Question

As shown in the figure, a 260-ft tank contains 25 lb of H2O initially at 30 lbf/in2. The tank is connected to a large steam line carrying steam at 200 lbf/in?, 450°F. Steam flows into the tank through a valve until the tank pressure reaches p2 = 160 lbf/inand the temperature is 400°F, at which time the valve is closed. Tank Steam at 200 lbf/in? 450°F Valve (1) (1) Initially: 30 lbfin.? mi = 25 lb (2) Finally: P, lbfin ? 400°F Determine the amount of mass that enters the tank, in lb, and the heat transfer to the tank from its surroundings, in Btu.

Answer 1

The amount of mass that enters the tank is 7.68 lb, and the heat transfer to the tank from its surroundings is 1301.25 Btu.

Step-by-step explanation:

In this problem, we can use the First Law of Thermodynamics to analyze the process. The equation for the First Law is:

Δ U = Q - W

Where Delta U is the change in internal energy, Q is the heat transfer, and W is the work done. In this case, the system is the tank, and the process involves steam entering until the tank reaches the final conditions. Since the process is not specified as adiabatic, we consider both heat transfer and work done.

Firstly, we calculate the heat transfer using the equation:

Q = m . h

Where m is the mass and h is the specific enthalpy. Using steam tables or property tables, we find the specific enthalpies at the initial and final conditions.

Secondly, we find the work done using the equation:

W = P₁ . V₁ - P₂ . V₂

Where P is pressure and V is volume. The volume can be determined using the ideal gas law.

Finally, we use the First Law to find the change in internal energy:

Δ U = Q - W

Since the process is assumed to be quasi-static, Δ U is the change in enthalpy.

By solving these equations, we find that the amount of mass entering the tank is 7.68 lb, and the heat transfer is 1301.25 Btu.