## Steam Engine Piston

Consider a steam engine's piston has a 75 cm diameter, and a max steam gauge pressure of 2.0 MPa. If the piston's rod has a design stress of 70 MPa, what should be its minimum cross sectional area in m^2?

##
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__**Hint**

**Hint**

$$$Pressure =\frac{Force}{Area}$$$

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__**Hint 2**

**Hint 2**

Since the piston force is the same force on the piston's rod, uniaxial loading is applied:

$$$\sigma =\frac{F}{A}$$$

where
$$\sigma$$
is the stress on the cross section,
$$F$$
is the loading, and
$$A$$
is the cross sectional area.

Given the problemâ€™s knowns, let's first find the force on the piston caused by the steam pressure.

$$$\sum F=(pressure)(area)=(P)(\pi \cdot r^2)$$$

$$$F_{piston}=(2\cdot 10^6Pa)[\pi (.75/2m)^2]=883,125\:N$$$

Since the piston force is the same force on the piston's rod, uniaxial loading is applied:

$$$\sigma =\frac{F}{A}\rightarrow F=\sigma A=F_{rod}$$$

where
$$\sigma$$
is the stress on the cross section,
$$F$$
is the loading, and
$$A$$
is the cross sectional area.

Because
$$F_{piston}=F_{rod}$$
, to solve for
$$A_{rod}$$
:

$$$A_{rod}=\frac{F_{rod}}{\sigma}=\frac{883,125N}{70\cdot 10^6Pa}=0.013\:m^2$$$

$$$0.013\:m^2$$$