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abruzzese [7]
3 years ago
9

A rigid tank contains 3 kg of water initially at 43.97% quality and at a temperature of 120°C. The water is heated until it reac

hes a pressure of 7 bar. Neglect changes in kinetic and potential energy.
a. At what temperature, in °C, will the water become saturated vapor?
b. Determine the heat transfer, in kJ.
c. The work of the cycle, in kJ

Engineering
1 answer:
makkiz [27]3 years ago
6 0

Explanation: see attachment below

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Its 0.001

0.01 x100 = 1mm
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A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi
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Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation =\frac{N_1 +N_2}{2}

Nm = \frac{500+750}{2} = 625 rpm

w =\frac{2\pi Nm}{60}

  =65.44 rad/sec

F_T = mw^2 e

F_T = mew^2

       = 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio =\frac{300}{428.36} = 0.7

also

transmission ratio = \frac{1}{[\frac{w}{w_n}]^{2} -1}

0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}

SOLVING FOR Wn

Wn = 42 rad/sec

Wn = \sqrt {\frac{k}{m}

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

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How is the fuel introduced into the Diesel engine?
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How would you use an inclined plane to increase speed?​
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One kilogram of "as received" yard trimmings is made up of approximately 620 g moisture, 330 g of decomposable organic matter (r
Evgesh-ka [11]

Answer:

Explanation:

(a) Given that 620g moisture and 330g decomposable organic matter in yard trimming is represented by C₁₂.₇₆H₂₁.₂₈O₉.₂₆N₀.₅₄

Given the atomic mass of Carbon C = 12, Hydrogen H = 1, Oxygen O = 16 and Nitrogen N = 14

1 mole of trimming = 12*12.76 + 1*21.28 + 16*9.26 + 14*0.54

=  153.12 + 21.28 + 148.16 + 7.56

= 330.12 g/mol

which means 1 kg of as received trimming has 330 g of decomposable that produce 1 mole of decomposable

The moles of methane produced will be given as

m = (4a + b -2c - 3d)/8

= (4*12.76 + 21.28 - 2*9.26 - 3*0.54)/8

= (51.04 + 21.28 - 18.52 - 1.62)/8

= 52.18/8

= 6.5225

(b) Volume of methane V is given as

V = (0.0224 m³ CH₄mol/CH₄) × (6.5225 mol CH₄/ kg)

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= 5805.025

≈ 5805 kJ/kg

5 0
3 years ago
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