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LUCKY_DIMON [66]
3 years ago
10

Problem #1: A centrifugal compressor has a mass flow rate of 1.6 kg/s while rotating at 11,000 rpm. The impellor is a radial imp

ellor meaning that the blades are strictly radial at exit (β2' = 0). The radius of the radial impeller at exit is 0.3 m with a blade width 0.01 m. Assume that the entrance flow is axial. For this compressor, the compressor efficiency is 83% and the slip coefficient is 0.9. (a) Clearly sketch the velocity triangles at the exit of the impellor. Include the flow slip and mark all the angles and velocities.

Engineering
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

a)  The slip coefficient is 0.9

b) Blade tip speed is 345.57 m/s

c) Stagnation temperature exiting the impellor is 416.84 k

d) Exit flow velocity is 84.88 m/s

e) Exit flow angle is 76.2°

f) Exit static temperature is 353.84 k

g) Impellor exit Mach number is 0.943

Explanation:

Flow at entry is axial α₁ = 0

Tagential velocity at entry V_{w1}= 0

Blade at exit is radial  β₂ = 0

μ₂ = V_{w2}

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A HSS152.4 × 101.6 × 6.4 structural steel [E = 200 GPa] section (see Appendix B for crosssectional properties) is used as a colu
Temka [501]

Answer:

(a) 126.66 kN (b) 31.665 kN (c) 258.49 kN (d) 506.64 kN

Explanation:

Solution

Given

A HSS152.4 × 101.6 × 6.4 structural steel is used as a column

Actual length of the column , L= 6 m

The elasticity modules, E = 200 GPa

The factor of safety with respect to failing buckling . F.S =2

Geometric properties  of structural steel shapes for, A HSS152.4 × 101.6 × 6.4

the moment of inertial about y axis Iy =4 .62 * 10^ 6 mm ^4

For

(a)  If the end condition is pinned - pinned

The effective  length factor, K =1

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 1* 6* 10 ^3)

= 253319.85N

= 253.32N

The maximum safe load , Pallow = 253.32 /2 = 126.66kN

hence, the maximum safe for the column is 126.66kN

(b)If the end condition is  fixed free-free

the effective length factor, K= 2

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 2 * 6 * 10 ^3)²

= 63329.96N

=63.33kN

The maximum safe load,  P allow = Pcr/F.S

P allow = 63.33/2

31.665 kN

Therefore the maximum safe for the column is 31.665 kN

(c) If the end condition is fixed- pinned

The effective  length factor K =0.7

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.7 * 6 * 10 ^3)²

=516979.2 8N

=516.98 kN

The maximum safe load,  P allow = Pcr/F.S

P allow = 516.98 kN/2

=258.49 kN

Therefore the maximum safe for the column is 258.49 kN

(d) If the end condition is fixed -fixed

The effective factor, K =0.5

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.5 * 6 * 10 ^3)²

=1013279.4 N

=1013.28 kN

The maximum safe load,  P allow = Pcr/F.S

P allow = 1013.28 / 2

= 506.64 kN

P allow = 506.64 kN

Therefore the maximum safe for the column is 506.64 kN

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A cartridge electrical heater is shaped as a cylinder of length L=200 mm and outer diameter D=20 mm. Under normal operating cond
Setler79 [48]

Answer:

When water is surrounding T_s = 34.17 degree C

When air surrounding T_S = 1434.7 degree C

from above calculation we can conclude that air is less effective than water  as heat transfer agent

Explanation:

Given data:

length  = 300 mm

Outer diameter  = 30 mm

Dissipated energy = 2 kw = 2000 w

Heat transfer coefficient IN WATER = 5000 W/m^2 K

Heat transfer coefficient in air  = 50 W/m^2 K

we know that q_{convection} =  P

From newton law of coding we have

q_{convection} =  hA(T_s -  T_{\infity})

T_s is surface temp.

T - temperature at surrounding

P = hA(T_s -  T_{\infity})[tex]\frac{P}{\pi hDL} =  (T_s -  T_{\infity})

solving for[/tex] T_s [/tex] w have

T_s = T_{\infty} + \frac{P}{\pi hDL}

T_s = 20 + \frac{2000}{\pi 5000\times 0.03\times 0.3}

T_s = 34.17 degree C

When air is surrounding we have

T_s = T_{\infty} + \frac{P}{\pi hDL}

T_s = 20 + \frac{2000}{\pi 2000\times 0.03\times 0.3}

T_s = 1434.7 degree C

from above calculation we can conclude that air is less effective than water  as heat transfer agent

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