Question

Steam enters a diffuser steadily at \(0.5 \mathrm{MPa}, 300^{\circ} \mathrm{C}\) and \(122 \mathrm{m} / \mathrm{s}\) at a rate of \(3.5 \mathrm{kg} / \mathrm{s}\). The inlet area of the diffuser is \((a) 15 \mathrm{cm}^{2}\) \((b) 50 \mathrm{cm}^{2}\) \((c) 105 \mathrm{cm}^{2}\) \((d) 150 \mathrm{cm}^{2}\) \((e) 190 \mathrm{cm}^{2}\)

Short answer

Answer: The inlet area of the diffuser is \(105 \ \mathrm{cm}^2\).

Step-by-step solution

Find the Density of Steam at the Inlet

To find the density of steam at the inlet, we can use the Ideal Gas Equation: \(PV=mRT\), where: - \(P\) is pressure - \(V\) is volume - \(m\) is the mass of the gas - \(R\) is the specific gas constant for steam (\(R = 461 \mathrm{J} / \mathrm{kg ⋅ K}\)) - \(T\) is the absolute temperature Rewriting the equation to find density (\(\rho = m/V\)) we get \(\rho = \frac{P}{RT}\). Converting the given pressure and temperature to SI units: - Pressure: \(0.5 \mathrm{MPa} = 0.5 \times 10^6 \mathrm{Pa}\) - Temperature: \(300^{\circ}\mathrm{C}=300+273.15=573.15\mathrm{K}\) Now, we can calculate the density of steam: \(\rho = \frac{0.5 \times 10^6 \mathrm{Pa}}{461\mathrm{J / kg \cdot K} \times 573.15\mathrm{K}}\)

Calculate the Density of Steam

Substitute the values into the formula for density: \(\rho = \frac{0.5 \times 10^6 \mathrm{Pa}}{461\mathrm{J / kg \cdot K} \times 573.15\mathrm{K}} = 1.857 \mathrm{kg} / \mathrm{m}^3\)

Apply the Continuity Equation

We can now use the continuity equation to find the area: \(A = \frac{m}{\rho v}\), where: - \(A\) is the area we want to find - \(m\) is the mass flow rate: \(3.5 \mathrm{kg} / \mathrm{s}\) - \(\rho\) is the density we found in step 2 - \(v\) is the velocity at the inlet: \(122 \mathrm{m} / \mathrm{s}\) Substituting the values, we get: \(A = \frac{3.5 \mathrm{kg} / \mathrm{s}}{1.857 \mathrm{kg} / \mathrm{m}^3 \times 122 \mathrm{m} / \mathrm{s}}\)

Calculate the Inlet Area

Find the area by plugging the values into the formula:\(A = \frac{3.5 \mathrm{kg/s}}{1.857 \mathrm{kg}/ \mathrm{m}^3 \times 122 \mathrm{m/s}} = 0.0105 \mathrm{m}^2\) To convert the area to cm², multiply by \(10^4\): \(A = 0.0105 \mathrm{m}^2 \times 10^4 = 105 \mathrm{cm}^2\) So, the inlet area of the diffuser is \(\boxed{105 \ \mathrm{cm}^2}\), which corresponds to option \((c)\).