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forsale [732]
2 years ago
9

A thermocouple, with a spherical junction diameter of 0.5 mm, is used for measuring the temperature of hot airflow in a circular

duct. The convection heat transfer coefficient of the airflow can be related with the diameter (D) of the sphere and the average airflow velocity (V) as h = 2.2(V /D)0.5, where D, h, and V are in m, W/m²•K, and m/s, respectively. The properties of the thermocouple junction are k = 35 W/m•K, rho = 8500 kg/m³, and cp = 320 J/kg•K.
If the maximum response time of the thermocouple to record 87 percent of the initially applied temperature difference is 5 s, determine the minimum airflow velocity that the thermocouple can be used.

Engineering
1 answer:
german2 years ago
6 0

Answer:

attached below

Explanation:

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What is this spray pattern defect most likely caused by:
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Answer:

fluid nozzle that is too large

6 0
1 year ago
A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind
Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

N_F=n_F\times \frac{N_S}{n_S} here,

n_F = the number of unseeded faults = 6

N_S = number of seeded faults = 30

n_s = number of seeded faults found = 15

So NF will be calculated as,

N_F=6\times \frac{30}{15}=12

And the estimate of faults remaining is  N_F-n_F = 12 - 6 = 6

8 0
2 years ago
Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an
Svet_ta [14]

Answer:

2.46 * 10⁵ W/m³

Explanation:

See attached pictures for detailed explanation.

6 0
3 years ago
Read 2 more answers
Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at
makkiz [27]

Answer:

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

Explanation:

The Young's module is:

E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }

E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}

Let assume that both specimens have the same geometry and load rate. Then:

E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}

The displacement rate for steel is:

\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}

\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

7 0
2 years ago
Read 2 more answers
Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:
Xelga [282]

Answer:

(a) Current density at P is J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\.

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\

where \textbf{a}_{\rho} and \textbf{a}_{\phi} unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation.  Therefore

J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\

(b) Total current flowing outward can be calculated by using the relation,

I=\int {\textbf{J} \, \textbf{ds}

where integral is calculated through the circular band given in the question. We can write the integral as below,

I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\

due to unit vector multiplication. Then,

I=10\int\(\rho^3z.dz.d\phi

where \rho=3,\ 0. Therefore

I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A

4 0
3 years ago
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