Chapter 5, Problem 5.36P

To determine

(a)

An expression for heat loss in terms of other three parameters.

Answer to Problem 5.36P

${\Pi }_1=\frac{{\stackrel{•}{Q}}_{loss}R}{A\Delta T}=const$

Explanation of Solution

Given information:

Heat loss ${\stackrel{•}{Q}}_{loss}$ depends on temperature difference $\Delta T$, surface area $A$ and $R$ value of the window

The unit of $R$ is $f{t}^{2}hF$

To find the relation between these variables,

Basically we have some steps to follow,

1. Find the number of variables $(n)$ given.

2. Find the dimensions of the given variables.

3. Find $j$ which is basically the number of different dimensions of given variables.

4. In other words $j$ is the number of variables which won’t form pi product.

5. Find the number of desired pi products $(k)$ which is given as, $k=n-j$

6. Finally write the dimensionless function obtained.

Calculation:

Write the function,

${\stackrel{•}{Q}}_{loss}=f\left(\Delta T,A,R\right)$

Count the number of variables $n$,

$n=4$

Find dimension of each variable,

$\begin{array}{l}{\stackrel{•}{Q}}_{loss}\to M{L}^2{T}^{-3}\\ \Delta T\to \Theta \\ A\to L^2\\ R\to T^3\Theta M^{-1}\end{array}$

Find $j$,

$j=3$

Find the number of pi products,

$k=n-j=4-3=1$

To obtain the equation,

${\Pi }_1=\Delta T^a{A}^b{R}^c{\stackrel{•}{Q}}_{loss}={\left(\Theta \right)}^a{\left(L^2\right)}^b{\left(T^3\Theta M^{-1}\right)}^c\left(M{L}^2{T}^{-3}\right)=M^0{L}^0{T}^0{\Theta }^0$

Equate exponents,

Length: $2b+2=0$

Mass: $-c+1=0$

Time: $3c-3=0$

Temperature: $a+c=0$

Therefore, we get

$a=-1,b=-1,c=1$

Therefore, the expression will be,

$\begin{array}{l}{\Pi }_1=\Delta T^{-1}{A}^{-1}{R}^1{\stackrel{•}{Q}}_{loss}\\ {\Pi }_1=\frac{{\stackrel{•}{Q}}_{loss}R}{A\Delta T}=const\end{array}$

Conclusion:

The expression for rate of heat loss is given as,

${\Pi }_1=\frac{{\stackrel{•}{Q}}_{loss}R}{A\Delta T}=const$.

To determine

(b)

What happens for rate of heat loss if the temperature difference doubles?

Answer to Problem 5.36P

$Q_{loss}\infty \Delta T$

Therefore, if temperature difference $\Delta T$ doubles, rate of heat loss $Q_{loss}$ also doubles.

Explanation of Solution

Given information:

Heat loss ${\stackrel{•}{Q}}_{loss}$ depends on temperature difference $\Delta T$, surface area $A$ and $R$ value of the window

The unit of $R$ is $f{t}^{2}hF$

To find the relation between these variables,

Basically we have some steps to follow,

1. Find the number of variables $(n)$ given.

2. Find the dimensions of the given variables.

3. Find $j$ which is basically the number of different dimensions of given variables.

4. In other words $j$ is the number of variables which won’t form pi product.

5. Find the number of desired pi products $(k)$ which is given as, $k=n-j$

6. Finally write the dimensionless function obtained.

Calculation:

Write the function,

${\stackrel{•}{Q}}_{loss}=f\left(\Delta T,A,R\right)$

Count the number of variables $n$,

$n=4$

Find dimension of each variable,

$\begin{array}{l}{\stackrel{•}{Q}}_{loss}\to M{L}^2{T}^{-3}\\ \Delta T\to \Theta \\ A\to L^2\\ R\to T^3\Theta M^{-1}\end{array}$

Find $j$,

$j=3$

Find the number of pi products,

$k=n-j=4-3=1$

To obtain the equation,

${\Pi }_1=\Delta T^a{A}^b{R}^c{\stackrel{•}{Q}}_{loss}={\left(\Theta \right)}^a{\left(L^2\right)}^b{\left(T^3\Theta M^{-1}\right)}^c\left(M{L}^2{T}^{-3}\right)=M^0{L}^0{T}^0{\Theta }^0$

Equate exponents,

Length: $2b+2=0$

Mass: $-c+1=0$

Time: $3c-3=0$

Temperature: $a+c=0$

Therefore, we get

$a=-1,b=-1,c=1$

Therefore, the expression will be,

$\begin{array}{l}{\Pi }_1=\Delta T^{-1}{A}^{-1}{R}^1{\stackrel{•}{Q}}_{loss}\\ {\Pi }_1=\frac{{\stackrel{•}{Q}}_{loss}R}{A\Delta T}=const\end{array}$

According to the above equation, we can say that,

$Q_{loss}\infty \Delta T$

Therefore, if temperature difference $\Delta T$ doubles, rate of heat loss $Q_{loss}$ also doubles.

Conclusion: