Engineering Fundamentals: An Introduction to Engineering, 4th ed.c

96 Chapter 4 Engineering Communication

ME 101

Mass of air inside the tank, m =?

Problem 3.1

A tank of compressed air

P = 20.8 MPa

V = 10 liters = 0.01 m^3

R = 287

T = 20°C = 293 K

Assuming ideal gas behavior

PV = m RT Eq(1)

where

P = 20.8 MPa = 20.8 X 10^6

V = 10 liters = 0.01 m^3

R = 287

T = 273 + 20 = 293 K

Substituting into Eq (1)

( 20.8 X 10^6 )(0.01 m^3 ) = m( 287 )(293 K)

and realizing that 1 J = 1 N · m,

m

12 Sept. 2001 ASGT. No. 1 Happy, Joe

1

1

GIVEN

FIND

SOLUTION

Any assisting diagrams

Always double-underline answers

and state units

Index answer

Calculations on the right

m = 2.473 Kg

air

N

m^2

N

m^2

J

Kg·K

J

Kg·K

J

Kg·K

■Figure 4.2 An example of engineering homework presentation for Example 4.1.

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