Question

Steam enters a horizontal \(15-\mathrm{cm}\)-diameter pipe as a saturated vapor at 5 bar with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) and exits at \(4.5\) bar with a quality of \(95 \%\). Heat transfer from the pipe to the surroundings at \(300 \mathrm{~K}\) takes place at an average outer surface temperature of \(400 \mathrm{~K}\). For operation at steady state, determine (a) the velocity at the exit, in \(\mathrm{m} / \mathrm{s}\). (b) the rate of heat transfer from the pipe, in \(\mathrm{kW}\). (c) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for a control volume comprising only the pipe and its contents. (d) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for an enlarged control volume that includes the pipe and enough of its immediate surroundings so that heat transfer from the control volume occurs at \(300 \mathrm{~K}\). Why do the answers of parts (c) and (d) differ?

Short answer

Find exit velocity using continuity; heat transfer using first law; entropy production rates using second law. Differences in entropy production stem from heat transfer temperatures.

Step-by-step solution

Determine the exit velocity

Use the steady-state flow equation, which states that the mass flow rate in equals the mass flow rate out. The volume flow rate is given by \ \( \text{mass flow rate} = \rho \times A \times V \ \) where \( \rho \) is the density, \( A \) is the cross-sectional area, and \( V \) is the velocity. For the initial conditions, find \( \rho_1 \) for saturated vapor at 5 bar. For the final conditions, use the quality and the specific volumes of saturated liquid and vapor at 4.5 bar to find \( \rho_2 \). Solve for the exit velocity \( V_2 \).

Calculate the rate of heat transfer

Use the first law of thermodynamics for an open system at steady state \ \( Q - W_s = \text{mass flow rate} \times (h_2 - h_1 + \frac{V_2^2 - V_1^2}{2}) \ \) where \( h \) is the specific enthalpy. Calculate the specific enthalpy at entry and exit, and then solve for the rate of heat transfer \( Q \).

Calculate the rate of entropy production in the pipe

Use the second law of thermodynamics for an open system \ \( \text{Entropy rate in} - \text{Entropy rate out} + \frac{Q}{T_s} = \text{Entropy production rate} \ \) Calculate the specific entropy at entry and exit to find the net entropy flow. Calculate the entropy change due to heat transfer, and solve for the entropy production rate in the pipe.

Calculate the rate of entropy production in the enlarged control volume

For the enlarged control volume, the only change is the temperature at which heat is transferred \ \( T_s = 300 \text{K} \ \). Use the same second law equation to find the new entropy production rate.

Explain the difference in entropy production rates

Entropy production differences arise from the different temperatures at which heat transfer occurs. High entropy production indicates irreversible processes and greater inefficiencies.

Key concepts

steady-state flow

In physical systems, many processes reach a point where their properties remain constant over time. This state is called *steady-state*. For steam flow in pipes, maintaining steady-state means that the mass flow rate entering and exiting the pipe remains constant. In other words, the amount of steam flowing into the pipe equals the amount flowing out, without any accumulation or depletion. This makes analysis simpler because we don't have to account for changes over time.
To determine the flow characteristics, we frequently utilize the steady-state mass flow rate equation:
\( \text{mass flow rate} = \rho \times A \times V \). This equation helps us track how fluid properties like velocity change between different points in the system.

mass flow rate

The concept of *mass flow rate* is essential in understanding fluid dynamics in pipes. It represents the mass of fluid moving through a cross-section of the pipe per unit of time, typically expressed in kilograms per second (kg/s). In our exercise, we calculate this to ensure continuity between the pipe's entrance and exit.
Given the steam enters as saturated vapor with known conditions (5 bar and a velocity of 10 m/s), we can determine the mass flow rate using the density of steam \( \rho \), the cross-sectional area \( A \), and the inlet velocity \( V \). At exit, using similar relationships but with different conditions (4.5 bar and 95% quality), we reconcile the mass flow to find the exit velocity.

first law of thermodynamics

The *first law of thermodynamics* is a crucial principle for energy analysis in engineering. Often referred to as the law of energy conservation, it states that energy cannot be created or destroyed in an isolated system. Instead, energy can only change forms. For an open system like our pipe:
\( Q - W_s = \text{mass flow rate} \times (h_2 - h_1 + \frac{V_2^2 - V_1^2}{2}) \)
Here, \( Q \) is the heat transfer rate, \( W_s \) is the work done by the system, \( h \) stands for specific enthalpy, and \( V \) represents velocity.
By analyzing the steam's enthalpy at the entrance and exit along with velocities, we solve for the heat transfer rate. This tells us how much energy is lost or gained by the system, helping to understand heat interactions in steady-state flow.

second law of thermodynamics

The *second law of thermodynamics* deals with the natural direction of energy processes and the concept of entropy, a measure of disorder or randomness. For an open system like our pipe,
\( \text{Entropy rate in} - \text{Entropy rate out} + \frac{Q}{T_s} = \text{Entropy production rate} \)
where \( Q \) is the heat transfer rate and \( T_s \) is the temperature at the system boundary. This law implies that the total entropy of a system and its surroundings tends to increase over time, indicating that real processes are irreversible and contribute to the universe's increasing disorder.
Calculating the specific entropy of steam at the entrance and exit, and the entropy associated with heat exchange, allows us to find the rate of entropy production, highlighting inefficiencies within the system.

entropy production

Finally, *entropy production* quantifies the irreversibilities within a thermodynamic process. High entropy production indicates greater inefficiencies. For our pipe, we calculate it using temperature-specific heat transfer and net entropy flow.
When considering different control volumes, such as just the pipe versus the pipe with its surroundings, the rate of entropy production differs because the temperature gradient changes. This difference underlines the effect of heat exchange temperatures on system efficiency.
Lower entropy production in the larger volume (with heat transferred at 300 K) compared to the isolated pipe (with heat transfer at 400 K) highlights lower irreversibilities in the broader context.