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bixtya [17]
3 years ago
11

You want a potof water to boil at 105 celcius. How heavy a

Engineering
1 answer:
ankoles [38]3 years ago
3 0

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

Patm = 101 kPa

to find out

How heavy a  lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

so

area will be here

area = \frac{\pi }{4} d^2    ..................1

here d is diameter

put the value in equation 1

area = \frac{\pi }{4} 0.15^2

area = 0.01767 m²

so net force is

Fnet = ( Psat - Patm ) × area

Fnet = ( 120.8 - 101 ) × 0.01767

Fnet = 0.3498 KN = 350 N

we know

Fnet = mg

mass = \frac{Fnet}{g}

mass  = \frac{350}{9.8}

mass = 35.7 kg

so 35.7 kg lid we put

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The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume mus = 0.18.
barxatty [35]

The omitted part of the question shown in bold format

The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume μs = 0.18 and μk = 0.16. What couple must be applied to the shaft to exert a 30-lb force on the clamped object?

Answer:

C = 0.967 in. lb

Explanation:

Given that:

The lead of the threaded shaft of the C-clamp is 0.05 in.

∴ the pitch of the screw = 0.05 in

the mean radius of the thread is r = 0.15 in.

Assuming:

(μs)= 0.18  which implies the  coefficient of the static friction

(μk) = 0.16  (coefficient of kinetic friction)

Force = 30-lb

What couple must be applied to the shaft to exert a 30-lb force on the clamped object?

To determine the Couple (C) that must be applied; we use the expression:

C = Fr  tan (\theta_k + \alpha)

where; F = force

r =  mean radius

\theta_k= angle of kinetic friction

\alpha = pitch angle

NOW, let's take then one after the other.

From the coefficient of the static friction and the kinetic friction; we can solve for their respective angles, so we have:

Angle of static friction (\theta_s)  = tan^{-1}(U_s)

(\theta_s) = tan^{-1}(0.18)

(\theta_s) = 10.204°

Angle of kinetic friction (\theta_k)  = tan^{-1}(U_k)

\theta_k = tan^{-1}(0.16)

\theta_k = 9.0903°

To determine the pitch angle(\alpha); we apply the expression:

(\alpha) = tan^{-1}(\frac{p}{2 \pi r} )

(\alpha) = tan^{-1}(\frac{0.05}{2 \pi 0.15} )

(\alpha) = tan^{-1}(0.0530516 )

(\alpha) = 3.0368°

Have gotten our parameters to solve for Couple (C); then we have:

C = Fr  tan (\theta_k + \alpha)

substituting our values; we have:

C = (30 × 0.15)  tan ( 9.0903 + 3.0368)

C = 4.5  × tan ( 12.1271)

C = 4.5  × 0.2148761968

C = 0.96669428854 in.lb

C = 0.967 in. lb

Therefore, 0.967 in. lb couple  must be applied to the shaft to exert a 30-lb force on the clamped object.

6 0
3 years ago
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