Question

A 3-ft' rigid tank initially contains saturated water vapor at \(300^{\circ} \mathrm{F}\). The tank is connected by a valve to a supply line that carries steam at 200 psia and \(400^{\circ} \mathrm{F}\). Now the valve is opened, and steam is allowed to enter the tank. Heat transfer takes place with the surroundings such that the temperature in the tank remains constant at \(300^{\circ} \mathrm{F}\) at all times. The valve is closed when it is observed that one-half of the volume of the tank is occupied by liquid water. Find (a) the final pressure in the tank, ( \(b\) ) the amount of steam that has entered the \(\tan \mathrm{k},\) and \((c)\) the amount of heat transfer.

Short answer

Based on the given problem and the step-by-step solution provided: (a) The final pressure in the tank is 133.11 psia. (b) The amount of steam that enters the tank is 92.689 lb. (c) The amount of heat transfer during the process is approximately -1837.2 Btu, indicating a heat loss to the surroundings.

Step-by-step solution

Determine the initial conditions

First, let's find the initial conditions of the tank. Since it is initially containing saturated water vapor at \(300^{\circ}F\), we can use steam tables (or software like EES) to determine the specific volume \(v_1\), pressure \(P_1\), and the enthalpy \(h_1\) for the saturated vapor at this temperature. From steam tables (or EES), at \(300^{\circ}F\) for saturated vapor: - \(v_1 = 4.1403 ft^3 / lb\) - \(P_1 = 66.22 psia\) - \(h_1 = 1188.4 Btu / lb\) The initial volume of the tank is given as \(3 ft^3\). Using the specific volume, we can find the initial mass of water vapor inside the tank: - \(m_1 = \frac{V_{tank}}{v_1} = \frac{3 ft^3}{4.1403 ft^3 / lb} = 0.7252 lb\)

Determine the mass of steam that enters the tank

Let's denote the mass of the steam that enters the tank as \(m_2\). At the end of the process, half of the tank's volume is occupied by liquid water and the other half by water vapor, both at the constant temperature of \(300^{\circ}F\). We can use the steam tables (or EES) to find the specific volume of the water and water vapor at the 50% point. From steam tables (or EES), at \(300^{\circ} F\), for saturated liquid and saturated vapor: - Specific volume of liquid \(v_{f} = 0.01613 ft^3 / lb\) - Specific volume of vapor \(v_{g} = 4.1403 ft^3 / lb\) Let's assume \(m_3\) represents half of the tank's volume occupied by liquid water, and \(m_4\) represents half of the tank's volume occupied by water vapor. So, we have: - \(m_3v_{f} = m_4v_{g} = \frac{1}{2}V_{tank}\) - \(m_3 = \frac{0.5V_{tank}}{v_{f}} = \frac{1.5 ft^3}{0.01613 ft^3 / lb} = 93.052 lb\) - \(m_4 = \frac{0.5V_{tank}}{v_{g}} = \frac{1.5 ft^3}{4.1403 ft^3 / lb} = 0.3626 lb\) Since the liquid and vapor inside the tank initially are at the same pressure, we can use the conservation of mass to find the mass of steam that enters the tank: - \(m_2 = m_3 + m_4 - m_1 = 93.052 lb + 0.3626 lb - 0.7252 lb = 92.689 lb\)

Calculate the final pressure and specific volume inside the tank

Now we need to find the final pressure inside the tank, which is denoted as \(P_2\). We can find \(P_2\) using the pressure-volume relationship in isobars, where the final pressure is the average of the initial pressure \(P_1\) and the incoming steam pressure (given as 200 psia). - \(P_2 = \frac{P_1 + 200psia}{2}= \frac{66.22 psia + 200 psia}{2} = 133.11 psia\) To find the specific volume of the mixture in the tank, denoted as \(v_2\), we will use the mass and specific volume of the water and water vapor inside the tank after the process: - \(v_2 = \frac{m_3v_{f} + m_4v_{g}}{m_2} = \frac{93.052 lb \cdot 0.01613 ft^3 / lb + 0.3626 lb \cdot 4.1403 ft^3 / lb}{92.689 lb} = 0.0501 ft^3 / lb\)

Estimate the amount of heat transfer

To find the amount of heat transfer (\(Q\)) during the process, we can use the enthalpy change and the conservation of energy principle: \(Q = m_1(h_2 - h_1) + m_2(h_2 - h_{incoming})\) Since we know that the final state of water vapor inside the tank is at a temperature of \(300^{\circ} F\) and pressure of \(133.11 psia\), we can find the enthalpy \(h_2\) using the steam tables (or EES). Similarly, we can find the enthalpy of the incoming steam \(h_{incoming}\) at \(400^{\circ} F\) and 200 psia. From the steam tables (or EES): - \(h_2 = 1190.2 Btu / lb\) (at \(300^{\circ} F\) and \(133.11 psia\)) - \(h_{incoming} = 1210.1 Btu / lb\) (at \(400^{\circ} F\) and 200 psia) Now we can substitute the values into the energy equation: \(Q = 0.7252 lb(1190.2 Btu / lb - 1188.4 Btu / lb) + 92.689 lb(1190.2 Btu / lb - 1210.1 Btu / lb) \approx -1837.2 Btu\)

Results:

(a) Final pressure in the tank: \(P_2 = 133.11 \ psia\) (b) Amount of steam entered: \(m_2 = 92.689 \ lb\) (c) Amount of heat transfer: \(Q \approx -1837.2 \ Btu\) (negative value indicates heat is lost to surroundings)