1. (a) Explain the physical process represented by the equation below. Whyis increasing? Hence define Equivalent Potential Temperature.LTWsdw s =Cp0e!deӨ(b) Thus, explain why the adiabatic liquid water content (LWC) is greaterthan the measured LWC inside a cumulus cloud.

This equation represents the adiabatic cooling process experienced by an air parcel rising in the atmosphere.In the given equation,L is the latent heat of condensation for waterT is temperaturews is the saturated mixing ratiocp is the specific heat capacityθ is the potential temperatureθe is the equivalent potential temperatureThe left-hand side of the equation represents the change in entropy due to the release of latent heat as the air parcel cools and condenses water vapour into liquid water. The right-hand side of the equation represents the change in entropy due to the expansion of the air parcel as it rises.As the parcel rises adiabatically, water vapor condenses releasing latent heat. This heat release causes the air temperature θ to increase at the dry adiabatic lapse rate.However, θe remains constant in this reversible process. θe is equivalent to potential temperature - it combines the effects of temperature change and latent heat release during phase changes. It is conserved for reversible adiabatic processes in the atmosphere.The adiabatic LWC represents the maximum amount of water that can condense out for an air parcel lifted reversibly. However, irreversible processes like entrainment, mixing and precipitation formation actually decrease the LWC inside a cloud.Therefore, the adiabatic LWC calculated from the parcel's properties exceeds what is measured inside the cloud. Real processes are not completely reversible and they limit the amount of condensate retained.a.The given equation represents the adiabatic cooling process experienced by an air parcel rising in the atmosphere.The left-hand side (LHS) of the equation represents the release of latent heat as the air parcel cools and condenses water vapour into liquid water. The right-hand side (RHS) of the equation represents the change in sensible heat due to the expansion of the air parcel as it rises.As the parcel rises adiabatically, water vapor condenses releasing latent heat. This heat release causes the air temperature θ to increase at the dry adiabatic lapse rate.However, θe remains constant in this reversible process. θe is equivalent to potential temperature - it combines the effects of temperature change and latent heat release during phase changes. It is conserved for reversible adiabatic processes in the atmosphere.b. The adiabatic LWC represents the maximum amount of water that can condense out for an air parcel lifted reversibly. However, irreversible processes like entrainment, mixing and precipitation formation actually decrease the LWC inside a cloud.Therefore, the adiabatic LWC calculated from the parcel's properties exceeds what is measured inside the cloud. Real processes are not completely reversible and they limit the amount of condensate retained.

Engineering

Mechanical Engineering

1. (a) Explain the physical process represented by the equation below. Why is increasing? Hence define Equivalent Potential Temperature. L T 0 Ws 0e dw s = Cp de ! 0 (b) Thus, explain why the adiabatic liquid water content (LWC) is greater than the measured LWC inside a cumulus cloud.

1. (a) Explain the physical process represented by the equation below. Why is increasing? Hence define Equivalent Potential Temperature. L T 0 Ws 0e dw s = Cp de ! 0 (b) Thus, explain why the adiabatic liquid water content (LWC) is greater than the measured LWC inside a cumulus cloud.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Q: In a closed container of 2 liters, there is 1 kg of wet water vapor at a temperature of 25°C. a.)…

Q: Consider an ideal gas that undergoes a process through which Volume ratio (Vol,/Vol, 0.50) and…

A: Volume ratio, V2V1=0.50 Pressure ratio, P2P1=3.01

Q: A 0.8-m diameter sphere contains helium at 20 bar and 28°C. How many cylindrical drums 12 cm in…

A: Data Provided:- Diameter .8m initial pressure 20 bar initial temp. 28 C

Q: (c) The saturation pressure and enthalpy of evaporation for refrigerant R134a at 25°C can be found…

A: Using R134 a refrigerant Table A-11 T (oC) P (kPa) 4 337.90 5…

Q: A closed piston-cylinder system consists of 3.0 moles of monatomic gas, initially at temperature…

Q: mass of hydrogen

A: m=0.19820Kg T2= 0.5478 K

Q: Q3: In a closed vessel with a volume of 50 dm3 there are 2 moles of an ideal monoatomic gas with cv,…

A: Given: Volume, V=50 dm3V=50 ×0.001 m3V=0.05 m3number of moles, n=2ideal monoatomic gas,Cv=12.471…

Q: and point out the range

A: The plot of polytropic process on P-V and T-S diagram for endothermic, volume decrease, heating and…

Q: Carbon dioxide (molar mass 44 kg/kmol) expands rever sibly in a perfectly thermally insulated…

Q: could you answer d part too please,(For state 1 evaluate the specific volume assuming the steam…

Q: Determine the missing properties, show it on a P-V diagram. T (°C) u (kJ/kg) P (kPa) 400 v (m²/kg)…

Q: 2. An ideal monoatomic gas at initial temperature of T, (in K) doubles its volume from initial…

Q: Hydrogen gas at 65 kPa, 200 °C is heated in a closed, rigid vessel until it reaches to 400 °C.…

Q: Find the missing properties and give the phase of the substance. (a) H2O s = 1.75 Btu/lbm R, P = 4…

Q: A 6m3 tank contains helium at 127 C and in evacuated from atmospheric pressure to a pressure of 740…

Q: Don't Use Chat GPT Will Upvote And Give Handwritten Solution Please

Q: The simplest equation of state for liquid water is the so-called stiffened-gas equation of state in…

A: Given:- γ= 4-7π=4×108Pac=4200kJ/kgK To find: a) Volumetric thermal expansion b) Isothermal…

Q: A/ Find the missing properties: P (kpa) 320 u (KJ/kg) 0.5 Phase description Material T('C) X R-134…

A: We are suppose to solve only one question. Please post other question as a separate question.

Q: 3. Find the coefficient of thermal expansion a and the isothermal com pressibility k from each of…

A: Given: As per the question

Q: A 0.8 m diameter sphere contains helium at 20 bar and 28 deg C. How many cylindrical drums, 15 cm in…

A: Given Data:- Diameter= 0.8m Pressure= 20 bar temp =28C

Q: 2.3 Consider 0.020 mol argon gas at 25°C with a constant volume molar heat capacity 12.47 J/ (K mol)…

Q: 025kg/s of Caton Dioxide (R 0.1889 kJ/kg-K, k 1.289) undergo a certain process in a close system…

Q: 90 90 bpB Vapor 80 + 80 70 70 Liquid 60 60 50 100% A 80 20 60 40 40 60 20 80 100% B 0% 0% b. What is…

Q: Determine, using complex numbers, the magnitude and direction of the resultant of the coplanar…

Q: For a certain gas, R = 0.277 kJ/kgm K and k = 1.384. (a) What are the value of cpand Cv? (b) What…

Q: Q3: In a closed vessel with a volume of 50 dm3 there are 2 moles of an ideal monoatomic gas with cv,…

A: Given data:-Volume of vessel(v)=50m3No. of moles=n=3specific heat of constant volumecv=12.471 Jk…

Q: An ideal gas inside a closed system is initially at atmospheric pressure and a temperature of 300K.…

A: Given:To find:Final temperature of the gas.

Q: Thermodynamics problem. Two kilograms of water at 125KPa are completely vaporized. Calculate the…

A: For water m = 2 kg P = 125 kPa x1 = 0, x2 = 1 It is required to determine V1 and V2

Q: Air is initially at 50 °C (dry-bulb temperature). Its specific volume is 0.95 m³/kg dr air. It is…

Q: 1 kg of Oxygen has initial volume of 2 x 106 cm3 is heated at Isobaric process from 298 K to 200°C.…

A: Given data: The mass of the oxygen is m=1 kg The initial volume is V1=2×106 cm3=2 m3 The initial…

Q: Calculate the Joule-Thomson coefficient (u=(aT/aP)„) and x, (@T / ƏP), coefficient for %3D steam at…

A: Temperature, T= 500∘C Pressure, P= 10 MPa To find: i) μ= ∂T∂PHii) κs= ∂T∂PS

Q: RT For a gas with the equation of state V= +(aT2 + b)P the P expression of is:

Concept explainers

Properties of Pure Substances

A substance with fixed chemical composition throughout is called a pure substance like water, air, and nitrogen. A pure substance does not have to be a single element or compound. A mixture of two or more phases of a pure substance is also a pure substance as long as the chemical composition is the same for all phases. These substances mainly have a constant/ uniform composition. The substances also have fixed boiling and melting points. A pure substance takes part in a chemical reaction to form predictable products.

Question

1. (a) Explain the physical process represented by the equation below. Why
is increasing? Hence define Equivalent Potential Temperature.
L
T
Ws
dw s =
Cp
0e
!
de
Ө
(b) Thus, explain why the adiabatic liquid water content (LWC) is greater
than the measured LWC inside a cumulus cloud.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1: sub part a

VIEW

Step 2: subpart b

VIEW

Solution

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

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

An ideal gas is adiabatically expanded along path AB from a temperature T=500 K to 300 K, and then isochorically heated along a path BC back to T=500 K. (i) Draw a pressure-volume diagram for this process and show that the ratio of pressures at points A and C obeys (3) PA PC where y is the adiabatic index of the ideal gas. (ii) Given that the ratio of pressures at points A and C is found to be PA 3.59 PC
Q3) Initially, 1kg air at 600 K and at 150 kPa is contained in a piston-cylinder.device. Now, the system is started to be heated during a quasi-equilibrium process until the temperature of air reaches to 900 K. During the process the pressure inside the cylinder is held constant. Air can be assumed as Ideal Gas under these conditions. [a) Calculate the heat provided to the air inside the cylinder during this process. [b) Sketch the P[kPa]-V[m³] diagram of the process. Indicate the direction of the process. No need to express the volume of the air on the diagram. Please do only a rough sketch. Ic) Calculate the work done by the air onto surrounding. I.
I need a clear answer by hand, not by keyboard | dybala
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 ||
Asap
Given: Propane, T = 18.8°C, h = 175 kJ/kg Find: u (kJ/kg) Your Answer:
2. Consider a metal rod with the equation of state L- Lo = Lo +a(T YA %3D Here Lo, Y, A, To, and a are constants,T is the temperature, L is the length, and F is the tension. Work is given by dW a) Using the first law and the expression for dU in terms of it partial derivatives with respect to T and L, show that for this system -FdL. (). -). G). (4). - se, b) By differentiating the first of these equations with respect to L and the second with respect to T show that ), -r(器). =F-T ƏT c) Using the above result and the equation of state, find an expression for (0/aL). Then, assuming that C. the heat capacity at constant length is a Constant. find an expression for the internal energy, U, as a function of T and L.
I'm not sure but You may need to use Clausius-Clapeyron equation ,Whatever just do it correctly.
Exercise 1 1: The specific volume of superheated water vapor at 0.6 MPa and 900°C is 0.90179 m³/kg in the steam tables, determine the specific volume also using: (a) The ideal-gas equation. :s) (b) The generalized compressibility chart. 'I (c) Can we consider the superheated vapor to behave as an ideal gas under the given temperature and pressure? Why?
ts Suppose you have a gas that obeys the following modified van der Waals equation: P(V – nb) = nRT %3D a) In 1-2 sentences, provide a physical interpretation for this equation. HINT: Compare this with the octual van der Waais equation. What does the missing term represent? b) For a gas obeying this modified van der Waals equation, derive an expression for the work done by a reversible and isothermal change in volume. c) For this modified van der Waals equation (with n and b as constants), find the following two partial derivatives: i) ii) T
QUESTIOΝ 1: A piston-cylinder arrangement contains 2 kg of steam originally at 200°C and x-0.9. The volume triples while thc temperature is held coustant. Pieten Calculate: (a) the linal pressure, in MPa, 7=200 C (c) the change in intermal energy, in kl.
Determine P (kPa) 400 the missing properties, show it on a P-V diagram. T (°C) u (kJ/kg) h (kJ/kg) X v (m³/kg) 0.465 ? ? ? Phase description