Week 4 lecture 5

Question 1

Equation for kinetic energy

Answer 1

0.5 m v^2

Question 2

Equation for potential energy

Answer 2

mgz

Question 3

Energy stored in the system, manifestations of it are temperature and phase.

Answer 3

Internal energy

Question 4

Equation for the change in energy of the system

Answer 4

(U + mgz + 0.5mv^2) - (U +mgz +0.5mv^2)

Assumption: we are dealing with a closed system of mass m

Question 5

Equations for delta U

Answer 5

1) U2 - U1
2) Q-W

U = Q-W is the first law of thermodynamics for closed systems

Question 6

Equation for enthalpy

Answer 6

H = U +PV

Used for open systems

Question 7

Equation for the energy exchange with the surroundings with mass flow

Answer 7

0 = Q - W + ΣUi + migzi + 0.5mivi^2- ΣUj + mjgzj +0.5mjvj^2

ΣUi + migzi + 0.5mivi^2 = sum for all inflows
ΣUj + mjgzj +0.5mjvj^2 = sum for all outflows

Question 8

Two elements of work

Answer 8

1. Shaft work
2. Flow work

Question 9

Equation for work using the two elements of work

Answer 9

Work = Shaft work + Flow work

Question 10

Equation for enthalpy change

Neglecting potential and kinetic energy

Answer 10

Q - Ws = deltaH

Ws = shaft work

Question 11

Equation for flow work

Answer 11

Wf = ΣPjVj - ΣPiVi

Question 12

All the other work exchanged between the process and surroundings excluding flow work

Answer 12

Shaft work

Question 13

The work which will be needed in order to maintain a continuous flow through a control volume

Answer 13

Flow work

https://www.hkdivedi.com/2016/07/flow-work-or-flow-energy-in.html

Question 14

In a nozzle or diffuser, the only work is _____

Answer 14

flow work

Question 15

What shaft work in a nozzle/diffuser is equal to

Answer 15

Ws = O

The change is potential energy is also negligible

Question 16

Energy balance over nozzle/diffuser

Answer 16

0 = Q + m[(h1-h2) + 0.5(V1^2 - V2^2)]

Normally for nozzles and diffusers the heat transfer rate is also relative to the enthalpy and kinetic energy thus the energy balance becomes:
0 = (h1-h2) + 0.5(v1^2 - v2^2)

Question 17

Mass flowing through state, steady flow devices is:

Answer 17

Constant

Question 18

The steady state, steady flow first law of turbines

Answer 18

0 = Q - Ws + m(h1-h2)

kinetic and potential energy changes have been neglected
When heat transfer between the turbine and surroundings is often small enough relative to the power and thalpy terms that is can also be neglected
Ws = m(h1-h2)

Question 19

Devices in which work is done on the substance, typically to increase the pressure and/or elevation

Answer 19

Compressors and pumps

Question 20

Steady rate, steady flow first law equation for compressors and pumps

Answer 20

0 = Q - Ws + m(h1-h2)

Ignoring heat transfer to the surroundings
Ws = m(h1-h2)

Question 21

Well insulated devices that allow energy exchange between hot and cold fluids without mixing the fluids

Answer 21

Heat exchangers

Question 22

The only work in a heat exchanger

Answer 22

Flow work

Question 23

What shaft work in a heat exchanger equals

Answer 23

Ws = 0

Question 24

Steady rate, steady flow first law equation for heat exchangers

Answer 24

0 = Q + Σmihi - Σmjhj

Question 25

Used to create a significant reduction in pressure by introducing a restriction into a line through which a gas or liquid flows

Answer 25

Throttling devices

Question 26

Steady rate, steady flow first law equation for throttling devices

Answer 26

0= (h1-h2)+1/2(v1^2-v2^2)

Assumptions:
- process is adiabatic
- potential energy is neglected
Neglecting kinetic energy: h1=h2