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tatuchka [14]
3 years ago
10

प्रहार का समरूपी भिन्नार्थक शब्द अर्थ के साथ ​

Engineering
1 answer:
Reika [66]3 years ago
3 0

Answer:

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Explanation:

djhidhfdiushfiudshfui

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Who is responsible for conducting a hazard assessment?
slava [35]

Answer:

The Employee

Explanation:

Because it is there responsibility

5 0
3 years ago
Heating of Oil by Air. A flow of 2200 lbm/h of hydrocarbon oil at 100°F enters a heat exchanger, where it is heated to 150°F by
irakobra [83]

Answer:

2062 lbm/h

Explanation:

The air will lose heat and the oil will gain heat.

These heats will be equal in magnitude.

qo = -qa

They will be of different signs because one is entering iits system and the other is exiting.

The heat exchanged by oil is:

qo = Gp * Cpo * (tof - toi)

The heat exchanged by air is:

qa = Ga * Cpa * (taf - tai)

The specific heat capacity of air at constant pressure is:

Cpa = 0.24 BTU/(lbm*F)

Therefore:

Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h

5 0
3 years ago
21. How long can food that requires time-temperature control be left in the danger zone?
dimulka [17.4K]

Answer: A maximum of 1 hour

Explanation:

Read your lesson buddy!!

8 0
3 years ago
Read 2 more answers
A mass of 5 kg of saturated liquid-vapor mixture of water is contained in a piston-cylinder device at 125 kPa. Initially, 2 kg o
arlik [135]
A because it is the best one
5 0
3 years ago
A furnace wall is to be built of 20-cm firebrick and building (structural) brick of same thickness. The thermal conductivities o
Norma-Jean [14]

Answer:

q=2313.04W/m^2

T=690.86°C

Explanation:

Given that

Thickness t= 20 cm

Thermal conductivity of firebrick= 1.6 W/m.K

Thermal conductivity of structural brick= 0.7 W/m.K

Inner temperature of firebrick=980°C

Outer temperature of structural brick =30°C

We know that thermal resistance

R=\dfrac{t}{KA}

These are connect in series

R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}

R=\dfrac{0.2}{1.6A}+\dfrac{0.2}{0.7A}\ K/W

R=\dfrac{23}{56A}\ K/W

Heat transfer

Q=\dfrac{\Delta T}{R}

Q=56A\times \dfrac{980-30}{23}\ W

So heat flux

q=2313.04W/m^2

Lets temperature between interface is T

Now by equating heat in both bricks

\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}

So T=690.86°C

6 0
3 years ago
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