Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 7, Problem 26P

(a)

To determine

The amount of coal consume during a 24 hour period.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The rate of required output of the power plant (W˙net,out) is 300MW.

The thermal efficiency of the plant (ηth) is 0.32.

The change in with respect with time is (Δt) is 24hours.

The value of heating coal (qHV) is 28000kJ/kg.

The ration of air-fuel is AF is 12kgair/kgfuel.

Calculation:

Write the expression for thermal efficiency of power plant.

  ηth=Wnet,outQin

Simplify the Equation (II) as per British thermal unit per hour to obtain the rate of heat transfer of the power plant (Q˙in).

  ηth=W˙net,outQ˙inQ˙in=W˙net,outηth

  Q˙in=(300MW)(0.32)=(300MW)(0.32)=937.5MW

Calculate the amount heat input to the power plant with respect to time.

  Qin=Q˙inΔt

  Qin=(937.5MJ/s)(24h)=(937.5MJ/s)(24h×3600sec1h)=81000000MJ=8.1×107MJ

Calculate the mass of the coal (mcoal).

  Qin=mcoal×qHVmcoal=QinqHV

  mcoal=(8.1×107MJ)(28000kJ/kg)=(8.1×107MJ)(28000kJ/kg×(103MJ/kg1kJ/kg))=2892857kg2.89×106kg

Thus, the amount of coal consume during a 24 hour period is 2.89×106kg_.

Calculate the rate of mass of coal consumed during this period.

  m˙coal=mcoalΔt

  m˙coal=(2.89×106kg)(24h)=(2.89×106kg)(24h×(3600sec1h))=33.48kg/s

(b)

To determine

The rate of air flowing through the furnace.

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculate the rate of air mass flowing through the furnace.

  m˙air=(AF)×m˙coal

  m˙air=(12kgair/kgfuel)×(33.48kg/s)=401.8kg/s

Thus, the rate of air flowing through the furnace is 401.8kg/s_.

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Chapter 7 Solutions

Fundamentals of Thermal-Fluid Sciences

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