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IgorLugansk [536]
3 years ago
7

Saturated steam coming off the turbine of a steam power plant at 40°C condenses on the outside of a 3-cm-outer-diameter, 35-m-lo

ng tube at a rate of 140 kg/h.
Determine the rate of heat transfer from the steam to the cooling water flowing through the pipe.
Engineering
1 answer:
Marina86 [1]3 years ago
5 0

Answer:

93.57 KJ/s

Explanation:

Rate of heat transfer from the steam to the cooling water = mass rate  ×  Heat of vaporization of water at 40 °C

H vaporization of water 40 °C  at saturation pressure = 2406.0 kJ / Kg

rate of heat transfer = 2406.0 kJ / Kg × 140 Kg / ( 60 × 60s) = 93.57 KJ/s

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It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
2 years ago
On aircraft equipped with fuel pumps, when is the auxiliary electric driven pump used?.
pochemuha
In an airplane equipped with fuel pumps, the auxiliary electric fuel pump is used in the event the engine-driven fuel pump fails.. hope this helped !
6 0
2 years ago
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
3 years ago
- WHEN YOU ARE TOWING A TRAILER:
zheka24 [161]

Answer:

And Im still going with B..

7 0
3 years ago
If the slotted arm rotates counterclockwise with a constant angular velocity of thetadot = 2rad/s, determine the magnitudes of t
astraxan [27]

Answer:

Magnitude of velocity=10.67 m/s

Magnitude of acceleration=24.62 ft/s^{2}

Explanation:

The solution of the problem is given in the attachments

3 0
3 years ago
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