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Chapter 4: Problem 44

Air enters a compressor operating at steady state at \(1 \mathrm{~atm}\) with a specific enthalpy of \(280 \mathrm{~kJ} / \mathrm{kg}\) and exits at a higher pressure with a specific enthalpy of \(1025 \mathrm{~kJ} / \mathrm{kg}\). The mass flow rate is \(0.2 \mathrm{~kg} / \mathrm{s}\). If the compressor power input is \(80 \mathrm{~kW}\), determine the rate of heat transfer between the compressor and its surroundings, in \(\mathrm{kW}\). Neglect kinetic and potential energy effects.

Short Answer

Expert verified
The rate of heat transfer is 229 kW.

Step by step solution

01

- Understand the First Law of Thermodynamics for Steady-State

The First Law of Thermodynamics for a control volume at steady state, where kinetic and potential energy effects are negligible, is given by: \[ \frac{dE_{cv}}{dt} = \frac{dm}{dt} (\text{inlet}) \times h_{in} - \frac{dm}{dt} (\text{outlet}) \times h_{out} + \frac{Q}{dt} - W_s \ \] Where: - \( \frac{dm}{dt} \) is the mass flow rate - \( h \) is specific enthalpy - \( \frac{Q}{dt} \) is the rate of heat transfer - \( W_s \) is the shaft work rate (power input).
02

- Plug In Given Values

Given values are: \( h_{in} = 280 \text{kJ/kg} \) \( h_{out} = 1025 \text{kJ/kg} \) \( \dot{m} = 0.2 \text{kg/s} \) \( W_s = 80 \text{kW} \)
03

- Apply the Mass Flow Rate

The rate of change of energy within the control volume is zero at steady state, so the equation simplifies to: \[ 0 = \dot{m} (h_{in} - h_{out}) + \dot{Q} - W_s \]
04

- Solve for Rate of Heat Transfer (\( \dot{Q} \))

Rearrange the equation to solve for \( \dot{Q} \): \[ \dot{Q} = W_s + \dot{m} (h_{out} - h_{in}) \] Substitute the given values: \[ \dot{Q} = 80 \text{kW} + 0.2 \text{kg/s} \times (1025 - 280) \text{kJ/kg} \] Calculate the difference in specific enthalpy: \[ 1025 - 280 = 745 \text{kJ/kg} \] Multiply by the mass flow rate: \[ 0.2 \text{kg/s} \times 745 \text{kJ/kg} = 149 \text{kW} \] Add the compressor power input: \[ 80 \text{kW} + 149 \text{kW} = 229 \text{kW} \] Therefore, the rate of heat transfer to the surroundings is \(229 \text{kW} \).

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Key Concepts

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

steady-state process
A steady-state process occurs when a system's properties do not change over time. In other words, at steady state, the mass, energy, and other properties remain constant. For a compressor, this means that air enters and exits at consistent rates, and so does the energy flow. This is why the rate of change of energy within the control volume is zero in our calculations. Understanding steady-state processes is crucial in analyzing and modeling many engineered systems, providing a simplified approach to otherwise complex problems.
enthalpy
Enthalpy is a thermodynamic property that represents the total heat content of a system. It combines the system's internal energy with the product of its pressure and volume. For incompressible fluids and ideal gases, enthalpy can be related directly to temperature. In the context of this exercise, enthalpy is pivotal because we calculate the work done using the difference in enthalpies between the inlet and outlet of the compressor. By finding this difference, we can determine how much additional energy (in the form of heat transfer) is required or released during the steady-state process.
control volume analysis
Control volume analysis is a fundamental approach in fluid dynamics and thermodynamics used to analyze physical processes. A control volume refers to a predefined region in space through which fluid flows. By applying the first law of thermodynamics to this volume, we can track energy transfer rates such as heat and work. In the problem given, we analyze the compressor as a control volume. This allows us to systematically account for the energies entering and exiting the system, including the mass flow and enthalpies of the air, as well as the work input and heat transfer with the surroundings.
compressor power input
The power input to a compressor is the work done on the air to increase its pressure. In this scenario, it's specified as an 80 kW input. This represents the rate at which work is supplied to the compressor. To determine the overall energy balance, this power input must be combined with the energy changes due to enthalpy differences and heat transfer. Essentially, the power input plus the energy added or removed from the system (enthalpy change) helps us calculate the total heat transfer. This is why the compressor power input is crucial in solving for the heat transfer rate, ensuring all energy flows are accounted for correctly.

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Most popular questions from this chapter

Air enters a compressor operating at steady state with a pressure of 1 bar, a temperature of \(20^{\circ} \mathrm{C}\), and a volumetric flow rate of \(0.25 \mathrm{~m}^{3} / \mathrm{s}\). The air velocity in the exit pipe is 210 \(\mathrm{m} / \mathrm{s}\) and the exit pressure is \(1 \mathrm{MPa}\). If each unit mass of air passing from inlet to exit undergoes a process described by \(p v^{1.34}=\) constant, determine the exit temperature, in \({ }^{\circ} \mathrm{C}\), and the diameter of the exit pipe, in \(\mathrm{mm}\).

Wind turbines and hydraulic turbines develop mechanical power from moving streams of air and water, respectively. In each case, what aspect of the stream is tapped for power?

Refrigerant \(134 \mathrm{a}\) enters the condenser of a refrigeration system operating at steady state at \(9 \mathrm{bar}, 50^{\circ} \mathrm{C}\), through a 2.5-cm-diameter pipe. At the exit, the pressure is 9 bar, the temperature is \(30^{\circ} \mathrm{C}\), and the velocity is \(2.5 \mathrm{~m} / \mathrm{s}\). The mass flow rate of the entering refrigerant is \(6 \mathrm{~kg} / \mathrm{min}\). Determine (a) the velocity at the inlet, in \(\mathrm{m} / \mathrm{s}\). (b) the diameter of the exit pipe, in \(\mathrm{cm}\).

In a turbine operating at steady state, steam enters at a pressure of \(4 \mathrm{MPa}\), specific enthalpy of the steam is \(3018.5 \mathrm{~kJ} / \mathrm{kg}\) and velocity is \(8 \mathrm{~m} / \mathrm{s}\). The steam expands and exits with a velocity of \(80 \mathrm{~m} / \mathrm{s}\) and a specific enthalpy of \(2458.6 \mathrm{~kJ} / \mathrm{kg}\). Pressure at the exit is \(0.09 \mathrm{MPa}\). The mass flow rate of steam is \(10 \mathrm{~kg} / \mathrm{s}\). Determine the power developed by the turbine. Neglect the effect of potential energy.

Steam with \(P_{1}=3.5 \mathrm{MPa}\) and \(T_{1}=350^{\circ} \mathrm{C}\) enters a converging-diverging nozzle operating at steady state. The velocity of steam at the inlet is \(8 \mathrm{~m} / \mathrm{s}\). At the exit of the nozzle, velocity is \(600 \mathrm{~m} / \mathrm{s}\) and pressure is \(P_{2}=1.6 \mathrm{MPa}\). There is negligible heat transfer during flow through the nozzle. Neglect the effect of change in potential energy. Determine the exit area of the nozzle if the mass flow rate is \(2.5 \mathrm{~kg} / \mathrm{s}\).
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