write what you understand in this example, simple answer EXAMPLE 5–10Bernoulli Equation for Compressible FlowDerive the Bernoulli equation when the compressibility effects are not negli-gible for an ideal gas undergoing (a) an isothermal process and (b) an isen-tropic process.SOLUTION The Bernoulli equation for compressible flow is to be obtainedfor an ideal gas for isothermal and isentropic processes.Assumptions 1 The flow is steady and frictional effects are negligible. 2 Thefluid is an ideal gas, so the relation P = pRT is applicable. 3 The specificheats are constant so that Plp* = constant during an isentropic process.Analysis (a) When the compressibility effects are significant and the flowcannot be assumed to be incompressible, the Bernoulli equation is given byEq. 5-40 asdP 4 V².v?+ gz = constant(along a streamline)(1)The compressibility effects can be properly accounted for by expressing p interms of pressure, and then performing the integration S dPlp in Eq. 1. Butthis requires a relation between P and p for the process. For the isothermalexpansion or compression of an ideal gas, the integral in Eq. 1 can be per-formed easily by noting that T = constant and substituting p = PIRT. ItgivesdP= RT In PPIRTSubstituting into Eq. 1 gives the desired relation,Isothermal process:RT In P ++ gz = constant(2)(b) A more practical case of compressible flow is the isentropic flow of idealgases through equipment that involves high-speed fluid flow such as nozzles,diffusers, and the passages between turbine blades. Isentropic (i.e.,reversible and adiabatic) flow is closely approximated by these devices, andit is characterized by the relation Plp* = C = constant, where k is the spe-cific heat ratio of the gas. Solving for p from Plp* = C gives p = C-/k pl./k.Performing the integration,P- +1- 1/k + 1dP = C(3)P - 1/k + 1Substituting, the Bernoulli equation for steady, isentropic, compressible flowof an ideal gas becomesk PIsentropic flow:+ gz = constant(4a)- 1+ gz, =+ gz(4b)A common practical situation involves the acceleration of a gas from rest(stagnation conditions at state 1) with negligible change in elevation. Inthat case we have z= z, and V, = 0. Noting that p = PIRT for ideal.......

Answered: Image /qna-images/answer/6fdd5566-d810-496f-bd56-686c896bd539.jpg

PLE 5–10 Bernoulli Equation for Compressible Flow the Bernoulli equation when the compressibility effects are not negli- for an ideal gas undergoing (a) an isothermal process and (b) an isen- process. TION The Bernoulli equation for compressible flow is to be obtained ideal gas for isothermal and isentropic processes. ptions 1 The flow is steady and frictional effects are negligible. 2 The s an ideal gas, so the relation P = pRT is applicable. 3 The specific are constant so that Plp* = constant during an isentropic process. is (a) When the compressibility effects are significant and the flow t be assumed to be incompressible, the Bernoulli equation is given by 40 as

PLE 5–10 Bernoulli Equation for Compressible Flow the Bernoulli equation when the compressibility effects are not negli- for an ideal gas undergoing (a) an isothermal process and (b) an isen- process. TION The Bernoulli equation for compressible flow is to be obtained ideal gas for isothermal and isentropic processes. ptions 1 The flow is steady and frictional effects are negligible. 2 The s an ideal gas, so the relation P = pRT is applicable. 3 The specific are constant so that Plp* = constant during an isentropic process. is (a) When the compressibility effects are significant and the flow t be assumed to be incompressible, the Bernoulli equation is given by 40 as

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter8: Natural Convection

Section: Chapter Questions

Problem 8.3P

See similar textbooks

Similar questions

A 0.6 m3 tank contains gas at a pressure of 5800 kPa gage, a temperature of 75 deg. Celsius and mass of 25 kg. A part of this gas was discharged and the temperature and pressure changed to 60deg. Celsius and 4130 kPa respectively. Heat was applied and the temperature was back to 75 deg. Celsius. Find: (a) the final mass in the tank in kg CHOICES 9.36 15.63 18.283 (b) the discharged mass in kg. CHOICES 6.716 15.63 9.36 (c) the final pressure of the gas in kPa gage CHOICES 4215.56 5135.88 4135.88
The same engineer decides to look into rates of cooling for liquids toexperiment with different cooling solutions for servers. She finds thatthe rate of cooling for one liquid can be modelled by the equation:y = 48 × 0.99t(0 ≤ t ≤ 80)where y is the temperature of the liquid in degrees Celsius and t is thetime in minutes.(i) State whether the type of reduction for this model is linear orexponential. Describe how reduction rate differs between linear andexponential functions. (ii) Calculate the temperature when t = 20. [3](iii) Write down the scale factor and use this to find the percentagedecrease in the temperature per minute. (iv) Use the method shown in Subsection 5.2 of Unit 13 to find the timeat which the temperature is 30◦(v) Determine the halving time of the temperature.
P Flag question For an ideal gas if the specific internal energy at a specific pressure and temperature of 20 °C is u=123.8 kJ/kg, what is the specific internal energy if the pressure is doubled while the temperature stays the same. Soloot ond
Write down the formula of Clausius inequality, explain all parameters in it and comment when the equality sign “=“ would apply
An oxygen gas R = 0.2598 KJ/kg°k and k = 1.395. If 4 kg of oxygen undergo a reversible non flow constant pressure process from initial volume =1.2 cubic meter and initial pressure = 690 kPa to a state where final temperature = 600°C. What is the initial temperature ? Determine the constant volume specific heat. need a complete solution, symbol, and cancellation:)
What is the value of x for air when below relationship is followed in adaibatic process: P^x V = Constant
5) Using Correlations to Estimate Values for Liquid Properties at Pressures of 2000 psia and 4000 psia Given T = 180°F P₁ = 2500 psia 10,SP R0.5P = 664 SCFISTB Yg.sp=0.56 Ygs = 0.60 POJAPI = 40 °API Pob = 39.5 lb/ft³ Po.2000 41.6 lb/ft³ % = 0,85
Use steam table, Saturated steam has a pressure of 2.546 Mpa. What are the temperature, specificinternal energy, specific volume, and specific entropy? Is the condition of the steamis wet, dry or superheated? Why?
1:09 O71VON 2_53057102948... -> Q/ A dry gas reservoir is initially at an average pressure of 5500 psia and temperature of 145F. The gas has a gravity 0.67. What will the average reservoir pressure be when 40% of the original gas(in Scf) has been remained in the reservoir. Assume the volume occupied by the gas in the reservoir remains constant. If the reservoir originally contained 1MM cuft of reservoir gas, how much gas has been produced at a final reservoir pressure of 600 psia.
-): A mass of air at 330 °C, contained in a cylinder expanded polytropically to Q4/ ( five times its initial volume and 1/8" its initial pressure which is 1 bar. Sketch the process on a p- v diagram. Calculate the value of the expansion index, and the work done per unit mass of air.
ription 3–34 Complete this table for H,O: T, °C Р, КРа v, m³/kg Phase description аpor 140 0.05 550 Saturated liquid 125 750 * (or other) 500 0.140 ||
2. What is the critical value of the pressure coefficient for an airplane flying at 200 mps in air at 10 deg C?