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Anvisha [2.4K]
3 years ago
10

In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini

tially at 295 K. There is negligible heat transfer between the contents of the tank and their surroundings. Modeling the metal part and water as incompressible with constant specific heats 0.5 kJ/kg K and 4.4 kJ/kg K, respectively, determine the final equilibrium temperature after quenching, in K.

Engineering
1 answer:
storchak [24]3 years ago
8 0

Answer:

attached below

Explanation:

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Sonja [21]

Answer:

sry but it's kinda hard

4 0
3 years ago
Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh
Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

R_a=e^{-\left ( \frac{T-r}{n}\right )B}

R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}

R_a=e^{-4.4}=0.01227

For B

R_b=e^{-\left ( \frac{300-336}{100}\right )3}

R_b=e^{1.08}=2.944

B is better for 100 hours

(b)For 750 hours

R_a=e^{-0.5866}=0.55621

R_b=e^{0.144}=1.154

So here B is more Reliable.

3 0
2 years ago
If superheated water vapor at 30 MPa iscooled at ​constant pressure​, it will eventually become saturated vapor, and with suffic
nirvana33 [79]

Answer:

False.

Explanation:

False. The pressure is above pressure at critical point (22.064 MPa.), the limit where pressure can prevent boiling.

3 0
2 years ago
How to make text take shape of object in affinity designer
Alina [70]

Answer:

To fit text to a shape in Affinity Designer, make sure you have your text selected. Then, grab the Frame Text Tool and click on the shape. A blinking cursor will appear within the shape, indicating that you can begin typing. The text you type will be confined to the boundaries of the shape.

Explanation:

6 0
3 years ago
A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t
gayaneshka [121]

Answer:

Following are the proving to this question:

Explanation:

\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}

using the energy equation for entry and exit value :

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}

where

\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}

         = (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\  V^{2}_{1}\\\\

         = \frac{1}{(2f (\frac{l}{D})  )} \  V^{2}_{1}\\

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\

\to 0+0+Z_0 = 0  +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g}   \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} )  \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to  \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})

L.H.S = R.H.S

7 0
2 years ago
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