What is temperature coefficient of resistance

An insulated 40 ft3 rigid tank contains air at 50 lbf/in2 and 120oF. A valve connected to the tank is now opened, and air is all

goldfiish [28.3K]

Answer:

The electrical work for the process is 256.54 Btu.

Explanation:

From the ideal gas equation:

n = PV/RT

n is the number of moles of air in the tank

P is initial pressure of air = 50 lbf/in^2 = 50 lbf/in^2 × 4.4482 N/1 lbf × (1 in/0.0254m)^2 = 344736.2 N/m^2

V is volume of the tank = 40 ft^3 = 40 ft^3 × (1 m/3.2808 ft)^3 = 1.133 m^3

T is initial temperature of air = 120 °F = (120-32)/1.8 + 273 = 321.9 K

R is gas constant = 8.314 J/mol.k

n = 344736.2×1.133/8.314×321.9 = 145.94 mol

The thermodynamic process is an isothermal process because the temperature is kept constant.

W = nRTln(P1/P2) = 145.94×8.314×321.9×ln(50/25) = 145.94×8.314×321.9×0.693 = 270669 J = 270669 J × 1 Btu/1055.06 J = 256.54 Btu

5 0

3 years ago

Read 2 more answers

Compared to arc welding, which of the following statements are true about<br> gas welding?

UNO [17]

Show than arc welding

4 0

3 years ago

Liquid flows at steady state at a rate of 2 lb/s through a pump, which operates to raise the elevation of the liquid 100 ft from

Greeley [361]

Answer:

D) 1.04 Btu/s from the liquid to the surroundings.

Explanation:

Given that:

flow rate (m) = 2 lb/s

liquid specific enthalpy at the inlet ( $h_{1}=40.09$ Btu/lb)

liquid specific enthalpy at the exit ( $h_{2}=40.94$ Btu/lb)

initial elevation ( $z_1=0ft$ )

final elevation ( $z_2=100ft$ )

acceleration due to gravity (g) = 32.174 ft/s²

$W_{cv}$ = 3 Btu/s

The energy balance equation is given as:

$Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0$

Since kinetic energy effects are negligible, the equation becomes:

$Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0$

Substituting values:

$Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\$

The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.

8 0

3 years ago

Which of these physical concepts is most important in civil engineering? A. energy transformations B. motion under gravity C. he

melamori03 [73]

Answer:

energy transformations

7 0

3 years ago

The settlement analysis for a proposed structure indicates that the underlying clay layer will settle 7.5 cm in 3 years, and tha

trapecia [35]

Answer:

A) For a double drainage system the time taken for total settlement will differ from that of a single drainage system, because in a double drainage case clay water escapes from both sides unlike in a single drainage case. hence water and air will be be expulsed faster for a double drainage making it attain total settlement faster.

B) for only single = 12 years

Explanation:

Given data :

ultimate clay settlement = 32 cm

settlement of clay in 3 years = 7.5 cm

Cv ( coefficient of consolidation ) = 1.5 x 10^-4 cm^2 /s

A) For a double drainage system the time taken for total settlement will differ from that of a single drainage system, because in a double drainage case clay water escapes from both sides unlike in a single drainage case. hence water and air will be be expulsed faster for a double drainage making it attain total settlement faster.

note : Total settlement is the same in both drainage system.

<u>B) Determine how long it will take for 7.5 cm of settlement to occur if there is only single </u>

applying the relation below

Tv = Cv t / H^2

where Tv = time factor , d = thickness of layer , H = drainage path

Given that Tv and Cv are constant in both cases

t ∝ H^2 hence

$\frac{t1}{t2} = \frac{H^21}{H^22}$ ------ ( 1 )

t1 = time for single drainage for d meters , t2 = time for double drainage for d meters

equation 1 can be rewritten as

t1/t2 = d^2 / (d/2)^2

∴ t1 = 4t2

given that t2 = 3 years ( value gotten from question )

t1 = 4 * 3 = 12 years

7 0

3 years ago