>T_1\)) in time \(t\) (\(\textrm{s}\)) by

\(\qquad\dfrac{Q}{t}=\lambda A\left(\dfrac{T_2-T_1}{x}\right),\)

where \(\lambda\) is the thermal conductivity in \(\textrm{Wm}^{-1}\textrm{K}\).

Problem in words

Suppose that the upper and lower sides of a metal plate connecting two containers are insulated and one end is maintained at a temperature \(T_2\) \((\textrm{K})\) (see the figure below)

The plate is assumed to be infinite in the direction perpendicular to the sheet of paper.

[Image: A laterally insulated metal plate]
A laterally insulated metal plate

1. Find a formula for \(T\).
2. If \(\lambda =205\) \((\textrm{Wm}^{-1}\textrm{K})\), \(T_1 = 300\) \((\textrm{K})\), \(A=0.004\) \((\textrm{m}^2)\), \(x=0.5\) \((\textrm{m})\), calculate the value of \(T_2\) required to achieve a heat flow of \(100\) \((\textrm{Js}^{-1})\).

Mathematical statement of the problem

1. Given \(\dfrac{Q}{t}=\lambda A\left(\dfrac{T_2-T_1}{x}\right)\) express \(T_2\) as the subject of the formula.
2. In the formula found in part (a) substitute \(\lambda =205,\ T_1 =300,\ A=0.004\), \(x=0.5\) and \(\dfrac{Q}{t}=100\) to find \(T_2\).

Mathematical analysis

1. \(\dfrac{Q}{t}=\lambda A\left(\dfrac{T_2-T_1}{x}\right)\)
Divide both sides by \(\lambda A\)
\(\qquad\dfrac{Q}{t \lambda A}=\dfrac{T_2-T_1}{x}\)
Multiply both sides by \(x\)
\(\qquad\dfrac{Qx}{t \lambda A}=T_2-T_1\)
\(\qquad\dfrac{Qx}{t \lambda A}+T_1=T_2\)
which is equivalent to
\(\qquad T_2=\dfrac{Qx}{t \lambda A}+T_1\)
2. Substitute \(\quad \lambda =205,\ T_1 =300,\ A=0.004 ,\ x=0.5\) and \(\dfrac{Q}{t}=100\) to find \(T_2\): \[T_2=\dfrac{100\times 0.5}{205\times 0.004}+300 \approx 60.9 +300=360.9\] So the temperature in container 2 is \(361\) \((\textrm{K})\) to 3 sig.fig.

Interpretation

The formula \(\ T_2=\dfrac{Qx}{t \lambda A}+T_1\) can be used to find a value for \(T_2\) that would achieve any desired heat flow. In the example given \(T_2\) would need to be about \(361\) \((\textrm{K})\) \((\approx 78^\circ\textrm{C})\) to produce a heat flow of \(100\) \((\textrm{Js}^{-1})\).

Exercise

The formula for the volume of a cylinder is \(V = \pi r^2h \). Find \(V\) when \(r = [value]\), \(h = [value]\). Round your answer to 3 significant figures and include the appropriate units.

\(V =\)
Feedback: We have \(r = [value]\) and \(h = [value]\). Substituting these values into the formula we find:\[\begin{aligned} V &=\pi r^2h\\ &= \pi \cdot ([value])^2 \cdot [value]\\ &= \pi \cdot [value] \cdot [value]\\ &= \pi \cdot [value]\\ &= [value]\\ &= [value] \text{ }_{\text{3.s.f}} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{aligned}\]

Exercise

If \(R = 5p^2\), find R when:

1. \(p=[value], R =\)
2. \(p=[value], R =\)
Feedback: 1. We have \(p = [value]\). Substituting these values into the formula we find: \(\begin{aligned} R &=5p^2\\ &= (5)([value]^2)\\ &= (5)([value])\\ &= [value]. \end{aligned}\) 2. We have \(p = [value]\). Substituting these values into the formula we find: \(\begin{aligned} R &=5p^2\\ &= (5)([value]^2)\\ &= (5)([value])\\ &= [value]. \end{aligned}\)

Exercise

For the following formulae, find y at the given values of x:

1. \(y = [value]\)

  1. \(x = [value] \text{ , } y =\)
  2. \(x = [value] \text{ , } y =\)
  3. \(x = [value] \text{ , } y =\)
2. \(y = [value]\)

  1. \(x = [value] \text{ , } y =\)
  2. \(x = [value] \text{ , } y =\)
  3. \(x = [value] \text{ , } y =\)
3. \(y = [value]\)

  1. \(x = [value] \text{ , } y =\)
  2. \(x = [value] \text{ , } y =\)
  3. \(x = [value] \text{ , } y =\)
Feedback: Substituting these values into the given equations we have: 1. We are given the equation \(y = [value].\) 1. For \( x= [value], \quad y = [value] = [value] = [value].\) 2. For \( x= [value], \quad y = [value] = [value] = [value].\) 3. For \( x= [value], \quad y = [value] = [value] = [value].\) 2. We are given the equation \(y = [value].\) 1. For \( x= [value], \quad y = [value] = [value] =[value].\) 2. For \( x= [value], \quad y = [value]= [value] = [value].\) 3. For \( x= [value], \quad y = [value]= [value] = [value].\) 3. We are given the equation \(y = [value].\) 1. For \( x= [value], \quad y = [value] = [value] = [value].\) 2. For \( x= [value], \quad y = [value] = [value] = [value].\) 3. For \( x= [value], \quad y = [value] = [value] = [value].\)

Exercise

If \(P = \dfrac{3}{QR}\), find \(P \text{ if } Q = [value] \text{ and } R = [value]:\)

(Give your answer as a decimal to three decimal places, note that trailing zeroes are required i.e. \(2.3\) must be written as \(2.300\).)

\(P =\)
Feedback: In this example we have \(Q = [value] \text{ and } R = [value]\). Substituting these values into the formula we find: \[P = \sqrt{\dfrac{3}{QR}} = \sqrt{\dfrac{3}{([value])([value])}} = \sqrt{\dfrac{3}{[value]}} = [value]_{\text{ 3.d.p}} \]

Exercise

If \(y = \sqrt{\dfrac{x}{z}}\), find \(y \text{ when } x = [value] \text{ and } z = [value]:\)

(give your answer as a decimal to three decimal places, note that trailing zeroes are required i.e 2.3 must be written as 2.300)

\(y =\)
Feedback: In this example we have \(x = [value] \text{ and } y = [value]\). Substituting these values into the formula and then using a calculator we find: \[y = \sqrt{\dfrac{x}{z}} = \sqrt{\dfrac{[value]}{[value]}} = [value] = [value]_{\text{ 3.d.p}} \]

Exercise

If \(M = \dfrac{\pi}{2r+s}\), find \(M \text{ when } r = [value] \text{ and } s = [value]:\)

(give your answer as a decimal to three decimal places, note that trailing zeroes are required i.e 2.3 must be written as 2.300)

\(y =\)
Feedback: In this example we have \(r = [value] \text{ and } s = [value]\). Substituting these values into the formula and then using a calculator we find: \[\begin{align} y & = \dfrac{\pi}{2r+s} = \dfrac{\pi}{[value]} \\ &= \dfrac{\pi}{[value]} = \dfrac{\pi}{[value]} \\ &= [value] \\ &= [value]_\text{ 3.d.p} \end{align}\]

Exercise

To convert a length measured in metres to one measured in centimetres, the length in metres is multiplied by 100.

Convert the following lengths to cm.

1. [value]
2. [value]
3. [value]
Feedback: 1. \([value] \cdot 100 = [value]\), hence [value] is equivalent to [value] 2. \([value] \cdot 100 = [value]\), hence [value] is equivalent to [value] 3. \([value] \cdot 100 = [value]\), hence [value] is equivalent to [value]

Exercise

To convert an area measured in [value] to one measured in [value], the area in [value] is multiplied by 10000.

Convert the following lengths to [value].

1. [value]
2. [value]
3. [value]
Feedback: 1. \([value] \cdot 10000 = [value]\), hence [value] is equivalent to [value] 2. \([value] \cdot 10000 = [value]\), hence [value] is equivalent to [value] 3. \([value] \cdot 10000 = [value]\), hence [value] is equivalent to [value]

Exercise

To convert an area measured in [value] to one measured in [value], the area in [value] is multiplied by 1000000.

Convert the following lengths to [value].

1. [value]
2. [value]
3. [value]
Feedback: 1. \([value] \cdot 1000000 = [value]\), hence [value] is equivalent to [value] 2. \([value] \cdot 1000000 = [value]\), hence [value] is equivalent to [value] 3. \([value] \cdot 1000000 = [value]\), hence [value] is equivalent to [value]

Exercise

If \(\eta = \dfrac{4Q_{p}}{\pi d^{2}Ln}\), find \(\eta\) if:

\(Q_p = [value] \text{, } d = [value] \text{, } L = [value] \text{ and } R = [value].\)

(Give your answer as a decimal to three decimal places, note that trailing zeroes are required i.e. \(2.3\) must be written as \(2.300\).)

\(P =\)
Feedback: In this example we have \( Q_p = [value] \text{, } d = [value] \text{, } L = [value] \text{ and } R = [value]\). Substituting these values into the formula we find: \[\begin{align} \eta & = \dfrac{4Q_{p}}{\pi d^{2}Ln} = \dfrac{4\cdot[value]}{\pi \cdot[value]^2\cdot[value]\cdot[value]} \\ & = \dfrac{4 \cdot [value]}{\pi\cdot[value] \cdot [value]\cdot[value]} = \dfrac{[value]}{\pi \cdot[value]} \\ & = [value] \\ & = [value]_{\text{ 3.d.p}} \end{align}\]

Exercise

The moment of inertia of an object is a measure of its resistance to rotation. It depends upon both the mass of the object and the distribution of mass about the axis of rotation. It can be shown that the moment of inertia, J, of a solid disc rotating about an axis through its centre and perpendicular to the plane of the disc, is given by the formula

\(\qquad J = \dfrac{1}{2}Ma^2\)

where M is the mass of the disc and a is its radius. Find the moment of inertia of a disc of mass [value] and diameter [value]. The SI unit of moment of inertia is [value] [value].

\(J =\)
Feedback: The diameter of the spinning disk is [value] and therefore its radius is \(\dfrac{[value]}{2} = [value]\). The moment of inertia of the spinning disk is given by the formula \(J = \dfrac{1}{2}Ma^2\) and we have \(M = [value]\) and \(a = [value]\). Substituting these values into the formula we find: \[J = \dfrac{1}{2}Ma^2 = \dfrac{1}{2}\cdot [value] \cdot [value]^2 = \dfrac{1}{2}\cdot [value] \cdot [value] = \dfrac{1}{2}\cdot [value] = [value]\]

Exercise

Transpose, and then fully simplify, the given formulae to make the given variable the subject.

1. \([value]\) for \(x\),
2. \([value]\) for \(x\),
3. \([value]\) for \(y\),
4. \([value]\) for \(x\),
Feedback: 1. Adding [value] to both sides of the equation gives: \[[value]\] We can then divide both sides of the equation by [value] and simplify: \[[value]\] 2. Subtracting [value] from both sides of the equation gives: \[[value]\] We can then divide both sides by [value] giving: \[[value]\] 3. Subtracting[value] from both sides of the equality gives: \[[value]\] Then dividing both sides of the equation by [value] gives: \[[value]\] 4. Adding [value] to both sides of the equation gives: \[[value]\] Then dividing both sides of the equation by [value] gives \[[value]\]

Exercise

Transpose the formula [value] for [value]
Feedback: Since we are trying to make [value] the subject of the formula we can divide both sides by [value] which gives: \[\dfrac{[value]}{[value]} = \dfrac{[value]}{[value]}\] This simplifies to \[[value]\]

Exercise

Transpose [value]

1. for \(x\),
2. for \(y\),
Feedback: For both parts raise both sides of the equation to the power of [value] to give: \[[value]\] 1. To then make x the subject of the equation subtract [value] from both sides giving: \[[value]\] 2. To make y the subject of the equation first subtract x from both sides giving: \[[value]\] Then divide both sides of the equation by [value] giving: \[[value]\]

Exercise

Transpose the formula [value] for [value]
Feedback: Since we are trying to make [value] the subject of the formula we first, to both sides of the equation, [value] \([value]\) which gives: \[[value]\] We then subtract 17 from both sides of the equation giving: \[[value]\]. Finally we divide both sides of the equation by [value] giving: \[[value]\]

Exercise

When a ball is dropped from rest onto a horizontal surface it will bounce before eventually coming to rest after a time \(T\) where \[T=\dfrac{[value]v}{g} (\dfrac{[value]}{1-e})\] where \(v\) is the speed immediately after the first impact, and \(g\) is a constant called the acceleration due to gravity. Transpose this formula to make \(e\), the coefficient of restitution, the subject.

\(e\) =
Feedback: We have the equation: \[T=\dfrac{[value]v}{g} (\dfrac{[value]}{1-e})\] First multiply both sides by g: \[\implies Tg = [value]v (\dfrac{[value]}{1-e})\] Then divide both sides by \([value]v\): \[\implies \dfrac{Tg}{[value]v} = \dfrac{[value]}{1-e}\] Multiply both sides by (1-e) and then expand out the brackets: \[\implies \dfrac{Tg}{[value]v}(1-e) = [value] \implies \dfrac{Tg}{[value]v} - \dfrac{Tg}{[value]v}e = [value]\] Add \(\dfrac{Tg}{[value]v}e\) to both sides: \[\implies \dfrac{Tg}{[value]v} = [value] + \dfrac{Tg}{[value]v}e\] Divide both sides by \(\dfrac{Tg}{[value]v}\) and simplify the fraction: \[\implies 1 = \dfrac{[value]}{\dfrac{Tg}{[value]v}} + e \implies 1 = \dfrac{[value]v}{Tg} + e\] Finally subtracting \(\dfrac{[value]v}{Tg}\) from both of the equation makes \(e\) the subject of the formula: \[\implies e = 1 - \dfrac{[value]v}{Tg}\]

Exercise

Transpose \(q = A_{1}\sqrt{\dfrac{[value]gh}{(A_1/A_2)^{2}-[value]}}\text{ for }A_{2}\)

\(A_{2}\) =
Feedback: We have the equation \[q = A_{1}\sqrt{\dfrac{[value]gh}{(A_{1}/A_{2)}^{2}-[value]}}\] First divide the equation by \(A_{1}\) and then square both sides \[\dfrac{q}{A_{1}} = \sqrt{\dfrac{[value]gh}{(A_1/A_2)^{2}-[value]}} \implies \dfrac{q^2}{A_{1}^2} = \dfrac{[value]gh}{(A_1/A_2)^{2}-[value]}\] Multiply both sides of the equation by \((\dfrac{A_{1}}{A_{2}})^{2}-[value]\) \[\dfrac{q^2}{A_{1}^2}((\dfrac{A_{1}}{A_{2}})^{2}-[value]) = [value]gh\] Expand out both sets of brackets \[\dfrac{q^2}{A_{1}^2}(\dfrac{A_{1}}{A_{2}})^{2}-\dfrac{[value] }{A_{1}^2}= [value]gh \implies \dfrac{q^2A_{1}^2}{A_{1}^2A_{2}^2}-\dfrac{[value]}{A_{1}^2} = [value]gh\] The \(A_{1}^{2}\) in the numerator and denominator cancel each other out \[\dfrac{q^2}{A_{2}^2}-\dfrac{[value]}{A_{1}^2} = [value]gh\] Add to both sides \(\dfrac{[value]}{A_{1}^2}\) \[\dfrac{q^2}{A_{2}^2} = [value]gh + \dfrac{[value]}{A_{1}^2}\] Multiply both sides by \(A_{2}^2\) \[q^2 = A_{2}^2([value]gh + \dfrac{[value]}{A_{1}^2})\] Divide both sides of the equation \(([value]gh + \dfrac{[value]}{A_{1}^2})\) \[A_{2}^2 = \dfrac{q^2}{[value]gh+\dfrac{[value]}{A_{1}^2}}\] Multiply \([value]gh\) by \(\dfrac{A_{2}^2}{A_{2}^2}\) and then simplify to make one fraction \[A_{2}^2 = \dfrac{q^2}{\dfrac{[value]ghA_{1}^2}{A_{1}^2}+\dfrac{[value]}{A_{1}^2}} = \dfrac{q^2}{\dfrac{[value]ghA_1^2+[value]}{A_{1}^2}} = \dfrac{q^2A_{1}^2}{[value]ghA_1^2+[value]}\] Taking the square root of both sides of the equation gives the final answer \[A_{2} = \sqrt{\dfrac{q^2A_{1}^2}{[value]ghA_1^2+[value]}}\]

Exercise

Make \(x\) the subject of

1. \(y = \dfrac{[value]+x}{1-[value]x}\), \(x =\)
2. \(y = \sqrt{\dfrac{x-[value]}{x+1}}\), \(x =\)
Feedback: 1. We start with the equation \[y = \dfrac{[value]+x}{1-[value]x}\] First multiply both sides by \(1-[value]x\) and expand the brackets \[\implies y(1-[value]x) = [value]+x \implies y - [value]xy = [value] + x\] Add \([value]xy\) to both sides of the equation \[\implies y = [value] + x + rxy\] Subtract \([value]\) \[\implies y - [value] = x + [value]xy\] Factorise x out of the right-hand side of the equation \[\implies y - [value] = x(1 + [value]y)\] Divide both sides of the equation by \(1 + [value]y\) to make x the subject of the equation \[\implies x = \dfrac{y-[value]}{1 + [value]y}\] 2. We start with the equation \[y = \sqrt{\dfrac{x-[value]}{x+1}}\] First square both sides of the equation \[\implies y^2 = \dfrac{x-[value]}{x+1}\] multiply both sides of the equation by \(x+1\) and expand the brackets \[\implies y^2(x+1) = x-[value] \implies xy^2+y^2 = x-[value]\] Subtract \(xy^2\) from both sides \[y^2 = x-[value]-xy^2\] Add \([value]\) to both sides of the equation \[[value] + y^2 = x - xy^2\] Factorise \(x\) out of the right-hand side of the equation \[[value] + y^2 = x(1 - y^2)\] Divide both sides of this equation by \(1 - y^2\) to make \(x\) the subject of the equation \[x = \dfrac{[value] + y^2}{1 - y^2}\]

Exercise

In the design of orifice plate flowmeters, the volumetric flowrate, \(Q\) \((\textrm{m}^{2}\textrm{s}^{-1})\), is given by \[Q = CO\sqrt{\dfrac{2g\Delta h}{1 - O^{2}/P^{2}}}\] where \(C\) is a dimensionless discharge coefficient, \(\Delta h \textrm{ (m)}\) is the head difference across the orifice plate and \(O\) \((\textrm{m}^{2})\) is the area of the orifice and \(P\) \((\textrm{m}^{2})\) is the area of the pipe.

1. Rearrange the equation to solve for the area of the orifice, \(O\), in terms of the other variables \(O =\)
2. A volumetric flowrate of [value] \(\textrm{cm}^{3}\textrm{s}^{-1}\) passes through a [value] \(\textrm{cm}\) inside diameter pipe. Assuming a discharge coefficient of [value], calculate the required orifice diameter, so that the head difference across the orifice plate is [value] \(\textrm{mm}\).
(Give your answer as a decimal to three decimal places, note that trailing zeroes are required i.e. \(2.3\) must be written as \(2.300\).)
Orifice diameter =
Feedback: 1. We have the equation \[Q = CO\sqrt{\dfrac{2g\Delta h}{1-O^{2}/P^{2}}}\] First divide by \(CO\) \[\implies \dfrac{Q}{CO} = \sqrt{\dfrac{2g\Delta h}{1-O^{2}/P^{2}}}\] Square both sides \[\implies \dfrac{Q^{2}}{C^{2}O^{2}} = \dfrac{2g\Delta h}{1-O^{2}/P^{2}}\] Multiply through by \((1-O^{2}/P^{2})\), expand out the brackets and simplify \[\implies \dfrac{Q^{2}}{C^{2}O^{2}}(1-O^{2}/P^{2}) = 2g\Delta h \implies \dfrac{Q^{2}}{C^{2}O^{2}}-\dfrac{Q^{2}}{C^{2}O^{2}}(\dfrac{O^{2}}{P^{2}}) = 2g\Delta h \implies \dfrac{Q^{2}}{C^{2}O^{2}}-\dfrac{Q^{2}}{C^{2}P^{2}} = 2g\Delta h\] Multiply through by \(C^2P^2\) \[\implies \dfrac{Q^{2}P^{2}}{O^2}-Q^2 = 2g\Delta hC^{2}P^{2}\] Add \(Q^2\) to both sides \[\implies \dfrac{Q^{2}P^{2}}{O^2} = Q^2+2g\Delta hC^{2}P^{2}\] Multiply by \(O^2\) \[\implies Q^{2}P^{2} = O^2(Q^2+2g\Delta hC^{2}P^{2})\] Divide by \((Q^2+2g\Delta hC^{2}P^{2})\) \[\implies O^2 = \dfrac{Q^{2}P^{2}}{Q^2+2g\Delta hC^{2}P^{2}}\] We then take the square root of both sides of the equation \[\implies \sqrt{O^2} = \sqrt{\dfrac{Q^{2}P^{2}}{Q^2+2g\Delta hC^{2}P^{2}}}\] Since Q, O and A take positive values this can be simplified to transpose the equation in terms of \(O\) \[\implies O = \dfrac{QP}{\sqrt{Q^2+2g\Delta hC^{2}P^{2}}}\] 2. We first need to convert these input values so to SI units. \(Q = [value] \textrm{ cm}^{3}\textrm{s}^{-1} = [value] \textrm{ m}^{3}\textrm{s}^{-1}\) diameter of pipe \(= [value] \textrm{ cm} = [value] \textrm{ m} \implies P = \pi \dfrac{[value]^2}{4} = [value]...\textrm{ m}\) \(C = [value]\) \(\Delta h = [value] \textrm{ mm} = [value] \textrm{ m}\) \(g = 9.81 \textrm{ m}\textrm{s}^{-2}\) We can now substitute these values into our equation for \(O\) \[O = \dfrac{QP}{\sqrt{Q^2+2g\Delta hC^{2}P^{2}}} = \dfrac{[value]\cdot [value]...}{\sqrt{[value]^2+2\cdot 9.81\cdot [value]\cdot [value]^{2}\cdot ([value]...)^{2}}} = [value]\] To finally get the orifice diameter from this value we transpose the formula for the area of a circle \(\text{area} = \pi \cdot (\dfrac{diameter}{2})^2\) to get \[\text{orifice diameter} = 2 \sqrt{\dfrac{O}{\pi}} = 2 \sqrt{\dfrac{[value]}{\pi}} = [value]\]