Answer
$μ=12.3^{∘}\text{C/MPa}$
$μ=4.9^{∘}\text{C/MPa}$
Work Step by Step
(a) The enthalpy of steam at $3 \mathrm{MPa}$ and $300^{\circ} \mathrm{C}$ is $h=2994.3\ \mathrm{~kJ} / \mathrm{kg}$. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as $$
\mu=\left(\frac{\partial T}{\partial P}\right)_h \equiv\left(\frac{\Delta T}{\Delta P}\right)_{b=29943\ \mathrm{kJNg}}
$$ Considering a throttling proeess from $3.5\ \mathrm{MPa}$ to $2.5\ \mathrm{MPa}$ at $h=2994.3\ \mathrm{~kJ} / \mathrm{kg}$, the Joule-Thomson coefficient is determined to be $$
\mu=\left(\frac{T_{3.5\ \mathrm{MPa}}-T_{2.5\ \mathrm{MPa}_2}}{(3.5-2.5) \mathrm{MPa}}\right)_{b=29943\ \mathrm{~kJ} / \mathrm{kg}_{\mathrm{g}}}=\frac{(306.3-294)^{\circ} \mathrm{C}}{(3.5-2.5) \mathrm{MPa}}=12.3^{\circ} \mathrm{C} / \mathrm{MPa}
$$ (b) The enthalpy of steam at $6\ \mathrm{MPa}$ and $500^{\circ} \mathrm{C}$ is $h=3423.1\ \mathrm{~kJ} / \mathrm{kg}$. Approximating differentials by differenoes about the specified state, the Joule-Thomson coefficient is expressed as $$
\mu=\left(\frac{\partial T}{\partial P}\right)_h \equiv\left(\frac{\Delta T}{\Delta P}\right)_{b=34231\ \mathrm{kN} / \mathrm{kg}}
$$ Considering a throttling process from $7.0\ \mathrm{MPa}$ to $5.0\ \mathrm{MPa}$ at $h=3423.1\ \mathrm{~kJ} / \mathrm{kg}$, the Joule-Thomson coefficient is determined to be$$
\mu=\left(\frac{T_{7.0\ \mathrm{MP}_2}-T_{5.0\ \mathrm{MPa}}}{(7.0-5.0) \mathrm{MPa}}\right)_{h=34231\ \mathrm{~kJ} / \mathrm{kg}}=\frac{(504.8-495.1)^{\circ} \mathrm{C}}{(7.0-5.0) \mathrm{MPa}}=4.9^{\circ} \mathrm{C} / \mathrm{MPa}
$$
