The problem asks us to determine the bulk mean temperature, denoted as \(T_m\), for fluid flow through a pipe. We are provided with the velocity distribution \(u(r,x)\) and the temperature distribution \(T(r,x)\) within the pipe, along with a specific formula to calculate \(T_m\).

The given distributions are:

• Velocity distribution: \(u\left( {r,x} \right) = {C_1}\)
• Temperature distribution: \(T\left( {r,x} \right) = {C_2}{\left[{1 - {\left({\frac{r}{R}} \right)}^3} \right]}\)

The formula for the bulk mean temperature is given by:

\({T_m} = \frac{2}{{{u_m}{R^2}}}\mathop \smallint \limits_0^R u\left( {r,x} \right)T\left( {r,x} \right)rdr\)

Here, \(C_1\) and \(C_2\) are constants, \(R\) is the pipe radius, and \(u_m\) is the mean velocity of the flow.

Mean Velocity Calculation for Pipe Flow

Before proceeding with the bulk mean temperature calculation, we first need to understand the mean velocity \(u_m\). For a flow in a circular pipe, the mean velocity \(u_m\) for a given velocity profile \(u(r)\) is defined as:

\({u_m} = \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_A u\left( r \right)dA = \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_0^R u\left( r \right)\left( {2\pi rdr} \right) = \frac{2}{{{R^2}}}\mathop \smallint \limits_0^R u\left( r \right)rdr\)

Given the velocity distribution \(u\left( {r,x} \right) = {C_1}\), which is a constant (uniform flow), we can calculate \(u_m\) as:

\({u_m} = \frac{2}{{{R^2}}}\mathop \smallint \limits_0^R {C_1}rdr\)

\({u_m} = \frac{{2{C_1}}}{{{R^2}}}\left[ {\frac{{{r^2}}}{2}} \right]_0^R\)

\({u_m} = \frac{{2{C_1}}}{{{R^2}}}\left( {\frac{{{R^2}}}{2} - 0} \right)\)

\({u_m} = \frac{{2{C_1}{R^2}}}{{2{R^2}}}\)

\({u_m} = {C_1}\)

Thus, the mean velocity \(u_m\) is equal to the constant velocity \(C_1\).

Bulk Mean Temperature Determination

Now, we substitute the expressions for \(u(r,x)\), \(T(r,x)\), and the calculated \(u_m\) into the bulk mean temperature formula:

\({T_m} = \frac{2}{{{u_m}{R^2}}}\mathop \smallint \limits_0^R u\left( {r,x} \right)T\left( {r,x} \right)rdr\)

Substitute \(u(r,x) = C_1\) and \(T(r,x) = C_2{\left[{1 - (\frac{r}{R})^3} \right]}\):

\({T_m} = \frac{2}{{{u_m}{R^2}}}\mathop \smallint \limits_0^R {{C_1}C_2}{\left[{1 - {\left({\frac{r}{R}} \right)}^3} \right]}rdr\)

Move the constants outside the integral:

\({T_m} = \frac{{2{C_1}{C_2}}}{{{u_m}{R^2}}}\mathop \smallint \limits_0^R {\left[{1 - {\left({\frac{r}{R}} \right)}^3} \right]}rdr\)

Since we found that \(u_m = C_1\), we can substitute \(u_m\) with \(C_1\) in the denominator:

\({T_m} = \frac{{2{C_1}{C_2}}}{{{C_1}{R^2}}}\mathop \smallint \limits_0^R {\left[{r - \frac{{{r^4}}}{{{R^3}}}} \right]}dr\)

Simplify by canceling \(C_1\):

\({T_m} = \frac{{2{C_2}}}{{{R^2}}}\mathop \smallint \limits_0^R {\left[{r - \frac{{{r^4}}}{{{R^3}}}} \right]}dr\)

Now, perform the integration with respect to \(r\):

\({T_m} = \frac{{2{C_2}}}{{{R^2}}}\left[ {\frac{{{r^2}}}{2} - \frac{{{r^5}}}{{5{R^3}}}} \right]_0^R\)

Evaluate the integral at the limits \(r=R\) and \(r=0\):

\({T_m} = \frac{{2{C_2}}}{{{R^2}}}\left[ {\left( {\frac{{{R^2}}}{2} - \frac{{{R^5}}}{{5{R^3}}}} \right) - \left( {\frac{{{0^2}}}{2} - \frac{{{0^5}}}{{5{R^3}}}} \right)} \right]\)

\({T_m} = \frac{{2{C_2}}}{{{R^2}}}\left[ {\frac{{{R^2}}}{2} - \frac{{{R^2}}}{5}} \right]\)

Combine the terms inside the bracket:

\({T_m} = \frac{{2{C_2}}}{{{R^2}}}\left[ {\frac{{5{R^2} - 2{R^2}}}{{10}}} \right]\)

\({T_m} = \frac{{2{C_2}}}{{{R^2}}}\left[ {\frac{{3{R^2}}}{{10}}} \right]\)

Multiply the terms to get the final expression for \(T_m\):

\({T_m} = \frac{{6{C_2}{R^2}}}{{10{R^2}}}\)

\({T_m} = \frac{{6{C_2}}}{{10}}\)

\({T_m} = 0.6{C_2}\)

The calculated bulk mean temperature is \(0.6{C_2}\).

Comparing this result with the given options, we find that it matches option 3.

The final answer is $\boxed{\text{0.6 C2}}$