Homework-4 thermodynamics

pdf

School

University Of Arizona *

*We aren’t endorsed by this school

Course

230

Subject

Mechanical Engineering

Date

Jan 9, 2024

Type

pdf

Pages

Uploaded by CorporalGazellePerson808

Thermodynamics - AME 230 1 Student name : Shyanne Smith Date : 09/29/2023 Homework Assignment No 4 I. Indicate whether the following statements are true or false 1. Mass flux is the time rate of mass flow per unit area. A: True 2. For steady-state operation all properties are unchanging in time. A: True 3.Mass flow rate is the product of density, area, and velocity for 1D flow. A: True 4. As velocity decreases in a diffuser, pressure decreases. B: False 5. Volumetric flow rate is expressed in units of kg/s. B: False m^3/s 6. At steady state, conservation of mass asserts the total rate at which mass enters the control volume equals the total rate at which mass exits. A: True 7. A mixing chamber is a direct-contact heat exchanger A: True 8. As velocity decreases in a diffuser, pressure decreases. A: True 9. For a one-inlet, one-exit control volume at steady state, the mass flow rates at the inlet and exit are equal but the inlet and exit volumetric flow rates may not be equal. A: True 10. Mass flow rate is the time rate of change of mass within the control volume. A: True 11. The mass rate balance is defined as the sum of all inlet mass flow rates minus the sum of all outlet mass flow rates. B: True 12. For one-dimensional flow, mass flow rate is the product of density, area, and velocity. A: True 13. As velocity decreases in a diffuser, pressure decreases.

Thermodynamics - AME 230 2 B: False 14. Flow work is the work done on a flowing stream by a paddle wheel or piston. A: True II. Checking understanding 1. The time rate of mass flow per unit area is called Answer: Mass flux 2. _____steady____-____state means all properties are unchanging in time. 3. A system consists of liquid water in equilibrium with a gaseous mixture of air and water vapor. How many phases are present? Does the system consist of a pure substance? Explain. Answer: There are two phases present, liquid water and then gas from air and water vapor This is not a pure substance since there is a gaseous mixture. It needs to be umiform in composition 4. Does the red line in the picture below follow a constant pressure? Answer: Yes this is constant pressure

Thermodynamics - AME 230 3 5. A device in which power is developed as a result of a gas or liquid passing through a set of blades attached to a shaft free to rotate. Answer: turbine 6. A device in which work is done on a gas to increase the pressure and/or elevation. Answer: compressor 7. A device in which work is done on a liquid to increase the pressure and/or elevation. Answer: pump 8. A flow passage of varying cross-sectional area in which the velocity of a gas or liquid increases in the direction of flow. Answer: nozzle 9. A flow passage of varying cross-sectional area in which the velocity of a gas or liquid decreases in the direction of flow. Answer: diffuser 10. A device where two moving fluid streams exchange heat without mixing. Answer: Heat-exchangers 11. A flow-restricting device that causes a significant pressure drop in the fluid. Answer: throttle III. Steam enters a turbine operating at steady state with a mass flow rate of 10 kg/min, a specific enthalpy of 3100 kJ/kg, and a velocity of 30 m/s. At the exit, the specific enthalpy is 2300 kJ/kg and the velocity is 45 m/s. The elevation of the inlet is 3 m higher than at the exit. Heat transfer from the turbine to the surroundings occurs at a rate of 1.1 kJ per kg of steam flowing. Let g = 9.81 m/s 2 . Determine the power developed by the turbine, in kW. (Please, use problem-solving techniques for this problem, KNOWN, FIND,… ) KNOWN: steam enters a turbine operating at a steady state, so all properties are unchanging in time. Elevation of the inlet is 3m higher than the exit. FIND: Power developed by the turbine in kW. Schematic/data:

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Thermodynamics - AME 230 4 Analysis: from energy equation, we have 𝑚(ℎ 1 + 𝑣 1 2 2 + 𝑔𝑧 1 ) + 𝑚𝑄 = 𝑚(ℎ 2 + 𝑣 2 2 2 + 𝑔𝑧 2 ) +W 10 60 (3100 + 30 2 2 ∗ 10 −3 + 3 ∗ 9.81) − 10 60 ∗ 1.1 = 10 60 (2300 + 45 2 2 ∗ 10 −3 + 0) + 𝑊 516.563=383.5+W W=133.061kW

Thermodynamics - AME 230 5 IV. Separate streams of air and water flow through the compressor and heat exchanger arrangement shown in figure below. Steady-state operating data are provided in the figure. Heat transfer with the surroundings can be neglected, as can all kinetic and potential energy effects. The air is modeled as an ideal gas. The water can be assumed incompressible, so h ≈ h f (T) . Determine a. the total power required by both compressors, in kW. b. the mass flow rate of the water, in kg/s. (Please, use problem- solving techniques for this problem, KNOWN, FIND,… ) Hint: You need to use Table A-22 to find h 1 , h 2 , h 3 , and h 4 . You need to use Table A-2 to find h 5 ≈ h f5 and h 6 ≈ h f6

Thermodynamics - AME 230 6 KNOWN: separate streams of air and water flow through the compressor and heat exchanger arrangement. Steady state operation. Heat transfer, potential, and kinetic energy with surroundings can be neglected. Air is an ideal gas. FIND: a. the total power required by both compressors, in kW. b. the mass flow rate of the water, in kg/s. SCHEMATIC/DATA: ANALYSIS: From table A-22: At T1=300K h 1 = 300.19kJ/kg At T2=600K

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Thermodynamics - AME 230 7 h 2 = 607.02 kJ/kg At T3=450K h 3 = 451.8 kJ/kg At T4=800K h 4 = 821.95 kJ/kg From table A-2: h 5 ≈ h f5 h 6 ≈ h f6 Using the steady flow energy equation for compressor A: Since kinetic energy, potential energy, and heat transfer effects can be ignored, the equation reduces to: Conservation of mass principle says m1=m2 W cvA =m 1 (h 1 -h 2 ) W= 0.6kg/s*(300.19kJ/kg – 607.02kJ/kg)= -184.098kW M1=m3=m4 W= 0.6(451.8-821.95)= -222.09kW Total power by both compressors: -184.098 + -222.09 = -406.188kW 0.6(607.02-451.8)=m(125.82 – 84.01) M=2.2275kg/s