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2 years ago

6

A cylindrical tank is 50 inches long, has a diameter of 16 inches and contains 1.65 lbm of water. Calculate the density of water

in Ibm per cubic feet.

1 answer:

CaHeK987 [17]2 years ago

8 0

Answer:

<em>0.285 lbm per cubic feet</em>

<em></em>

Explanation:

length of tank = 50 inches = 50/12 feet = 4.17 feet

diameter of tank = 16 inches = 16/12 feet = 1.33 feet

weight of water = 1.65 lbm

density of water = ?

We know that the density of a substance is given as

ρ = w/v

where ρ is the density in Ibm per cubic feet

w is the weight in lbm

v is the volume in cubic feet

Volume of a cylinder = $\frac{\pi d^{2} l}{4}$

where d is the diameter

l is the length

volume = $\frac{3.142*1.33^{2}* 4.17}{4}$ = 5.79 cubic feet

Therefore, the density of water will be

ρ = w/v = 1.65/5.79 = <em>0.285 lbm per cubic feet</em>

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<u>Explanation:</u>

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2 years ago

An ideal Stirling cycle filled with air uses a 75 F energy reservoir as a sink. The engine is designed to so that the maximum ai

algol13

Answer:

Explanation:

For the given stirling circle with air as the working fluid, temperature, pressure, and volume i^th stage are $T_t, P_i$ and $V_i$ respectively

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substitute 535 for $T_L$ , 2Btu for $W_{net}$ , 5Btu for $Q_{in}$ to find $T_H$

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Hence ,the amount of air conditioned in the engine is 0.3785 Ibm

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$P_1=\frac{mRT_1}{V_1}$

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$0.3704 psia ft^3/Ibm.R for R,\\\\ 891.7R for T_1 \ \ and \ \ 0.06ft^3 for V_1$

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sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is $0.44 \frac{W}{m^{2} \times K}$ , the average convection heat transfer coefficient over the entire plate is $0.293 \frac{W}{m^{2} \times K}$ and the total heat flux transfer to the plate is $61.6 KJ$ .

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.

Hot carbon dioxide exhaust gas

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$c_p = 1.02 \frac{kJ}{Kg \times K}$

$m= 231 \times 10^{7} \frac{N \times s }{m^2}$

υ = $21.8 \times 10^{6} \frac{m^2}{s}$

$k = 32.5 \times 10^{3} \frac{W}{m \times K}$

$\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}$

$Pr = 0.725$

Apart from these other data arr given below,

$v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C$

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

$Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

where h= Average heat transfer coefficient

L= Length of a plate

k= Thermal Conductivity of carbon dioxide

Re = Reynold's Number

Pr = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

is referred as local convection heat transfer coefficient.

To find convection heat transfer coefficient at 1 m from leading edge,

$Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

Here, first we have to find Re and Pr,

$Re = \frac{r \times v \times L}{m}$

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Pr number is take from physical property data and Pr is 0.725.

Putting value of Re and Pr in main equation,

we get

$Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

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Finally,

$h = \frac{0.44}{1.5}$

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(C) Total heat flux transfer to the plate is found out by,

$Q = h \times (T_g - T_s)$

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8 0

2 years ago

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WARRIOR [948]

Answer:

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Explanation:

H₀: The die is not biased

Ha: The die is biased

We can apply binomial distribution and determine whether the die is biased or not. (we can also perform z-test, it will provide similar results)

We know that a binomial distribution is given by

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p < α

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Since the p value is less than the α value therefore, we reject the null hypothesis so we have evidence to conclude that the die is biased.

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