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3 years ago

What is the lehr and what purpose does it serve?

Engineering

1 answer:

levacccp [35]3 years ago

5 0

Explanation:

Step1

Lehr is the long open or closed insulated space for glass manufacturing. Lehr must be large enough to keep the cooling of glass uniform. The function of Lehr is the same as an annealing process in metallurgy.

Step2

Lehr decrease the cooling and temperature variation in glass production. Uneven temperature creates the internal stress in the glass. Lehr reduces the internal stress in the glass product. So, the main purpose of the Lehr is to reduce the internal stress and keep the cooling uniform.

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Compute the volume percent of graphite, VGr, in a 3.1 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Arte-miy333 [17]

Answer:

The volume percent of graphite is 9.9%

Explanation:

Given;

weight percent of graphite, C = 3.1wt%

density of ferrite, $\rho _f$ = 7.9 g/cm³

density of graphite, $\rho _g$ = 2.3 g/cm³

Determine the mass fraction;

$W_f = \frac{C_g - C_0}{C_g -C_f} \\\\W_f =\frac{100 - 3.1}{100-0}\\\\W_f = 0.969\\\\\\W_g = \frac{C_0 - C_a}{C_g -C_f} \\\\W_g =\frac{3.1-0}{100-0}\\\\W_g = 0.031$

Determine the volume fraction;

$V = \frac{W_g/ \rho_g}{W_f / \rho_f \ + \ W_g / \rho_g} *100 \%\\\\V = \frac{0.031/ 2.3}{0.969 / 7.9 \ + 0.031 / 2.3}*100\%\\\\V = 9.9\%$

Therefore, the volume percent of graphite is 9.9%

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3 years ago

Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24°C and leaves at 0.8 MPa and 60°C. The mass flow rate o

Leni [432]

Answer:

(a) The power input to the compressor: $\dot{W}=73.07 kJ/s = 73.07 kW$

(b) The volume flow rate of the refrigerant at the compressor inlet: $\dot{v}=0.209 m^{3}/s$

Explanation:

(a)

We need to check the values of enthalpy (as we have an open system) for both states, being the inlet, state 1 and the outlet, state 2. We will know these values by checking the vapor charts of R134a, I used the ones found in Thermodynamics of Cengel, 7th edition.

Then, our values are:

$h_{1}=235.92kJ/kg\\h_{2}=296.81kJkg$

Now we proceed to know the work with the following expression:

$\dot{W}=\dot{m}(h_{2}-h_{1})$

Now we replace values and our result is:

$\dot{W}=73.07 kJ/s = 73.07 kW$

(b)

To know the volume rate at the compressor inlet, we need to know the specific volume in that phase, as we have that is saturated and at -24°C, we can read our table:

$\nu=0.1739m^{3}/kg$

With our specific volume and the mass rate, we can calculate the volume rate:

$\dot{v}=\nu * \dot{m}\\\dot{v}=0.209 m^{3}/s$

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4 years ago