Chapter 5, Problem 5.10P

To determine

(a)

The dimension $\left\{MLT\theta \right\}$ of a quantity.

Answer to Problem 5.10P

The dimension $\left\{MLT\theta \right\}$ of the following quantity is = $\left\{M{L}^{-2}{T}^{-2}\right\}$

Explanation of Solution

Given Information:

$\rho u\frac{\partial u}{\partial x}$

$\rho u\frac{\partial u}{\partial x}$

$\left\{\rho \right\}=\left\{M{L}^{-3}\right\}$

$\left\{u\right\}=\left\{L{T}^{-1}\right\}$

$\left\{\frac{\partial u}{\partial x}\right\}=\left\{T^{-1}\right\}$

$\left\{\rho u\frac{\partial u}{\partial x}\right\}=\left\{\rho \right\}\left\{u\right\}\left\{\frac{\partial u}{\partial x}\right\}$

$\left\{\rho u\frac{\partial u}{\partial x}\right\}=\left\{M{L}^{-3}\right\}\left\{L{T}^{-1}\right\}\left\{T^{-1}\right\}$

$=\left\{M{L}^{-2}{T}^{-2}\right\}$

Conclusion:

The dimension $\left\{MLT\theta \right\}$ of the following quantity is = $\left\{M{L}^{-2}{T}^{-2}\right\}$.

To determine

(b)

The dimension $\left\{MLT\theta \right\}$ of the following quantity.

Answer to Problem 5.10P

The dimension $\left\{MLT\theta \right\}$ of the following quantity is = $\left\{M{L}^1{T}^{-2}\right\}$.

Explanation of Solution

Given Information:

${\displaystyle \underset{1}{\overset{2}{\int }}\left(p-p_0\right)}dA$

${\displaystyle \underset{1}{\overset{2}{\int }}\left(p-p_0\right)}dA$

$\left\{p\right\}=\left\{M{L}^{-1}{T}^{-2}\right\}$

$\left\{A\right\}=\left\{L^2\right\}$

$\left\{{\displaystyle \underset{1}{\overset{2}{\int }}\left(p-p_0\right)}dA\right\}=\left\{{\left|\left(p-p_0\right)A\right|}_1^2\right\}$

$=\left\{\left(p-p_0\right)\right\}\left\{A\right\}$

$=\left\{M{L}^{-1}{T}^{-2}\right\}\left\{L^2\right\}$

$=\left\{M{L}^1{T}^{-2}\right\}$

Conclusion:

The dimension $\left\{MLT\theta \right\}$ of the following quantity is = $\left\{M{L}^1{T}^{-2}\right\}.$.

To determine

(c)

The dimension $\left\{MLT\theta \right\}$ of the following quantity.

Answer to Problem 5.10P

The dimension $\left\{MLT\theta \right\}$ of the following quantity is = $\left\{M{L}^{-3}{T}^{-2}\right\}.$

Explanation of Solution

Given Information:

$\rho c_p\frac{{\partial }^{2}T}{\partial x\partial y}$

$\rho c_p\frac{{\partial }^{2}T}{\partial x\partial y}$

$\left\{\rho \right\}=\left\{M{L}^{-3}\right\}$

$\left\{c_p\right\}=\left\{L^2{T}^{-2}{\Theta }^{-1}\right\}$

$\left\{\frac{{\partial }^{2}T}{\partial x\partial y}\right\}=\left\{\frac{\Theta }{L\times L}\right\}$

$=\left\{L^{-2}\Theta \right\}$

$\left\{\rho c_p\frac{{\partial }^{2}T}{\partial x\partial y}\right\}=\left\{M{L}^{-3}\right\}\left\{L^2{T}^{-2}{\Theta }^{-1}\right\}\left\{\frac{\Theta }{L\times L}\right\}$

$=\left\{M{L}^{-3}{T}^{-2}\right\}$

Conclusion:

The dimension $\left\{MLT\theta \right\}$ of the following quantity is = $\left\{M{L}^{-3}{T}^{-2}\right\}.$.

To determine

(d)

The dimension $\left\{MLT\theta \right\}$ of the following quantity.

Answer to Problem 5.10P

The dimension $\left\{MLT\theta \right\}$ of the following quantity is = $\left\{ML{T}^{-2}\right\}.$

Explanation of Solution

Given Information:

${\displaystyle \iiint \rho \frac{\partial u}{\partial t}}dxdydz$

${\displaystyle \iiint \rho \frac{\partial u}{\partial t}}dxdydz$

$\left\{\rho \right\}=\left\{M{L}^{-3}\right\}$

$\left\{\frac{\partial u}{\partial t}\right\}=\left\{\frac{L{T}^{-1}}{T}\right\}=\left\{L{T}^{-2}\right\}$

$\left\{dxdydz\right\}=\left\{L\times L\times L\right\}=\left\{L^3\right\}$

$\left\{{\displaystyle \iiint \rho \frac{\partial u}{\partial t}}dxdydz\right\}=\left\{M{L}^{-3}\right\}\left\{L{T}^{-2}\right\}\left\{L^3\right\}$

$=\left\{ML{T}^{-2}\right\}$

Conclusion:

The dimension $\left\{MLT\theta \right\}$ of the following quantity is = $\left\{ML{T}^{-2}\right\}.$.