Oil system cleaning products should not use solvents problem

Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−

sergeinik [125]

Answer:

Power output, $P_{out} = 178.56 kW$

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, $T = 350^{\circ}C$

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, $\dot{m} = 0.1 kg.s^{- 1}$

Diameter of exhaust pipe, $d_{h} = 15 cm = 0.15 m$

Pressure at exhaust, P' = 50 kPa

temperature, T' = $100^{\circ}C$

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

$\dot{m} = \frac{Av_{i}}{V_{1}}$

$0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}$

$0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}$

$v_{i} = 3.986 m/s$

Now, eqn for compressible fluid:

$\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}$

Now,

$\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}$

$\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}$

$\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}$

$v_{e} = 19.33 m/s$

Now, the power output can be calculated from the energy balance eqn:

$P_{out} = -\dot{m}W_{s}$

$P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}$

$P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW$

4 0

3 years ago

25 points and brainliest is it A, B, C, D

Bogdan [553]

Answer:

SIR IT IS D HOPE THIS HELPS (☞ﾟヮﾟ)☞☜(ﾟヮﾟ☜)

Explanation:

4 0

3 years ago

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Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be

bonufazy [111]

Answer:

T =272.497degrees

%=34.08%

Explanation:

Please see attachment

4 0

4 years ago

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One good way to improve your gas Milage is to ___.

VashaNatasha [74]

Answer: B

Explanation:

One good way to improve your gas mileage is to accelerate smoothly and directly to a safe speed.

Hope this helps!

5 0

3 years ago

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On a nonprecision approach, what is the maximum acceptable descent rate during the final stages of the approach (below 1,000 ft.

Dovator [93]

Answer: For non-precision approaches, the maximum acceptable descent rate acceptable should be one that ensures the aircraft reaches the minimum descent altitude at a distance from the threshold that allows landing in the touch down zone. Otherwise, a decent rate greater than 1000fpm is unacceptable.

Explanation: For non-precision approaches, a descent rate should be used that ensures the aircraft reaches the minimum decent altitude at a distance from the threshold that allows landing in the touchdown zone (TDZ) . On many instrument approach procedures, this distance is annotated by a visual descent point (VDP) If no VDP is annotated, calculate a normal descent point to the TDZ. To determine the required rate of descent, subtract the TDZ elevation (TDZE) from the final approach fix (FAF) altitude and divide this by the time inbound. For illustration, if the FAF altitude is 1,000 feet mean sea level (MSL), the TDZE is 200 feet MSL and the time inbound is two minutes, an 400 fpm rate of descent should be applied.

A descent rate greater than approximately 1,000 fpm is unacceptable during the final stages of an approach (below 1,000 feet AGL). Operational experience and research shows that this is largely due to a human perceptual limitation that is independent of the airplane or helicopter type. As a result, operational practices and techniques must ensure that descent rates greater than 1,000 fpm are not permitted in either the instrument or visual portions of an approach and landing operation.

3 0

4 years ago