Concept explainers
(a)
The critical temperature of air.
The critical pressure of air.
The critical density of air.
(a)
Answer to Problem 37P
The critical temperature of air is
The critical pressure of air is
The critical density of air is
Explanation of Solution
Determine the stagnation temperature of ideal gas.
Here, the static temperature of ideal gas is
Determine the stagnation pressure of ideal gas.
Here, the static pressure of ideal gas is
Determine the density of the ideal gas.
Here, the pressure of the ideal gas is
Determine the critical temperature at the throat of nozzle.
Here, the stagnation temperature of ideal gas is
Determine the critical pressure at the throat of nozzle.
Here, the stagnation pressure of ideal gas is
Determine the critical density at the throat of nozzle.
Here, the stagnation density of ideal gas is
Conclusion:
From the Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of universal gas constant, specific heat of pressure, and the specific heat ratio of air at
Substitute
Substitute 200 kPa for
Substitute 317.0 kPa for
Substitute
Thus, the critical temperature of air is
Substitute
Substitute
Thus, the critical density of air is
(b)
The critical temperature of helium.
The critical pressure of helium.
The critical density of helium.
(b)
Answer to Problem 37P
The critical temperature of helium is
The critical pressure of helium is
The critical density of helium is
Explanation of Solution
Conclusion:
From the Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of universal gas constant, specific heat of pressure, and the specific heat ratio of helium at
Substitute
Substitute 200 kPa for
Substitute 213.3 kPa for
Substitute
Thus, the critical temperature of helium is
Substitute
Thus, the critical pressure of helium is
Substitute
Thus, the critical density of helium is
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Chapter 17 Solutions
Thermodynamics: An Engineering Approach
- The fluid is heated from 125 degrees Fahrenheit to 225 degrees Fahrenheit. Consider an ideal gas with thefollowing characteristics: R = 85 ft-lbf/lbm-RCp = 0.35 + 0.000325T BTU/lbm-R If the heating is at constant volume, compute for (a) the change in internal energy, (b) the change in enthalpy,and (c) the change in entropy. If the heating is at constant pressure, compute for (d) the change in entropy, and (e) the value of k at 160 degrees Celsius. If the fluid undergoes an isentropic process, determine (f) non-flow work and (g) steady-flow work. (For item f and g, use the value of k at 160 degrees Celsius)arrow_forwardSteam at 500kPa and a quality of 90 percent occupied a rigid vessel of volume 0.3m2. Calculate the mass, internal energy, and enthalpy of the steam.arrow_forwardAt a point in an airflow the pressure, temperature, and velocity are 1 atm, 320 K, and 1000 m/s. Calculate the total temperature and total pressure at this point.arrow_forward
- Helium enters a nozzle at 0.5 MPa, 600 K, and a velocity of 120 m/s. Assuming isentropic flow, determine the pressure and temperature of helium at a location where the velocity equals the speed of sound. What is the ratio of the area at this location to the entrance area?arrow_forwardA prototype boiler that contains 10 kg of saturated steam at a pressure of 482.5kPa. (a) Determine the amount of heat which must be rejected in order to reduce the quality to 70%. (b) What will be the pressure and temperature of the steam at this new state? Compute for the (c) heat, (d) internal energy, (e) entropy, and (f) enthalpy at this new state.arrow_forwardHelium enters a nozzle at 0.5 MPa, 600 K, and a negligible entrance velocity. Assuming isentropic flow, determine the pressure and temperature of helium at a location where the velocity equals the speed of sound. What is the ratio of the area at this location to the entrance area?arrow_forward
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- Three cubic meter of air at a pressure of 20 kPag and 20 0 C is compressed atconstant volume to a pressure of 425 kPag. determine the following: a. change in internal energy, change in enthalpy and change in entropyarrow_forwardFor the specific volume of wet steam, SV=(1-Xv)SV (liq) + XvSV (vapor). Entropy is also calculated this way. If a tank initially has 5kg of wet steam with mass of vapor =1 kg at 100 kPa, and it is heated such that saturated vapor remains in the tank. Assuming that the process is in constant volume, what will be the entropy change of the steam (Kj/K)?arrow_forwardConsider 9 kg/s of saturated liquid at 30 atm. Calculate its entropy in kW/karrow_forward
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