Question

A pressure cooker cooks a lot faster than an ordinary pan by maintaining a higher pressure and temperature inside. the lid of a pressure cooker is well sealed, and steam can escape only through an opening in the middle of the lid. a separate metal piece, the petcock, sits on top of this opening and prevents steam from escaping until the pressure force overcomes the weight of the petcock. the periodic escape of the steam in this manner prevents any potentially dangerous pressure buildup and keeps the pressure inside at a constant value. determine the mass of the petcock of a pressure cooker whose operation pressure is 100 kpa gauge and has an opening cross-sectional area of 4 mm2. assume an atmospheric pressure of 101 kpa.

Answer 1

Answer:

$m=40.816\ gm$

Explanation:

Given;

The operational pressure of the cooker, $P=10^5\ Pa (gauge)$

area of the opening, $a=4\ mm^2$

• We know that the gauge pressure is measured with respect to the atmospheric pressure. As the atmospheric pressure acts on all the components of the system being observed so, we only need to counter the gauge pressure.

As the petcock sits on the opening of the cross-section so, it balances the built-in pressure due to its weight.

Hence:

$m.g=P\times a$

where:

$m=$ mass of the petcock

$g=$ acceleration due to gravity

$m\times 9.8=10^5\times 4\times 10^{-6}$

$m=40.816\ gm$