Question

Steam enters a long, horizontal pipe with an inlet diameter of \(D_{1}=16 \mathrm{cm}\) at \(2 \mathrm{MPa}\) and \(300^{\circ} \mathrm{C}\) with a velocity of \(2.5 \mathrm{m} / \mathrm{s}\). Farther downstream, the conditions are \(1.8 \mathrm{MPa}\) and \(250^{\circ} \mathrm{C},\) and the diameter is \(D_{2}=14 \mathrm{cm} .\) Determine (a) the mass flow rate of the steam and ( \(b\) ) the rate of heat transfer.

Short answer

Answer: The mass flow rate of the steam is 50.84 kg/s, and the rate of heat transfer is approximately -3936.88 kJ/s, where the negative sign indicates heat being lost from the steam to the surroundings.

Step-by-step solution

Calculate properties of steam at the inlet and outlet conditions

First, we must determine the properties of steam at the given inlet and outlet conditions. We will need to look up the values of the specific volume (v), enthalpy (h), and internal energy (u) in the steam tables for the given pressures and temperatures. For the inlet conditions, \(P_1 = 2 \mathrm{MPa}\) and \(T_1 = 300^\circ \mathrm{C}\): - \(v_1= 0.0993 \mathrm{m}^3/\mathrm{kg}\) - \(h_{1}= 3038.2 \mathrm{kJ}/\mathrm{kg}\) - \(u_{1}= 2725.2 \mathrm{kJ}/\mathrm{kg}\) For the outlet conditions, \(P_2 = 1.8 \mathrm{MPa}\) and \(T_2 = 250^\circ \mathrm{C}\): - \(v_2= 0.1118 \mathrm{m}^3/\mathrm{kg}\) - \(h_{2}= 2960.9 \mathrm{kJ}/\mathrm{kg}\) - \(u_{2}= 2659.9 \mathrm{kJ}/\mathrm{kg}\)

Determine mass flow rate of the steam

To determine the mass flow rate of the steam in the pipe, we will use the continuity equation. Since the velocity and diameter at the inlet and outlet are given, the equation is: \(\dot{m} = \frac{A_1 V_1}{v_1} = \frac{A_2 V_2}{v_2}\) First, calculate the cross-sectional areas at the inlet and outlet: - \(A_1 = \frac{\pi D_{1}^2}{4} = \frac{\pi (0.16 \mathrm{m})^2}{4} = 0.0201 \mathrm{m}^2\) - \(A_2 = \frac{\pi D_{2}^2}{4} = \frac{\pi (0.14 \mathrm{m})^2}{4} = 0.0154 \mathrm{m}^2\) Now, plug in the values and solve for the mass flow rate: \(\dot{m} = \frac{0.0201 \mathrm{m}^2 \times 2.5 \mathrm{m}/\mathrm{s}}{0.0993 \mathrm{m}^3/\mathrm{kg}} = 50.84 \mathrm{kg} / \mathrm{s}\) The mass flow rate of the steam is \(50.84 \mathrm{kg} / \mathrm{s}\).

Determine the rate of heat transfer

The conservation of energy equation for a steady-flow process is given by: \(\dot{Q} = \dot{W} + \dot{m}(h_2 - h_1)\) Since there is no work done in this process, \(\dot{W} = 0\), and we can find the heat transfer rate as follows: \(\dot{Q} = \dot{m}(h_2 - h_1)\) Plug in the values for mass flow rate and enthalpy at the inlet and outlet conditions to calculate the heat transfer rate: \(\dot{Q} = 50.84 \mathrm{kg} / \mathrm{s} (2960.9 \mathrm{kJ}/\mathrm{kg} - 3038.2 \mathrm{kJ}/\mathrm{kg}) = -3936.88 \mathrm{kJ} / \mathrm{s}\) The rate of heat transfer is approximately \(-3936.88 \mathrm{kJ} / \mathrm{s}\), where the negative sign indicates heat is being lost from the steam to the surroundings.