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tatiyna
2 years ago
9

Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 76 vol% carbide

particles in a metal matrix. Assume thermal conductivities of 30 and 67 W/m-K for the carbide and metal, respectively.
Engineering
1 answer:
Georgia [21]2 years ago
8 0

Answer:

The answer is below

Explanation:

Given that:

Volume of carbide (V_{C}) = 76% = 0.76, Volume of Nickel (V_{M}) = 100% - 76% = 24% = 0.24, thermal conductivities of carbide (E_{C}) = 30 W/m-K and thermal conductivities of meta. (E_M) = 67 W/m-K

a) The maximum thermal conductivity is given by:

Max = E_CV_c+E_mV_m=(0.76*30) + (0.24*67) = 38.88\ W/mK

b) The minimum thermal conductivity is given by:

Min = \frac{E_ME_C}{E_MV_C+E_CV_M}=\frac{30*67}{(0.76*67)+(0.24*30)}=34.58\ W/mK

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Air at T1 = 32°C, p1 = 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate
RUDIKE [14]

Answer:

4.5kg/min

Explanation:

Given parameters

T_1 = 32^0 C,  m_1 = 3 kg/min, T_2 = 7^0 C ,T_3 = 17^0

if we take  

The mass flow rate of the second stream = m_2(kg/min)

The mass flow rate of mixed exit stream = m_3 (kg/min)

Now from mass conservation

m_3 = m_2 + m_1

m_3 = m_2 + 3 (kg/min)

The temperature of the mixed exit stream given as

T_3m_3 = T_2m_2 +T_1m_1\\\\17 ( 3 + m_2) = 7 \times m_2 + 32 \times 3\\\\51 + 17 m_2 = 7 m_2 + 96\\\\10 m_2 = 96 - 51\\\\m_2 = 4.5 kg/min\\\\\\\\

Therefore the mass flow rate of second stream will be 4.5 kg/min.

7 0
3 years ago
Two resistors, A and B, individually connect to a 9V battery. A student notices that resistor A is warmer than resistor B. Which
dybincka [34]

Answer:

Resistor B

Explanation:

Since resistance is the opposition to the flow of current in a circuit,

first let assume the two resistors are connected in parallel to the voltage, recall that when connection is in parallel, the different amount of current pass through the resistors depending on the value with the small resistor having  a lower resistance effect hence higher current will pass through

The energy dissipated in each resistor can be calculated as

E=\frac{1}{2}IR^{2}t.

from the formula we can conclude that the energy value will be higher for the resistor with small resistance value. hence more heating effect which will cause it to be warm.

Also when connected individually the current flow from the voltage source will pass through the resistor which when we calculate the energy dissipated, the resistor with smaller value will be higher because it will draw more current which will in turn lead to a heating effect and cause the resistor to be warm. Hence we can conclude that the resistance B has greatest resistance value.

4 0
2 years ago
Design an op amp circuit to average the input of six sensors used to measure temperature in restaurant griddles for a large fast
Marina CMI [18]

Answer:

See the attached file for the design.

Explanation:

Find attached for the explanation.

3 0
3 years ago
) An ammonia nitrogen analysis performed on a wastewater sample yielded 30 mg/L as nitrogen. If the pH of the sample was 8.5, de
Delicious77 [7]

Answer:

NH_4^+ = 2.5 mg/lt

Explanation:

Given data:

Ammonia Nitrogen 30 mg/L

pH = 8.5

-log[H +] = 8.5

[H +] = 10^{-8.5}

NH_4 ^{+} ⇄ H^{+} + NH_3

Rate constant is given as

K_a = \frac{[H^{+}] [NH_3]}{NH_4^{+}} ...........1

K_a = 5.6 \times 10^{-10}

Total ammonia as NItrogen is given as 30 mg/l

\%NH_4^{+} = \frac{ [NH_4] \times 100}{[NH_4^{+}] + [NH_3]}

                    = \frac{100}{\frac{NH_4^+}{NH_4^+} +\frac{NH_3^+}{NH_4^+}}

                    = \frac{100}{1+ \frac{NH_3^+}{NH_4^+}} .....2

from equation 1 we have

\frac{NH_3^+}{NH_4^+} =\frac{K_a}{[H^+]} = \frac{5.6\times 10^{-10}}{10^{8.5}}

plug this value in equation 2 we get

\%NH_4^{+} = 84.96 \%

Total ammonia as N = 30 mg/lt

NH_4^+ = \frac{84.96}{100} \times 30 = 25.5 mg/lt

7 0
3 years ago
An industrial system is rated at 600 v the system includes a partially exposed terminal block thats mounted on a bulkhead a new
Nastasia [14]

Answer:

10 ft or 3.048 m (for unqualified person)

1.09 ft or 0.33 m (for qualified person)

Explanation:

OSHA has prescribed "Minimum Approach Distance - MAD" to ensure safety of personnel working with electrical power lines.

For a unqualified person:

The minimum approach distance is 10 ft or 3.048 m

For a qualified person:

For a voltage in the range of 300V to 750V (Phase to Ground or Phase to Phase) the minimum approach distance is 1.09 ft or 0.33 m

Since 600V lies in the above range hence it is applicable on 600V.

OSHA has developed a full list of minimum approach distance according to the level of voltages (50V to 72000V)

7 0
2 years ago
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