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ladessa [460]
3 years ago
7

Q5. A hypothetical metal alloy has a grain diameter of 2.4 x 10-2 mm. After a heat treatment at 575°C for 500 min, the grain dia

meter has increased to 7.3 x 10-2 mm. Compute the time required for a specimen of this same material (i.e., d0 = 2.4 x 10-2 mm) to achieve a grain diameter of 5.5 x 10-2 mm while being heated at 575°C. Assume the n grain diameter exponent has a value of 2.2. (10 points)
Engineering
1 answer:
alexdok [17]3 years ago
8 0

Time is 244.89 minutes

<u>Explanation:</u>

<u />

Given:

Hypothetical diameter, d₀ = 2.4 X 10⁻² mm

Increase in diameter, d = 7.3 X 10⁻² mm

d₀ = 2.4 X 10⁻² mm

Time, t = 500 min

To solve K:

K = \frac{d^n - d_o^n}{t}

On substituting the value:

K = \frac{(7.3 X 10^-^2)^2^.^2 - (2.4 X 10^-^2)^2^.^2}{500} \\\\\\K = 5.8 X 10^-^6 mm^2^.^2/min

From the value of K, t can be calculated as:

t = \frac{d^2^.^2 - d_o^2^.^2}{K} \\\\t = \frac{(5.5 X 10^-^2)^2^.^2 - (2.4 X 10^-^2)^2^.^2}{5.8 X 10^-^6} \\\\t = 244.89 min

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4. The process is stated to be reversible

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P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

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m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

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b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

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=5.8491(246.82-219.9)

=5.8491(26.91)

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