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fredd [130]
4 years ago
11

Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A

ir enters the heat exchanger at 95 kPa and 20°C at a rate of 0.65 m3/s. The combustion gases (cp = 1.10 kJ/kg·°C) enter at 160°C at a rate of 0.95 kg/s and leave at 95°C. Determine the rate of heat transfer to the air and its outlet temperature.
Engineering
1 answer:
liberstina [14]4 years ago
3 0

Answer:

Explanation:

https://www.chegg.com/homework-help/air-cp-1005-kj-kg-c-preheated-hot-exhaust-gases-cross-flow-h-chapter-7-problem-145p-solution-9780077366742-exc

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Explanation:

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5 0
3 years ago
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You have a motor that runs at 1.5 amps from a 12 volt power source, how many watts of power is it using?
Leviafan [203]

Answer:

18 Watts

Explanation:

For this problem, we simply need to understand the relationship of power to voltage and current.  This relationship is derived from Ohm's law:

Power = Voltage * Current

Given this equation, we can say the following to find the power consumption of the motor:

Power = 12volts * 1.5amps

Power = 18 Watts

Hence, the motor is consuming 18 Watts of power.

Cheers.

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3 years ago
Here you go!!!!!!!!!!!!!!!!!1
sweet [91]

Answer:

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Explanation:

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3 years ago
At a ski resort, water at 40 °F is pumped through a 3-in.-diameter, 2001-ft-long steel pipe from a pond at an elevation of 4286
deff fn [24]

Answer:

P = 24.38 hp

Explanation:

Given data:

diameter of steel pipe = 3 inc

length of pipe = 2001 ft

elevation of pond = 4286 ft

elevation of snow making machine = 4620 ft

Rate of water = 0.26 ft^3/s

Applying bernouli equation on bioth side

\frac{P}{\rho} + \frac{v_1^2}{2g} + z_1 + h_p =\frac{P_2}{\rho} + \frac{v_2^2}{2g} + z_2 + \frac{flv^2}{2gD}.....1

where P_2 = 182 psi

P_1 = 0, v_1 = 0

V =V_2 = \frac{Q}{\frac{\pi}{4} D^2}= \frac{0.26}{\frac{\pi}{4} \times (3/12)^2} = 5.30 ft/s

calculation fro friction factor

from standard table we have

\frac{\epsilon}{D} = \frac{0.00015}{(3/12)} = 6\times 10^{-4}

Re =\frac{VD}[\nu} = \frac{5.30 \times (3/12)}{1.66 \times 10^{-5}} = 7.96 \times 10^4

so for calculated R and\frac{\epsilon}{D}, F value = 0.0212

from equation 1

h_p = \frac{P_2}{\rho} + Z_2 -Z_1 +(1 + f\frac{l}{D}) \frac{V^2}{2g}

h_p = \frac{182 \times 144/ lb /ft^2}{62.4} + 4620 - 4286 + (1 + 0.0212 \frac{2001}{(3/12)}) \frac{5.30}{2\times 32.2}

h_p = 826.76ft so that

P =  \rho Q h_p = 62.4 \times 0.26 \times 826.76 = 13413.38 ft lb/s = 24.38 hp

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3 years ago
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Answer:

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