Fundamentals of Momentum, Heat and Mass Transfer
Fundamentals of Momentum, Heat and Mass Transfer
6th Edition
ISBN: 9781118804292
Author: WELTY
Publisher: DGTL BNCOM
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Chapter 6, Problem 6.8P
Interpretation Introduction

Interpretation:

The expression for the change in temperature between given stations 1 and 2 is to be determined in terms of A1,A3,v1,v3,cV, and θ .

Concept Introduction:

The general energy balance equation according to the first law of thermodynamics written for a control volume is:

  δQdtδWdt=C.S(e+Pρ)ρ(vn)dA+tC.VeρdV+δWμdt   ...... (1)

The general continuity equation to be used for a controlled volume is:

  C.Sρ(vn)dA+tC.VρdV=0   ...... (2)

The general equation written for the overall linear momentum balance for a controlled volume is:

  F=C.Svρ(vn)dA+tC.VρvdV   ...... (3)

Here, C.S is the control surface over which the integral dA is taken, C.V is the control volume over which the integral dV is taken, ρ is the density of the fluid, v is the velocity vector, n is the direction of the vector v , δQ/dt is the rate of heat added (positive) or removed (negative) from the system, δW/dt is the rate of work done by (positive) or on (negative) the system, and δWμ/dt is the rate of work done to to overcome the viscous effect at the control surface. The product vn is scalar defined as:

  vn=|v||n|cosθ

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The power out of an adiabatic steam turbine is 5 MW and the steam enters turbine at 2 MPa and velocity of 50 m/s, specific enthalpy (h) of 3248 kJ/kg. The elevation of the inlet is 10 m higher than at the datum. The vapor mixture exits at 15 kPa and a velocity of 180 m/s, specific enthalpy (h) of 2361.01 kJ/kg. The elevation of the exit is 6 m higher than at the datum. Let g = 9.81 m/s². Assuming the ideal gas model and R = 0.462 KJ/(kg.K). The steam specific heat ratio is 1.283. Calculate:
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