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Kazeer [188]
3 years ago
14

The constant-pressure specific heat of air at 25°C is 1.005 kJ/kg·°C. Express this value in kJ/kg·K, J/g·°C, kcal/ kg·°C, and Bt

u/lbm·°F. Round the final answers to three decimal place
Engineering
1 answer:
Dmitry [639]3 years ago
6 0

Answer:

a) C_v = 1.005 KJ/kgK

b) C_v = 1005.000 J/kgC

c) C_v = 0.240 kcal/kgC

d) C_v = 0.240 Btu/lbmF

Explanation:

Given:

- constant-pressure specific heat C_v = 1.005 KJ/kgC

Find C_v in units of:

a) kJ/kg·K

b) J/g·°C

c) kcal/ kg·°C

d) Btu/lbm·°F

Solution:

a) C_v is Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass. Usually, it's the heat in Joules needed to raise the temperature of 1 gram of sample 1 Kelvin or 1 degree Celsius. Hence,

C_v = 1.005 KJ/kgK

b)

                           C_v = 1.005 KJ/kgC * ( 1000 J / KJ)

                           C_v = 1005.000 J/kgC

c)

                           C_v = 1.005 KJ/kgC * ( 0.239006 kcal / KJ)

                           C_v = 0.240 kcal/kgC

d)

C_v = 1.005 KJ/kgC * ( 0.947817 Btu / KJ) * ( kg / 2.205 lbm)*(Δ1 C / Δ1.8 F)

C_v = 0.240 Btu/lbmF

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VMariaS [17]

Answer:

hello your question is incomplete attached below is the complete question

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B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

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4 years ago
What is capillary action?
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2 years ago
An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.
saul85 [17]

Answer:

a. L_o  = 40 psf

b. L ≈ 30.80 psf

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d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, L_o  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)

For the school, K_{LL} = 2

Therefore, we have;

L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, W_d = b × W_{D + L} =

∴  W_d =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, W_d = 812.8 ft./lb

d. For the uniformly distributed load, we have;

V_{max} = 812.8 × 26/2 = 10566.4 lbs

M_{max} =  812.8 × 26²/8 = 68,681.6 ft-lbs

v_{max} = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

P_L = 1000 lb

W_{D} = 20 × 16 = 320 lb/ft.

V_{max} = 1,000 + 320 × 26/2 = 5,160 lbs

M_{max} =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

v_{max} = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

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dem82 [27]

Answer:

0.264 ; 0.079

Explanation:

Given that:

Sample size, n = 100

Probability of being active, p = 1% = 1/100 = 0.01

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that more than 1 user will be active

P(x > 1) = 1 - [p(x=0) + p(x = 1)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x > 1) = 1 - [0.366 + 0.370]

P(x > 1) = 0.264

2.)

Probability that more than 2 user will be active

P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185

P(x > 1) = 1 - [0.366 + 0.370 + 0.185]

P(x > 1) = 0.079

7 0
3 years ago
A man can swim at 4 ft/s in still water. He wishes to cross tje 40-ft wide river to point B, 30 ft downstream. If the river flow
iVinArrow [24]

Answer:

v \approx 4.472\,\frac{ft}{s}, t = 10\,s

Explanation:

Since man and river report constant speeds and velocities are mutually perpendicullar, the absolute speed of the man is calculated by the Pythagorean Theorem:

v = \sqrt{(4\,\frac{ft}{s} )^{2}+(2\,\frac{ft}{s} )^{2}}

v \approx 4.472\,\frac{ft}{s}

The required time to make the crossing is:

t = \frac{40\,ft}{4\,\frac{ft}{s} }

t = 10\,s

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3 years ago
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