#### To determine

**To calculate:**

The work done by the engine during a cycle using exercise 29

#### Answer

**W=1.88×103 lb.ft**

#### Explanation

**1) Concept:**

The work is calculated by using the formula W=∫abf(x)dx

**2) Given:**

The pressure and volume satisfies the equation PV1.4=k and the initial pressure P=160 lb/in2 and the initial and final volumes are V1=100 in3, V2=800 in3

**3) Calculation:**

Convert inch to feet by using the relation 12 inch = 1 feet, so 144 in2=1 ft2 and 1728 in3=1 ft3

At the start P=160 lb/in2=160·144 lb/ft2

V1=100 in3=1001728ft3, V2=8001728ft3 But

PV1.4=k So

k=160·144 10017281.4=2304010017281.4≈426.5 lb.ft

Therefore, from the equation PV1.4=k the value of P is

P=kV-1.4≈426.5V-1.4

Hence, from exercise 29 the work done when the volume expands from V1=100 in3 to V2=800 in3 is

W=∫10017288001728426.5V-1.4 dV

W=426.5∫10017288001728V-1.4 dV

W=426.5V-0.4-0.410017288001728

W=-1066.2517288000.4-17281000.4

W=-1066.252.160.4-17.280.4

W=-1066.251.36-3.12

W=-1066.25-1.766

W=1883.21

W=1.88×103lb.ft

**Conclusion:**

Thework done by the engine during a cycle is W=1.88×103 lb.ft