Ayo, how do I change my username on here?

Nuetrik [128]

Answer:

hi same here. hru btw

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3 0

2 years ago

Read 2 more answers

Determine the design moment strength for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compressio

SVETLANKA909090 [29]

This question is incomplete, the complete question is;

Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.

Use Fy=50 ksi and assume Cb=1.0 (if needed).

Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft

Explanation:

Given that;

section W 21 x 73 steel beam;

now from the steel table table for this section;

Zx = Sx = 151 in³

also given that; fy = 50 ksi and Cb = 1.0

QMn = 0.9 × Fy × Zx

so we substitute

QMn = 0.9 × 50 × 151

QMn = 6795 k-inch

we know that;

12inch equals 1 foot

so

QMn = 6795 k-inch / 12

QMn = 566.25 f-ft

Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft

7 0

2 years ago

A 6-pack of canned drinks is to be cooled from 23°C to 3°C. The mass of each canned drink is 0.355 kg. The drinks can be treated

Vesna [10]

Answer:

178 kJ

Explanation:

Assuming no heat transfer out of the cooling device, and if we can neglect the energy stored in the aluminum can, the energy transferred by the canned drinks, would be equal to the change in the internal energy of the canned drinks, as follows:

ΔU = -Q = -c*m*ΔT (1)

where c= specific heat of water = 4180 J/kg*ºC

m= total mass = 6*0.355 Kg = 2.13 kg

ΔT = difference between final and initial temperatures = 20ºC

Replacing by these values in (1), we can solve for Q as follows:

Q = 4180 J/kg*ºC * 2.13 kg * -20 ºC = -178 kJ

So, the amount of heat transfer from the six canned drinks is 178 kJ.

8 0

2 years ago

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Refrigerant 134a enters an air conditioner compressor at 4 bar, 208C, and is compressed at steady state to 12 bar, 808C. The vol

SCORPION-xisa [38]

Answer:

heat transfer rate is -15.71 kW

Explanation:

given data

Initial pressure = 4 bar

Final pressure = 12 bar

volumetric flow rate = 4 m³ / min

work input to the compressor = 60 kJ per kg

solution

we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is

h1 = 262.96 kJ/kg

v1 = 0.05397 m³/kg

h2 = 310.24 kJ/kg

and here mass balance equation will be

m1 = m2

and mass flow equation is express as

m1 = $\frac{A1\times V1}{v1}$ .......................1

m1 = $\frac{4\times \frac{1}{60}}{0.05397}$

m1 = 1.2353 kg/s

and here energy balance equation is express as

0 = Qcv - Wcv + m × [ ( h1-h2) + $\frac{v1^2-v2^2}{2}$ + g (z1-z2) ] ....................2

so here Qcv will be

Qcv = m × [ $\frac{Wcv}{m} + (h2-h1)$ ] ......................3

put here value and we get

Qcv = 1.2353 × [ {-60}+ (310.24-262.96) ]

Qcv = -15.7130 kW

so here heat transfer rate is -15.71 kW

6 0

2 years ago

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

hammer [34]

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

B) The physical properties are copper

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

3 0

2 years ago