In a double acting steam engine, the average pressure of steam is… - Vidyadip

Question

In a double acting steam engine, the average pressure of steam is $8 \times 10^4 \mathrm{Nm}^{-2}$. The length of the stroke is 0.8 m and area of cross section of the piston is $0.18 \mathrm{~m}^2$. If the piston makes 360 rpm , calculate the horse power of the engine.

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Answer

Here, $\text{P}=8\times10^4\text{Nm}^{-2},\text{L}=0.8\text{m}$ $\text{A}=0.18\text{m}^{2},\text{N}=360\text{rpm}=6\text{rps.}$ Horcse power $=\frac{2\text{P}\times\text{L}\times\text{A}\times\text{N}}{746}$ $=\frac{2\times8\times10^4\times0.8\times0.18\times6}{746}$ $=185.3\text{HP}$

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