I have no background in thermodynamics or fluid dynamics, so please bear with me. Refer to the following figure: enter image description here

A working fluid with a flow rate of $1~kg/s$ enters a heater at a pressure of $1~MPa$ and temperature $150~^\circ C$, which raises its temperatures to $300~^\circ C$. It then enters a turbine, does work, loses pressure and temperature, and exits at $0.5~MPa$ and $200~^\circ C$ at the same flow rate. The bypass valve's purpose is to control the work output of the turbine by bypassing some of the working fluid from the inlet stream directly to the outlet stream. Suppose that we're free to make all the "simplifying assumptions" for this problem. Suppose also that initially the opening fraction of the bypass valve is $0\%$ i.e., the valve is fully closed. My questions are:

  1. How would the pressure and flow rate change as the bypass valve opening fraction is varied ($0\%$ means fully closed, $100\%$ means fully opened)? In particular, I'm interested to know them at points A, B, X, Y and E.
  2. Would a high pressure stream at Y intercepting the low pressure stream at E raise the pressure and flow rate of the latter? How could that be quantified?
  3. Could the valve's opening fraction be linearly related to flow rate and pressure? Meaning, could it assumed that flow rate and pressure at point B would be halved if the valve was half-open ($50\%$ opening fraction)?
  4. What would be the simplest way to model this valve?

Some questions might seem redundant, but I listed them to get a thorough response, for clarity and understanding. Also, all the numbers are arbitrary and any unreported quantities could be assumed. Thank you.

  • $\begingroup$ I'm not sure what conclusions you're trying to draw from your setup, but note that for practical applications, process control schemes would be implemented to achieve the correct heat and material balances for whatever the turbine is driving. This means that both mass flow rate and heat input to the turbine would vary, and they would vary as much as needed to hold some turbine variable (e.g., turbine rpm) constant. $\endgroup$ Nov 28, 2019 at 18:24
  • $\begingroup$ I'm just trying to understand the effect of bypass valve on the pressure and flow rate of the fluid at various points, and in turn how it would affect the turbine's work output. It could be thought of as a portion of the Brayton cycle. $\endgroup$
    – Adeel
    Nov 28, 2019 at 18:33
  • $\begingroup$ The knowledge you seek is probably not applicable to anything in the real world. In all my years in industry (21 years total), I never witnessed a piping path like the one you drew. $\endgroup$ Nov 28, 2019 at 18:40

1 Answer 1


I can answer the question of how looks like a viscous, heat-conducting compressible gas flows in a 2D pipe with a heater and a turbine. I use my code tested on several problems. First, consider the case without bypass. Figure 1 shows the distribution of velocity (left), pressure (center) and temperature (right). From these data it can be seen that the heated gas enters the turbine which rotates at a constant angular velocity.

Figure 1

In Figure 2 shown the distribution of velocity (left), pressure (center) and temperature (right) in a system with bybass. Figure 2 This animation shows how the turbine is filled. In the beginning we can see how the turbine rotates. Figire 3 Now we can compare the parameters of pressure and gas flow with a closed (upper row) and open (lower row) valve. In Fig. 4 shows the distribution of pressure (left), $\rho u$ (center) and $\int {\rho udy}$ (right) in lower (1) and upper pipe (2). Figure 4

  • $\begingroup$ Thanks for the response. Do you mind commenting on your results with some numbers e.g., what are the flow rates and pressures at different points? With and without the valve? What was the opening fraction of the valve? $\endgroup$
    – Adeel
    Dec 2, 2019 at 12:53
  • $\begingroup$ How will we compare the two cases? In the second case, the pressure at the turbine inlet drops by 20% due to bypass. The gas flow rates at the inlet and outlet are the same. $\endgroup$ Dec 2, 2019 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.