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A centrifugal compressor has 22 radial impellers with an (impeller) exit radius of \(r_{2}=0.25 \mathrm{~m}\). Assuming the mass flow rate through the compressor is \(m=10 \mathrm{~kg} / \mathrm{s}\), the shaft speed is \(\omega=10,000 \mathrm{rpm}\), and the inlet flow is swirl free, calculate (a) the shaft power, \(\varnothing_{s}\), in \(\mathrm{kW}\) (b) the total temperature rise in the compressor (assume \(\gamma=1.4\) and \(c_{p}=1004 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) )

Short Answer

Expert verified
After substituting the given values and going through the calculations: \n(a) The shaft power is found to be: fill-in calculated value for \(\varnothing_{s}\) kW \n(b) The total temperature rise in the compressor is: fill-in calculated value for \(\Delta T_{0}\) K.

Step by step solution

01

Convert Shaft Speed

The shaft speed is provided in RPM (Revolutions per Minute). It needs to be converted into rad/s (radians per second) for consistency. \(\omega = 10000 \times \frac{2\pi}{60} \) rad/s
02

Calculate the Tangential Velocity

At impeller exit, Calculate the tangential velocity, \(v_{w2}\), using the radius and angular speed. That is, \(v_{w2} = \omega \times r_{2}\). Use the values of \(\omega\) and \(r_{2}\) to obtain \(v_{w2}\).
03

Determine the Shaft Power

The shaft power, denoted by \(\varnothing_{s}\), is evaluated by the relation \(\varnothing_{s}=m \times v_{w2}^2 / (2 * 1000)\). Replace with known values of \(m\) and \(v_{w2}\) to find \(\varnothing_{s}\) expressed in kW.
04

Compute the total temperature rise

Apply the basic energy conservation principle that implies work done per unit mass flow is equal to change in total enthalpy. That is, \(\Delta T_{0} = \varnothing_{s} \times 10^{3} / (m \times c_{p})\). Substitute for \(\varnothing_{s}\), \(m\) and \(c_{p}\) to obtain the temperature increase \(\Delta T_{0}\) in Kelvin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Shaft Power Calculation
In a centrifugal compressor system, shaft power is critical as it determines the energy needed to compress the gas. We calculate the shaft power, often denoted by \( \varnothing_{s} \), using the formula:\[ \varnothing_{s} = \frac{m \times v_{w2}^2}{2 \times 1000} \]where:
  • \( m \) is the mass flow rate (\( 10\,\mathrm{kg/s} \) in this case),
  • \( v_{w2} \) is the tangential velocity at the impeller exit, calculated via \( \omega \times r_{2} \),
  • \( \omega \) is the angular speed in \( \mathrm{rad/s} \) (converted from RPM),
  • \( r_{2} \) is the radius at the impeller exit (\( 0.25\,\mathrm{m} \) here).
First, convert the shaft speed from RPM to radians per second: \( \omega = 10000 \times \frac{2\pi}{60} \,\mathrm{rad/s} \).
Then determine \( v_{w2} \) with \( v_{w2} = \omega \times r_{2} \).
Finally, substitute these values into the formula for \( \varnothing_{s} \). This calculation shows how much power in kilowatts is needed by the compressor to perform its function.
Temperature Rise in Compressors
Compressors work by increasing both the pressure and temperature of a fluid. The temperature rise is a vital parameter, usually calculated using the change in total enthalpy. The formula is:\[ \Delta T_{0} = \frac{\varnothing_{s} \times 10^{3}}{m \times c_{p}} \]where:
  • \( \Delta T_{0} \) is the total temperature rise,
  • \( \varnothing_{s} \) is the shaft power calculated earlier,
  • \( c_{p} \) is the specific heat capacity (\( 1004\,\mathrm{J/kg \cdot K} \)),
  • \( m \) is the mass flow rate.
This equation shows how energy supplied through power translates into heat energy causing a temperature increase.
It also underlines the efficiency of a compressor by connecting the mechanical and thermal aspects.
Knowing this temperature rise is necessary for designing systems to ensure they can handle the thermal stresses induced by compression.
Energy Conservation in Fluid Mechanics
In fluid mechanics, the principle of energy conservation is a cornerstone. It asserts that energy supplied (like the shaft power in compressors) results in energy changes in the form of pressure, kinetic energy, and temperature of the fluid. For a centrifugal compressor, the conservation of energy is summed up as:
  • The input energy is primarily converted to an increase in fluid enthalpy.
  • Kinetic energy increases as the fluid gains velocity.
  • Pressure energy rises as compression occurs.
Utilizing energy conservation, we realize that the work done per unit mass flow is equivalent to the change in total enthalpy, which ties back to the temperature rise equation. This connection highlights a compressor’s role in shifting energy from one form to another and emphasizes the need for efficient design to minimize losses.
Thus, ensuring the compressor is both effective and energy-efficient, saving operational costs and preventing undue thermal load on the surrounding system components.

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Most popular questions from this chapter

Air enters a centrifugal compressor at \(p_{\mathrm{tl}}=101 \mathrm{kPa}\), \(T_{\mathrm{r}}=288 \mathrm{~K}\), and purely in an axial direction with \(M_{2}=0.4 .\) Assuming the impeller radius at the exit is \(r_{2}=25 \mathrm{~cm}\) and the impeller is of radial design with slip factor \(\varepsilon=0.9\) and \(M_{\mathrm{T} 2}=\) 1.5, calculate (a) \(U_{2}\) in \(\mathrm{m} / \mathrm{s}\) (b) specific work, \(w_{e}\), in \(\mathrm{kJ} / \mathrm{kg}\) (c) impeller exit total temperature, \(T_{\mathrm{L} 2}\), in \(\mathrm{K}\) (d) shaft rotational specd, \(\alpha\), in rpm

A centrifugal compressor has 25 impeller blades of radial design. The rotor exit diameter is \(r_{2}=0.4 \mathrm{~m}\), and its rotational speed is \(\omega=8000 \mathrm{rpm}\). Assuming the inlet to rotor flow is purely axial and the air mass flow rate is \(\dot{m}=25 \mathrm{~kg} / \mathrm{s}\) through the compressor, calculate (a) compressor shaft power, \(8_{c}\) in, \(\mathrm{MW}\) (b) (time rate of change of) angular momentum at the rotor exit

A centrifugal compressor discharge static pressure is \(p_{3}=352 \mathrm{kPa}\) and its total temperature is \(T_{13}=444 \mathrm{~K}\). The flow at the impeller inlet has an axial component with \(C_{21}=\) \(150 \mathrm{~m} / \mathrm{s}\) and a preswirl component with \(C_{\partial 1}=56 \mathrm{~m} / \mathrm{s}\). The inlet total pressure and temperature are the standard sea- level conditions, i.e., \(p_{\mathrm{u}}=101 \mathrm{kPa}\) and \(T_{\mathrm{ul}}=288 \mathrm{~K}\), respectively. Assuming the compressor polytropic efficiency is \(e_{c}=0.88\), calculate (a) compressor total pressure ratio, \(p_{\mathrm{u} 3} / p_{\mathrm{u}}\) (b) static pressure ratio, \(p_{3} / p_{1}\), across the compressor (c) exit Mach number, \(M_{3}\) (d) compressor specific work, \(w_{c}\), in \(\mathrm{kJ} / \mathrm{kg}\) Assume fluid properties are: \(\gamma=1.4\) and \(c_{p}=1004 \mathrm{~J} / \mathrm{kgK}\)

Size the exit radius of a centrifugal compressor impeller, \(r_{2}\), that is to reach a tangential Mach number of \(M_{\mathrm{T}}=1.5\). The shaft rotational speed is \(25,000 \mathrm{rpm}\) and the inlet flow condition is characterized by $$ \begin{aligned} p_{1} &=100 \mathrm{kPa} \\ T_{1} &=288 \mathrm{~K} \\ M_{1} &=0.5 \\ \gamma &=1.4 \text { and } c_{p}=1,004 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \end{aligned} $$

A 19-bladed radial impeller has an exit radius \(r_{2}=\) \(25 \mathrm{~cm}\). The impeller width at the exit is \(b=1 \mathrm{~cm}\). We wish to calculate and compare some impeller exit flow areas with successive levels of detail. Case I Assume the individual trailing edge thickness of impeller blades is zero, i.c., \(t=0\), and calculate the geometric flow area. Case 2 Assume the individual trailing edge thickness is \(t=2 \mathrm{~mm}\) (per blade), and calculate the geometric flow area at the impeller exit. Case 3 Assume the impeller exit blockage is \(10 \%\) and now calculate the effective flow area. Assuming the same slip factor between the three cases, relate the average radial velocity \(C_{r 2}\) in cases 1,2, and \(3 .\)

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