FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 4, Problem 4.20CU
To determine
The mass flow rate does not depend on which property of the fluid.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
4.5. Values of pressure (kPa) measured at various stations of a zero-reaction gas
turbine stage, all at the mean blade height, are shown in the table given below.
Stagnation pressure
Nozzle entry 414
Nozzle exit 400
Static pressure
Nozzle exit
Rotor exit
207
200
The mean blade speed is 291 m/s, inlet stagnation temperature 1100 K, and the flow
angle at nozzle exit is 70° measured from the axial direction. Assuming the
magnitude and direction of the velocities at entry and exit of the stage are the same,
and the axial velocity remains constant, determine the total to total efficiency of
the stage. Assume a perfect gas with Cp = 1.148 kJ/(kg °C) and y = 1.333.
Question 19
The figure below shows a turbine-driven pump that provides water to a mixing chamber located dz = 25 m
higher than the pump, where mass flow rate is 50 kg/s. Steady state operating data for the turbine and
pump are labeled on the figure. Heat transfer from the water to its surroundings occurs at a rate of 2 kW.
For the turbine, heat transfer with the surroundings and potential energy effects are negligible. Kinetic
energy effects at all numbered states can be ignored.
Steam
P3 = 30 bar
T3 = 400°C
3
Turbine
▼
P4= 5 bar
T₁ = 180°C
h₂ = 417.69 kJ/kg
Ocv = 2 kW
Pump
dz
+Mixing chamber
Saturated liquid water
m₁, P₁ = 1 bar
Derive the 7 general property equation of the following thermodynamics processes:
a. ISOMETRIC PROCESS
Chapter 4 Solutions
FUND OF ENG THERMODYN(LLF)+WILEYPLUS
Ch. 4 - Prob. 4.1ECh. 4 - Prob. 4.2ECh. 4 - Prob. 4.3ECh. 4 - Prob. 4.4ECh. 4 - Prob. 4.5ECh. 4 - Prob. 4.6ECh. 4 - Prob. 4.7ECh. 4 - Prob. 4.8ECh. 4 - Prob. 4.9ECh. 4 - Prob. 4.10E
Ch. 4 - Prob. 4.11ECh. 4 - Prob. 4.12ECh. 4 - Prob. 4.13ECh. 4 - Prob. 4.14ECh. 4 - Prob. 4.15ECh. 4 - Prob. 4.1CUCh. 4 - Prob. 4.2CUCh. 4 - Prob. 4.3CUCh. 4 - Prob. 4.4CUCh. 4 - Prob. 4.5CUCh. 4 - Prob. 4.6CUCh. 4 - Prob. 4.7CUCh. 4 - Prob. 4.8CUCh. 4 - Prob. 4.9CUCh. 4 - Prob. 4.10CUCh. 4 - Prob. 4.11CUCh. 4 - Prob. 4.12CUCh. 4 - Prob. 4.13CUCh. 4 - Prob. 4.14CUCh. 4 - Prob. 4.15CUCh. 4 - Prob. 4.16CUCh. 4 - Prob. 4.17CUCh. 4 - Prob. 4.18CUCh. 4 - Prob. 4.19CUCh. 4 - Prob. 4.20CUCh. 4 - Prob. 4.21CUCh. 4 - Prob. 4.22CUCh. 4 - Prob. 4.23CUCh. 4 - Prob. 4.24CUCh. 4 - Prob. 4.25CUCh. 4 - Prob. 4.26CUCh. 4 - Prob. 4.27CUCh. 4 - Prob. 4.28CUCh. 4 - Prob. 4.29CUCh. 4 - Prob. 4.30CUCh. 4 - Prob. 4.31CUCh. 4 - Prob. 4.32CUCh. 4 - Prob. 4.33CUCh. 4 - Prob. 4.34CUCh. 4 - Prob. 4.35CUCh. 4 - Prob. 4.36CUCh. 4 - Prob. 4.37CUCh. 4 - Prob. 4.38CUCh. 4 - Prob. 4.39CUCh. 4 - Prob. 4.40CUCh. 4 - Prob. 4.41CUCh. 4 - Prob. 4.42CUCh. 4 - Prob. 4.43CUCh. 4 - Prob. 4.44CUCh. 4 - Prob. 4.45CUCh. 4 - Prob. 4.46CUCh. 4 - Prob. 4.47CUCh. 4 - Prob. 4.48CUCh. 4 - Prob. 4.49CUCh. 4 - Prob. 4.50CUCh. 4 - Prob. 4.51CUCh. 4 - Prob. 4.1PCh. 4 - Prob. 4.2PCh. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Prob. 4.5PCh. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - Prob. 4.11PCh. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Prob. 4.16PCh. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Prob. 4.22PCh. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Prob. 4.29PCh. 4 - Prob. 4.30PCh. 4 - Prob. 4.31PCh. 4 - Prob. 4.32PCh. 4 - Prob. 4.33PCh. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - Prob. 4.36PCh. 4 - Prob. 4.37PCh. 4 - Prob. 4.38PCh. 4 - Prob. 4.39PCh. 4 - Prob. 4.40PCh. 4 - Prob. 4.41PCh. 4 - Prob. 4.42PCh. 4 - Prob. 4.43PCh. 4 - Prob. 4.44PCh. 4 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - Prob. 4.48PCh. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - Prob. 4.51PCh. 4 - Prob. 4.52PCh. 4 - Prob. 4.53PCh. 4 - Prob. 4.54PCh. 4 - Prob. 4.55PCh. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Prob. 4.58PCh. 4 - Prob. 4.59PCh. 4 - Prob. 4.60PCh. 4 - Prob. 4.61PCh. 4 - Prob. 4.62PCh. 4 - Prob. 4.63PCh. 4 - Prob. 4.64PCh. 4 - Prob. 4.65PCh. 4 - Prob. 4.66PCh. 4 - Prob. 4.67PCh. 4 - Prob. 4.68PCh. 4 - Prob. 4.69PCh. 4 - Prob. 4.70PCh. 4 - Prob. 4.71PCh. 4 - Prob. 4.72PCh. 4 - Prob. 4.73PCh. 4 - Prob. 4.74PCh. 4 - Prob. 4.75PCh. 4 - Prob. 4.76PCh. 4 - Prob. 4.77PCh. 4 - Prob. 4.78PCh. 4 - Prob. 4.79PCh. 4 - Prob. 4.80PCh. 4 - Prob. 4.81PCh. 4 - Prob. 4.82PCh. 4 - Prob. 4.83PCh. 4 - Prob. 4.84PCh. 4 - Prob. 4.85PCh. 4 - Prob. 4.86PCh. 4 - Prob. 4.87PCh. 4 - Prob. 4.88P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- 1) One of the common devices used in Engineering is a nozzle to transfer fluids such as air. Assume a steady-state flow and air as an ideal gas with the following parameters: Inlet: Mass flow = 2.1 kg/s V1 = 0.72 m³/kg T1 = 600 K A = 240 cm? Outlet: V2 = 2.7 m³/kg A2 = 40 cm2 a) What is the air pressure at the outlet. b) What is the velocity of the airflow at the inlet. c) What is the velocity of air at the outlet.arrow_forward17. I need the Equations for the nozzle of gas turbine engine of an airplane. (Balance of energy and mass) everythingarrow_forward4 STEAM P₁ = 40 BAR (AV), P₂ 3 P4 = Ask: = 70m3 min 20 Bar T₁ = 500°C T2= P3= 20 Bar SATURATED VAPOR 400°C T3 = 500°C KINETIC AND POTENTIAL ENERGY = STEADY STATE = 0,6 Bar TURBINE 1 TOTAL POWER iN kW RATE OF HEAT TRANSFER Ø GENERATED By а (data) TURBINE 2 REHEATER dQ/dt BOTH IN ки TURBINES IN THE RE HEATERarrow_forward
- Q2. The following particulars relate to a two row velocity compounded impulse wheel which forms a first stage of a combination turbine. Steam velocity at nozzle outlet = 579.12m/s Mean blade velocity = 115.82m/s Nozzle outlet angle = 16° Outlet angle first row of moving blades = 18° Outlet angle fixed guide blades = 22° Outlet angle, second row of moving blades = 36° Steam flow rate = 2.4 kg/s The ratio of the relative velocity at outlet to that at inlet is 0.84 for all blades. Determine for each row of moving blades the following ⚫ The velocity of whirl • The tangential thrust on blades • The axial thrust on the blades • The power developed What is the efficiency of the wheel as a whole?arrow_forwardQUESTION 1 Match the laws accordingly. | 1st law of thermodynamics A. No Heat Engine can be more efficient than a Reversible Heat Engine operating between the same temperature limits 2nd law of thermodynamics (foward engine) B. The internal energy of a perfect gas is a function of the absolute temperature only c. When a system undergoes a complete cycle, the net Heat supplied plus the net Work input is zero. Joules law 3rd law of thermodynamics 2nd law of thermodynamics (reverse engine) D. It is impossible to construct a device that operating in a cycle will produce no effect other than the transfer of heat from a cooler to a hotter body. E. A pure crystalline substance at absolute zero temperature is in perfect order, and its entropy is zero Second law of thermodynamics (Carnot) F. It is impossible for a heat engine to produce a net work output in a complete cycle if it exchanges heat only with a single energy reservoir.arrow_forwardTHERMODYNAMICSarrow_forward
- As shown in the figure, air with a volumetric flow rate of 17,000 ft3/min enters an air-handling unit at 80oF, 1 atm. The air-handling unit delivers air at 45oF, 1 atm to a duct system with three branches consisting of two 26-in.-diameter ducts and one 50-in. duct. The velocity in each 26-in. duct is 19 ft/s. Assume steady state operation and ideal gas behavior for the air. Determine the mass flow rate of air entering the air-handling unit, in lb/s. The volumetric flow rate in each 26-in. duct, in ft3/min. The velocity in the 50-in. duct, in ft/s.arrow_forwardThe displacement volume of an internal combustion engine is 3 liters. The processes within each cylinder of the engine are modeled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by p1 = 95 kPa, T1 = 228 degree C, and V1 = 3.17 liters. Determine:a.) the net work per cycle, in kJb.) the power developed by the engine, in kWc.)the thermal efficiency, if the cycle is executed 1000 times per minarrow_forwardStep by step solution and answer please thank youuarrow_forward
- At the beginning of the compression process of an air standard Otto cycle, p1 = 1 bar, T1 = 300 K. The maximum temperature in the cycle is 2250 K and the compression ratio is 9.8. The engine has 4 cylinders and an engine displacement of Vd = 2.2 L. Determine per cylinder: a) the volume at state 1.b) the air mass per cycle.c) the heat addition per cycle, in kJ.arrow_forward1. A single-acting, single-cylinder air compressor is rated at 4.25 (m)^3/min of air. The suction conditions are 1 atm and 27 C and discharge pressure is 1034 kPa. The compression process follows the equation PV^1.35 = C. Determine the power, in kW, required to compress the air. a. -30.22 kW b. -22.87 kW c. -82.27 kW d. -27.22 kWarrow_forwardThe following are examples of a steady-state flow except: a. running a gas through a fan b. filling up an evacuated tank c. deflating a helium balloon d. boiling water in a pressure cookerarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
The Laws of Thermodynamics, Entropy, and Gibbs Free Energy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8N1BxHgsoOw;License: Standard YouTube License, CC-BY